Physics 1: University Physics for Scientists & Engineers - East LA ...

Physics 1: University Physics for Scientists & Engineers Please note, this is a work in progress, and as such, will undergo lots of modification until the end of the semester. Most notably, the page breaks, which I want to place at strategic places (so as not to cut off something important into 2 pages), but trying to do it now will only fail as I add and remove lines, so I will do that only at the end. In the meantime, keep this in mind.

• Chapter 1: Physics and Measurement o

• Chapter 2: Motion in One Dimension o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 • Chapter 3: Vectors o Ex.1 Ex.2 • Chapter 4: Motion in Two Dimensions o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 • Chapter 5: The Laws of Motion o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 • Chapter 6: Circular Motion and Other Applications of Newton’s Laws o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 • Chapter 7: Energy and Energy Transfer o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 • Chapter 8: Potential Energy o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 • Chapter 9: Linear Momentum and Collisions o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 • Chapter 10: Rotation of a Rigid Object About a Fixed Axis o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10 • Chapter 10.9: Rolling Motion o Ex.1 Ex.2 Ex.3 • Chapter 11: Angular Momentum o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10 • Chapter 12: Static Equilibrium and Elasticity o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10 • Chapter 15: Oscillatory Motion o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10 • Final Exam Study Guide o Ex.1 Ex.2 Ex.3 Ex.4 Ex.5 Ex.6 Ex.7 Ex.8 Ex.9 Ex.10 o Ex.11 Ex.12 Ex.13 Ex.14 Ex.15 Ex.16 Ex.17 Ex.18 Ex.19 Ex.20 Comment [as1]: Notes for Monday, June 12, 2006 begin here

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I. Chapter 1: Measurements II. Chapter 2: Motion in One Dimension

(return to top) (return to top)

A. Purpose: B. Definitions (and symbols):

1. Distance (d) (scalar) is the total length of space that an object travels. 2. Displacement (D) (vector) is the length of space between your origin and your destination. 3. Speed (s) (scalar) the rate of movement of a particle. a. Average Speed = the ratio of the distance covered in a certain time i.

S=

d t

b. Instantaneous Speed = how fast you are going at any particular instant. It is also known as the magnitude of the instantaneous velocity.

r

i. S(t ) =| v | 4. Velocity (v) (vector) the rate of displacement of a particle a. Average Velocity = the ratio of displacement covered in a certain time i.

vX =

Δx Δt

b. Instantaneous Velocity = the velocity of an object at an instant in time. i.

vx = lim

Δt →0

Δx dx = Δt dt

5. Acceleration (a) (vector) the rate at which velocity changes a. Average Acceleration = the ratio of velocity covered in a certain time i.

r r r v f − vi Δvr = a≡ t Δt

b. Instantaneous Acceleration = the acceleration of an object at an instant in timed i.

r dv r a(t) = dt

C. Example 1: 1. A particle begins traveling 5 feet to the right and then stops and travels to the left for 2 feet, traveling the entire distance in 3 seconds. Assuming average velocity, analyze the problem. a. What is the distance (d) and displacement (D)? r i. d = 7 ft D = 3 ft (or 3ft right, 3 ft east, +3ˆi ft) b. What is the average speed ( s ) and instantaneous speed (s)? i.

r 7 s =| v |= = 2.33 ft/s 3

s=2.33 ft/s (if constant)

c. What is the average velocity ( v ) and instantaneous velocity (v)? r r i. v = 1iˆ ft/s v ≅ 2.33 ft/s

d. What is the average acceleration ( a ) and instantaneous acceleration (a)?

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i.

r v f − vi (−2.33) − (2.33) a= = = −1.55 ft/s 2 t 3

ii.

draw the path of the particle

Comment [as2]: Notes for Wednesday, June 14, 2006 begin here.

Chapter 2 (Motion in One Dimension)

iii.

Regarding average velocity – since the displacement is 3 feet in 3 seconds, the average velocity is 1 ft/s. this would mean that a particle traveling at 1ft/s directly towards the goal (as opposed to the other particle which went forward 5ft and then returned 2 feet), will arrive at the same time as the particle traveling the longer route. As for instantaneous velocity, there is not enough information to figure this out (we don’t know if velocity is constant or changing).

D. Example 2:

1. A car goes west for 40 miles in 2/5 hours and then stops for ½ hour. It then goes east for 70 miles in 7/10 hours. Assume the car is going at constant speed and its initial & final speeds are not zero. a. What is the average speed of the car in the first stage, second stage, & the whole trip? i.

s1 =

40 = 100mph 2/5

s2 =

70 = 100mph 7 / 10

stot =

110 = 68.75mph 1 2 7 + + 2 5 10

b. What is the average velocity of the car in the first stage, second stage, & whole trip? i.

r v1 = −100 mph

r v 2 = 100 mph

r 30 v tot = = 18.75 mph 1.6

c. What is the average acceleration of the car in the first stage, second stage, & whole trip?

i.

r v f (1) − vi(1) 0 − (−100) v f (2) − vi(2 ) 100 − 0 r a = a1 = = = 250mph 2 = = 142.86mph 2 2 t1 t 2/5 7 / 10 r v f (2) − vi(1) 100 − (−100) atot = = = 125mph 2 1.6 ttot

d. Draw the path of the particle & include all relevant data

e. Graph x vs. t, v vs. t, s vs. t, & a vs. t & show the geometrical meanings of the average values.

E. Example 3: Page 3 of 59

Chapter 2 (Motion in One Dimension) 1. A particle is traveling on a path given by the equation x(t) = (t 2 − 4)(t + 5) from 0 ≤ t ≤ 6 seconds. Fully analyze the particle’s path, its speed, velocity and acceleration. a. Step 1: figure out the significant markers related to the position function (at the beginning, end and when the function equals zero). To figure out how far the particle will travel and when it will cross the origin (t=0).

i. ii.

x(t) = t 3 + 5t 2 − 4t − 20 x(0) = −20 x(6) = 352 (These will be our initial & final positions)

iii.

x(t) = 0

→

0=t 3 + 5t 2 − 4t − 20

→

t=2

(this is the time the particle will be at the origin) b. Step 2: derive the position function to get the velocity function and its significant markers (again at the beginning, end, and when the function equals zero) to determine its velocity at the beginning and end of its path. Where the function equals zero, the particle has no velocity and could either continue on its path, or change direction.

i. ii.

x '(t) = v(t) = 3t 2 + 10t − 4 v(0) = −4 v(6) = 164 (These will be our initial & final velocities)

iii. iv.

v(t) = 0 → 0 = 3t 2 + 10t − 4 → t = 0.361 (velocity is zero here) x(0.361) = −20.745 (This means at 0.361 seconds, the position will be at either a local

max or local min, IF the sign changes. In this case, the sign changes from negative to positive, meaning x(0.361) is a local minimum. This also tells us our problem will be broken down into two stages, the one with the negative velocity (up to t=0.361) and the one with the positive velocity (from t=0.361 and beyond). c. Step 3: derive the velocity function to get the acceleration function and its known values.

i. ii.

x ''(t ) = v'(t) = a(t) = 6t + 10 a(0) = 10 a(6) = 46 (These will be our initial & final accelerations)

d. Step 4: find the average speed for stage 1, stage 2, and the average of both i.

372.745 .745 = 66.1 m/s = 2.06 m/s S2 = (6 − .361) .361 372.745 − (−.745) Stot = = 62.25 m/s 6

S1 =

e. Step 5:r find the average velocity for stage 1, stage 2, and the average of both r i. v1 = −2.06 m/s v2 = 66.1 m/s

r 372 vtot = = 62 m/s 6

f.

Step 6: find the average acceleration for stage 1, stage 2, and the average of both

i.

r 0 − (−4) = 11.08 m/s 2 a1 = 0.361 r 164 − (−4) a tot = = 28 m/s 2 6

r 164 − (0) = 29.08 m/s 2 a2 = 6 − 0.361

g. Draw a picture of the particle’s motion and graph the x, v, s, & a functions.

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Chapter 2 (Motion in One Dimension)

F. Linear Kinematics Equations (for objects undergoing constant acceleration) 1. Basic Velocity Function a. v f = vi + at = this is the fundamental velocity equation 2. Other Kinematic Equations a. v 2f = vi2 + 2a(x f − xi ) velocity, acceleration, & position 1 b. x f = xi + (vi + v f )t position, velocity & time 2 1 c. x f = xi + vi t + at 2 position, velocity, time & acceleration 2 3. Freefall Kinematic Equations – since acceleration is -9.8 m/s² we can use these: a. v f = vi − 9.8t velocity & time b. v 2f = vi2 − 19.6(x f − xi ) c.

x f = xi + vi t − 4.9t

2

position & velocity position, velocity, & time

G. Example 4:

1. Car A is going at 75mph when it realizes its closing in on the car in front of him (car B). Car B in front of him is 200ft away and is moving at 65mph when car A hits the brakes, decelerating at 2ft/s². Car B also notices this and 1 second later, accelerates at 3ft/s². a. Is there an accident and if yes, what are their velocities upon impact? i. Step 1 – convert mph to ft/s and find out the velocity & position after 1 second

b)

75 mi 1h 5280 ft = 110 ft / s h 3600 s 1mi v f = 110 − 4(1)2 = 106

c)

car 1 x f = xi + vit +

d)

car 2 x f = 200 + 95.3 3(1) + 0 = 295.33 ft

a)

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65 ×

5280 = 95.3 3 ft / s 3600

1 2 1 a t = 0 + 1 1 0 + ( − 4 )(1) 2 = 1 0 8 ft 2 2

Comment [as3]: Notes for Monday, June 19, 2006 begin here

Chapter 2 (Motion in One Dimension) ii.

Step 2 – determine the position equations for both vehicles from 1 second onward.

car 1 → x f = xi + vi t +

a)

1 2 at 2

x f = 108 + 106t − 2t 2

1 3 car 2 → x f = 295.3 3 + 95.33t − (3)t 2 x f = 295.33 + 95.33t + t 2 2 2 Step 3 – set both vehicles’ final positions ( x f ) to equal each other. If they have a solution,

b)

iii.

then a collision will occur.

3 2 t 2

a)

108 + 106t − 2t 2 = 295.3 3 + 95.33t −

t = 10.667 ± 16.152i

b)

A solution cannot have imaginary numbers, so there is no solution = they do not collide.

b. If they do not collide, what is the distance of their closest approach? i. Step 1 – find their closest approach by adding both position equations (making them one) and finding the resultant equation’s derivative. Then maximize the derivative to find its local max/min.

3 f (t) → 108 + 106t − 2t 2 = 295.3 3 + 95.33t + t 2 2 f (t) → 187.33 − 10.66t + 27 t 2

a)

f '(t) → 7t − 10.66

f '(t) = 0

7t − 10.66 = 0

t = 1.524s (time of max or min distance)

car 1 x f = 108 + 106(1.524) − 2(1.524)2 = 264.879 ft 3 car 2 x f = 295.3 3 + 95.33(1.524) − (1.524)2 x f = 444.104 2 329.137 − 264.879 = 179.225 ft (this is the closest they will ever get)

b) c) d) e)

time of closest approach is 2.524 seconds.

H. Example 5:

1. Train A is going to the left at 30m/s when he sees train B going to the right at 25m/s on the same tracks 160m away. Train A decelerates at 2m/s² and train B decelerates at 3m/s². Is there an accident and if so, what are their velocities upon impact? a. Draw a diagram showing the trains and their relevant values

b. Step 1- calculate the trains’ x positions and set them equal to each other (same x = collision). This will return the time at which they collide.

i. ii.

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1 (train A) x f = 160 − 30t + (2)t 2 2 1 (train B) x f = 0 + 25t − (3)t 2 2

= 160 − 30t + t 2

3 = 25t − t 2 2

Chapter 2 (Motion in One Dimension) iii.

3 160 − 30t + t 2 = 25t − t 2 2 t = 3.45s

→

5 2 t − 55t + 160 = 0 2

c. Step 2 – check the velocities of the trains upon impact. This tells us both the impact velocity and if validates the equation. If the velocities of the trains keep the same sign, the time of collision is valid

i.

(train A) v f = (−30) + (2)(3.45) = −23.1 (sign remains the same)

ii.

(train B) v f = (25) + (−3)(3.45) = 14.65 (sign remains the same)

d. In this case, both trains retain their signs. This means the time of collision is valid. If one of the trains changed sign, it would mean the train stopped before the aforementioned time which invalidates the equation. In this case, we need to find out when the sign-changing train has zero velocity because this is its final resting point. Once we figure that out, we simply treat the rest of the problem (with the other train) as a simple “train hitting the wall” equation with the “wall” being the train that stopped. To find out the time of the REAL collision, set the other train’s x position to the same and find the time and velocity of impact!

I. Vertical Motion Problems Including Freefall

1. In the case of freefall, we use the same equations, but now because we always know that g is 9.8, we use that for all vertical motion problems (except where acceleration is affected by something else).

J. Example 6:

1. A guy on a building 50m high throws a ball up with a velocity of 20m/s. The ball goes up, and then returns but keeps going to ground level where there is a lake. Upon hitting the lake, its acceleration changes to 2m/s down, taking 12 seconds to reach the floor. 2. Questions: a. What is the depth of the lake? b. What is the final velocity? 3. Solution Strategy: a. Step 1 – find out the time the ball hits the water

i.

1 y f = yi + vi t + at 2 2 0 = 50 + 20t − 4.9t 2 t = 5.83 (the time it hits the water)

b. Step 2 – find its velocity upon hitting the water i. v f = 20 − 9.8(5.83) v f = −37.134m / s c. Step 3 – now use that velocity, the time left in 12 seconds, and the known acceleration in water to figure out the depth of the lake and velocity upon impact.

1 Depth = Vi t + at 2 2 i.

ii.

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1 = (−37.134)(12 − 5.83) + (−2)(12 − 5.83)2 2 = (−37.134)(6.17) − (6.17)2 = −267.161m Lake v f = 37.13 + (2)(6.17) = 49.47m / s

Chapter 2 (Motion in One Dimension)

K. Example 7:

1. A rocket goes up for a constant acceleration for 8 seconds, then its engines fail and it goes into a freefall, hitting the ground with a velocity of 200ft/s.

2. Questions: a. What is the total time for this trip? b. What is the acceleration of the rocket? c. What is the maximum height of the rocket? 3. Solution Strategy: since we have multiple unknowns, we have to figure out a way to relate them to each other. a. Step 1 – figure out how the velocity relates to the height i. v f = vi + at = 0 + a(8) = 8a (This is the final velocity of the rocket during its firing stage. This will become the initial velocity of our rocket in its freefall stage)

ii.

y f = yi + vi t +

1 2 1 at = 0 + 0 + a(8)2 = 32 a 2 2

(This will be the final height of the rocket during its firing stage. This will become the initial height of our rocket in its freefall stage) b. Step 2 – we now have enough to calculate the descent since we have all variables but time. We can use the velocity squared function (the one that excludes time) to put it all together.

v 2f = vi2 + 2a(y f − yi ) i.

200 2 = (8a)2 + 2(−32)(0 − 32a) a = 13.68

Keep in mind, the “a” in this final equation is NOT the same as the “a” in our previous two equations. Since this equation is based on the freefall, the “a” is -32ft/s, whereas the previous “a” was based on the rocket’s propelled ascent. c. Step 3 – with our acceleration, we can now figure out the time at which the rocket has zero velocity (the top) and from there, get the position at the top ( y f )

i.

v f = vi + at

y f = yi + vi t + 1 2 at 2

⇔

y f = 32a + 8at − 16t 2

0 = 8(13.68) + (−32)t t = 3.42s

y f = 32(13.682) + 8(13.682)(3.42) + (−16)(3.42)2 y f = 625.02 ft

III. Chapter 3: Vectors A. Coordinate Systems

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1. Background - There are various systems for denoting vectors, such as Polar Coordinates, bearing (N, S, E, W) & magnitude or component vectors. a. Polar coordinates example: i. 10m @ 260° b. Bearing & magnitude example: i. 10m @ 10°W of N c. Component vectors example: i.

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10Cos(260°)iˆ −10Sin(20°)iˆ which is equivalent to: 10Sin(260°) ˆj 10Cos(20°) ˆj

Chapter 3 (Vectors)

B. Example 1:

1. A box has the following forces applied to it ( ax = −4N , ay = +2N ). Draw a picture of the box with its vectors and find out what is the resultant vector in polar format. 2. Solution Strategy: a. Step 1: add up the vectors i.

(−4)2 + (2)2 = 4.472

b. Step 2: find the angle i. Tan −1 ( −42 ) = −26° = 180 − 26 = 154°

C. Example 2:

1. A car goes 40 miles at 35° N of W in ⅔hr and then turns around and goes 30 miles at 60° E of N in 3/5hr.

2. Questions: since this is a two stage problem, we will find all relevant data for all stages) a. Find s1 , s2 , stot r r r b. Find v1 , v2 , vtot r r r c. Find a1 , a2 , atot d. Draw the path of the car

3. Solution Strategy: a. Step 1 – determine the unit vectors for both lines and add them up ur i. A = 40Cos(145°)iˆ + 40Sin(145°) ˆj = −32.766iˆ + 22.943 ˆj ii. iii.

ur B = 30Cos(22°)iˆ + 30 Sin(22°) ˆj = 27.816iˆ + 11.238 jˆ −32.766iˆ + 22.943 ˆj + 27.816iˆ + 11.238 ˆj

= −4.95iˆ + 34.183 ˆj

b. Step 2 – find the resultant vector’s magnitude and direction i. ii.

ur C = 4.95 2 + 34.1832 = 34.540 34.183 Tan −1 ( ) = 81.76° 4.95

c. Step 3 - find the velocity and speed for all three stages:

i. ii. iii.

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v1 =

x f 1 − xi1

=

40 − 0 = 60mph @35° N of W 2/3

t 30 − 0 v2 = = 50 mph @ 68° E of N 3/5 ur C 34.540 vtot = = = 27.268 @ 81.76° N of W t tot 2 / 3 + 3 / 5

iv.

Since the speed is the absolute value of the velocity, your speed for stage 1 and stage 2 are the absolute value of the velocity, but the third speed is NOT. Unlike the total velocity that is the total displacement over the total time, the total speed is the total distance over the total time.

v.

s1 = v1 = 60mph

s2 = v2 = 50mph

Chapter 3 (Vectors) vi.

stot =

dtot 30 + 40 = 55.26mph ttot 2 / 3 + 3 / 5

d. Step 4 – find out the acceleration of the car i. a1 = 0 Because neither the velocity nor the direction changed ii. a2 = 0 Because neither the velocity nor the direction changed

iv.

r v f − vi [50Cos(22°)iˆ + 50Sin(22°) ˆj] − [60Cos(145°)iˆ + 60Sin(145°)] atot = = 19 / 15 ttot ˆ ˆ v f − vi (46.359i + 18.730 j) − (−49.149iˆ + 34.415 ˆj) a3 = = ttot 1.26 95.508iˆ − 15.684 jˆ = 75.401iˆ − 12.382 ˆj (Total acceleration in vector components) 1.26 r atot = (75.401)2 + (−12.382)2 = 76.411 (Total acceleration)

v.

θ = Tan −1 ⎜

iii.

⎛ −12.382 ⎞ = −9.326° = 350.674° (theta of Total acceleration) ⎝ 75.401 ⎟⎠

IV. Chapter 4: Motion in Two Directions

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A. Background:

1. This chapter is really an extension of Chapter 2 (motion in one dimension) and Chapter 3(vectors). From chapter 2, we know that the motion of a particle is a function of time, from which we can infer its velocity, and acceleration. From Chapter 3,

B. Distance (d)(in 2 dimensions)

1. When measuring the distance of an object moving in two dimensions, we must include all distances traveled by the object. a. if the path is linear or has linear segments, calculating its distance is a matter of adding up its segments.

b. If the path is curved and we know the function that represents the curve, we can use the arc length integral to determine its distance from point “a” to point “b” of the function. b

i.

dcurve =

∫

1 + f '(x)2 dx

a

C. Displacement (D)(in 2 dimensions) 1. Displacement is the distance between two points. In vector form, this works for any vector, whether it be position, velocity, or acceleration. a. b. c. d. e.

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r r r r Displacement ( D) = rfinal − rinitial = (x f − xi )iˆ + (y f − yi ) ˆj = Δr Δr Average Velocity v ≡ Δt r dr dx ˆ dy ˆ Instantaneous Velocity v = i+ j = vx iˆ + vy ˆj = dt dt dt r v f − vi Δv Average Acceleration a ≡ = t f − ti Δt dv Instantaneous Acceleration a ≡ dt

Comment [as4]: Notes for Wednesday, June 21, 2006 begin here

Chapter 4 (Motion in Two Dimensions)

D. Example 1:

1. Problem: An object is traversing a path given by the equation y(x)=x³ from (-1, 1) to (2, 8) feet. It is traveling at such a rate that dx/dt = 2 ft/s and it makes the journey in 1.5 seconds. 2. Questions: a. Calculate the total distance traveled and the average speed. b. Calculate the total displacement and the average velocity. c. What is the initial & final velocity, initial & final speed, initial & final acceleration, and average acceleration?

3. Solution Strategy: a. Nota bene: a common mistake most people make is to confuse the position and the velocity vectors and thus get the wrong result and can’t figure out why…if your results are not what you expected (but sometimes by coincidence they might work until you get to the acceleration vector), be aware of which are needed for what! b. Step 1 – calculate the distance for this curve using the arc length integral, which you can use to calculate the average speed. 2

∫

i.

d=

1 + (3x 2 )2 dx = 10.178

ii.

10.178 s= = 6.79 ft s 1.5

−1

c. Step 2 – calculate the total displacement and average velocity r i. D = rf − ri = (2iˆ + 8 ˆj) − (−1iˆ − 1 ˆj) = 3iˆ + 9 ˆj

r r D 9 2 + 32 9.487 v= = = = 6.325 ft s 1.5 t 1.5 θ = Tan −1 (9 / 3) = 71.57°

ii.

iii. d. Step 3 – calculate the component-position function in order to derive the velocity and acceleration functions. Since y(t)=x³, you need to figure out x(t) because your velocity function depends on both the y position and the x position:

i.

a)

if

dx dx = 2 → x = ∫ dt = ∫ 2dt = 2t + c dt dt

This equation will give us the x-position of the particle as a function of time but we still need to figure out the constant “c.” we do this by remembering that at time = 0, x = -1 (see position graph) b)

when t=0, x=-1, so x(0)=-1

→ x(0)=2(0)+c=-1

c = −1

x(t) = 2t − 1

Now we know the x function and only need to recall the y function. c)

y(x) = x 3 This equation was given. It gives us the y-position as a function of x-position. We can make it a function of time by sticking the x function inside it

y(t) = y(x(t)) = (2t − 1)3 once we’ve done this, we’re ready to put both functions together. d)

r r(t) = x(t)iˆ + y(t) ˆj = (2t − 1)iˆ + (2t − 1)3 ˆj

This equation creates a component-position function that tells us where the x and y components of the particle are at any point in time. For example, at 1.5 seconds, it tells us the particle is at (2,8) which matches our position function. This function is used to arrive at the velocity since the velocity is a function of the absolute position over time.

ii.

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Once we have the position function r(t), we can derive it with respect to time to get the velocity function:

Chapter 4 (Motion in Two Dimensions) r r dr a) v = = 2iˆ + 3(2t − 1)2 (2) ˆj = 2iˆ + (24t 2 − 24t + 6) jˆ dt iii.

From here, we continue the process to get to the acceleration function:

r a = (48t − 24) ˆj

a)

e. Step 4 – with all necessary functions, calculate their values at the beginning and end of the trip and any other significant landmarks (for example, if and when the velocity and/or acceleration went to zero, we might want to know that in some cases) i. We can see that the initial and final positions were given (which is where we got our equations from, but if you want to check, plug them back into the component-position function:

ri = r(0) = (0 − 1)iˆ + (0 − 1)3 ˆj = − iˆ − ˆj r = r(1.5) = (2(1.5) − 1)iˆ + (2(1.5) − 1)3 ˆj = 2iˆ + 8 ˆj

a) b)

ii.

Comment [as5]: The velocity picture has an absolute placement which might get messed up if you move the whole thing around, so keep this in mind

f

Now we calculate initial and final velocity:

vi = v(0 ) = 2iˆ + (0 + 0 + 6) ˆj = 2iˆ + 6 ˆj v = v(1.5) = 2iˆ + (24(1.5)2 + 24(1.5) + 6) ˆj = 2iˆ + 96 ˆj

a) b)

f

This tells us the velocity at the beginning and the end of the particle’s trajectory.

iii.

Since the speed is the magnitude of the velocity vectors we get:

si = vi = 2iˆ + 6 ˆj s = v = 2iˆ + 96 jˆ

a) b)

iv.

f

f

Now we do the same for the acceleration:

ai = a(0) = (48(0) − 24) jˆ = −24 jˆ a = a(1.5) = (48(1.5) − 24) ˆj = 48 ˆj

a) b)

f

This tells us the acceleration at the beginning and end of the particle’s trajectory.

f.

Step 5 – this allows us to figure out the average acceleration:

i.

aavg =

v f − vi ttot

=

(2iˆ + 96 ˆj ) − (2iˆ + 6 ˆj ) 90 ˆ = j = 60 jˆ 1.5 1.5

E. Example 2:

1. Problem: a particle is going at a constant speed in a counterclockwise circular motion with a radius of 22ft. it starts at an angle of 25° and ends up at 287° in 10 seconds. 2. Questions: a. What is the average speed and average velocity? b. What is the instantaneous acceleration? c. What is the average acceleration and show that it points to the center when placed at the midpoint of the path.

3. Solution Strategy: a. Step 1 – find the average speed by calculating the distance i. d = rΔθ = 22[(287° − 25°) π 180 = 100.6 ft ii. s = d t = 100.6 10 = 10.06 ft / s b. Step 2 – find the velocity by completing the triangle to get the displacement (keep in mind that since both sides of the triangle are the same length, you have an isosceles triangle, which tells you two of its legs are the same angle (the

Page 12 of 59

Comment [as6]: Notes for Monday, June 26,2006 begin here

Chapter 4 (Motion in Two Dimensions) two legs that make angles with the leg of different length). We can use this to complete the triangle. i. Using the law of cosines

ii.

a)

c 2 = a 2 + b 2 + 2abCos(γ )

b)

c = 22 2 + 22 2 + 2(22)(22)Cos (98) = 33.21

Using the law of sines a)

Sin(a ) Sin(b ) = b r a v = D t = 33.2110 = 3.321 ft/s

Sin(41°) Sin(98°) = 22 x

x = 33.21

iii. c. Step 3 – the instantaneous acceleration is centripetal given its circular motion:

ac =

i.

v 2 3.3212 = = 0.501 ft/s 2 r 22

d. Step 4 – the average acceleration is (like in the previous example), is the result of the final minus initial acceleration over time. The trick here though is finding the final and initial acceleration! One thing to remember is that the velocity doesn’t change throughout this curve. i. The initial & final velocities require that we figure out their vectors. Since the velocity vector in a circular motion is the tangent (90° ) to the particle’s angular position, our calculations only require simple math (as long as you understand the previous statement) a) b)

ii.

Once we know the direction, we calculate the initial and final velocity by multiplying the speed by the direction to get a component (vector) of the velocity. a) b)

iii.

vθ initial = xθ initial + 90° = 25° + 90° = 115° vθ final = xθ final + 90° = 287° + 90° = 17°

v f = s f × vθ f = 10.06 × 17° = 10.06[Cos(17°)iˆ + Sin(17°) ˆj] = 9.620iˆ + 2.941 ˆj v = s × vθ = 10.06 × 115° = 10.06[Cos (115°)iˆ + Sin(115°) jˆ ] = −4.251iˆ + 9.117 jˆ i

i

i

from here, calculating the average acceleration vector is the same as other examples: a)

b) c)

r v f − vi (9.620iˆ + 2.941 ˆj ) − (−4.251iˆ + 9.117 ˆj ) 13.872iˆ − 6.176 jˆ a= = = 10 t tot 10 r ˆ ˆ a = 1.387i − 0.618 j r a = 1.387 2 − 0.618 2 = 1.518 (The magnitude in case we need it) ⎛ −.618 ⎞ = −24° or 336° Tan −1 ⎜ ⎝ 1.387 ⎟⎠

F. Projectile Motion

1. Background: Projectile motion is really an extension of our previous two-dimensional motion problems, except simplified. The initial velocity vector is broken down into horizontal and vertical components and these components are solved in their own kinematic equations. In most cases, the airtime of the particle will determine the trajectory of the particle. This tells us that we want to find out how long the ball is in the air (y-kinematics) and from that, we can figure out how far the ball will go (x-kinematics) since both are related by time (the same time, duh).

Page 13 of 59

Chapter 4 (Motion in Two Dimensions)

2. In this example, we can see a number of notable properties of a ball in projectile motion: a. The horizontal velocity does not change (if we ignore wind resistance) b. The vertical velocity is the same at the same y-position. So the catcher of the ball in this example feels the same velocity with which the thrower threw it.

c. The point at which vertical velocity is zero is the apex of the curve, many problems will require that you figure this out.

d. The only acceleration is vertical (g) which affects the velocity of the ball throughout the trip.

G. Example 3:

1. Problem: a basketball is thrown from a height of 6’10” towards a basket 25’ away and 10’ high. The initial velocity of the ball is 28ft/s and the angle is 42°. 2. Questions: a. Does the person make the basket? What is the final velocity? b. If not, what angle should they aim at? 3. Solution Strategy: a. First we figure out the components of the initial velocity i. 28Sin42° = 18.736 ii. 28Cos42° = 20.808 b. Now we calculate how long it takes the ball to travel 25 feet in the x direction. Since this time will be our limiting time, we will then use it to figure out our y altitude at that time. If it is exactly 10 ft, we have a basket. Any other value will mean the ball is over or under the basket, thus missing.

i.

x f = xi + vi t

ii.

y f = yi + vi t +

25 = 20.81t

1

2

at

2

t = 1.20s

y f = 6.83 3 + 18.736(1.2) − 16(1.2)2 = 6.276

iii. at x=25ft, y is only 6.3 feet, falling short of the basket. c. Since the ball fell short, we try to determine an angle that (with the same velocity) will give us the required (x=25, y=10) values. We set up both x and y equations and set them equal to each other

i. ii.

Page 14 of 59

10 = 6.83 3 + 28Sin(θ °)t − 16t 2 (The y-equation) 25 (The x-equation) 25 = 28Cos(θ °)t t= 28Cos(θ °)

Chapter 4 (Motion in Two Dimensions) 2

iii.

25 ⎛ ⎞ ⎛ 25 ⎞ 10 = 6.833 + 28Sin(θ ) ⎜ − 16 ⎜ ⎟ ⎝ 28Cos(θ ) ⎠ ⎝ 28Cos(θ ) ⎟⎠ 16(625) 3.166 = 25Tan(θ ) − (784)Cos 2 (θ ) 3.166 = 25Tan(θ ) − 12.755Sec 2 (θ ) → no solution

d. Since there is NO angle at which we can launch it, we now keep the angle the same and change the velocity to find out a velocity at which both would be the same(x=25, y=10).

i.

3.166 = 25Tan(42°) −

16(625) vi Cos 2 (42°) 2

v i = 30.596

H. Example 4:

1. Problem – during a soccer game, a penalty corner kick is awarded to the attacking team. The kicker, trying to get past the goalie, wants to kick the ball to his teammate who can then “head” it into the goal. This “header” is at an angle of 10° from the kicker and right down the middle of the field (32.5 meters). 2. Questions: 3. Solution strategy:

I. Kinematics of Circular Motion

1. Background: Centripetal (towards the center) acceleration occurs whenever a particle turns a corner or otherwise travels in a circular arc or a whole circle. Calculation of centripetal acceleration is the result of the direction changing (because while the object moves, its acceleration keeps pointing to the center, constantly changing in angle), while the speed usually remains constant; this is uniform circular motion. If both the speed and direction change, then the particle undergoes non-uniform circular motion. a. Centripetal acceleration:

ac = ω 2 r =

r r v ⋅v vvCos(0) v2 ⋅r = ×r= r ⋅r r×r r

this is used in uniform circular motion

b. Tangential acceleration: at =

dv dt

this is used in non-uniform circular motion along with the centripetal acceleration equation, as both are components of an object’s trajectory.

2. In uniform circular motion, since there is no tangential acceleration, centripetal acceleration is also the total acceleration. In non-uniform circular motion, the tangential component is simply tangential (at a 90° angle) to the radial component. The resultant vector is the total acceleration. Another way to denote total acceleration (for both

types of circular motion) is to use the θˆ, rˆ notation, but please note, that the rˆ unit is directed outwards, so if using it, you must reverse the sign of the centripetal vector. a. atot =

ac 2 + at 2

θ (atot ) =

at (for non-uniform circular motion) ac

b. atot = ac (− rˆ ) + at (θˆ )

J. Example 5:

1. Problem – A car starts out at an initial velocity of 6ft/s around a circle of 60ft radius. It accelerates at a rate 3t+1 ft/s².

Page 15 of 59

Comment [as7]: Notes for Wednesday, June 28, 2006 begin here

Chapter 4 (Motion in Two Dimensions) 2. Questions: What are the initial centripetal acceleration, initial tangential acceleration and initial total acceleration (including its angle)? b. What are the centripetal acceleration, tangential acceleration and total acceleration (including its angle) after 12 seconds?

a.

3. Solution Strategy: a. Figure out the centripetal and tangential components to calculate the resultant vector:

62 = .6 ft/s 60

i.

ac =

ii.

atot = .6 2 + 12 = 1.1662 ft/s

at = 3(0) + 1 = 1 ft/s ⎛ 1⎞ ArcTan ⎜ ⎟ = 59.04° ⎝ .6 ⎠

b. For question b, the challenge is figuring out the centripetal acceleration at 12 seconds. Since the object is speeding up as it moves, its centripetal component will grow dramatically larger over time. Remembering that velocity is the integral of acceleration, we simply integrate the acceleration function with a time of 12 seconds and add 6 seconds (because the initial velocity was 6 ft/s). From there, it's the same as the previous equations: 12

i.

v(12) = ∫ (3t + 1)dt + 6 = 234 ft/s

ac =

0

v 2 (234)2 = = 912.6 ft/s 2 r 60

at = 3(12) + 1 = 37 ft/s 2 ii.

atot = 912.6 2 + 37 2 = 913.35

⎛ 37 ⎞ = 2.32° ArcTan ⎜ ⎝ 912.6 ⎟⎠

K. Relative Velocity and Relative Acceleration

1. Background: this topic comes from the idea that observations of movement are often related to the reference frame of the observer. For instance, a person walking down the moving walkway between Caesars Palace and the Mirage would appear to be going much faster to a stationary observer while an observer standing on the moving walkway would only see him going at a normal walking pace. This relativity of speed forms the basis of calculation of speeds where we would want to figure out the movement of an object (an airplane) in a moving medium (air), relative to a static position (the ground). Just like the speed of the walkway walker relative to the ground is simply the addition of the speed of the belt and the speed of the walker, we could calculate an airplane’s speed relative to the ground by adding its velocity relative to the air with the velocity of the air relative to the ground.

L. Example 6:

1. Problem – An airplane is headed from Seattle WA to Miami FL just in time for the hurricane season. The distance between the 2 cities is 2400 miles and the angle is 40° S of E. A wind rises up during the flight with a speed of 70mph @ 80° N of W 2. Questions: a. If the plane wants to make the trip in 4 hours, what should their average airspeed and direction be?

b. If the pilot forgets to account for the wind, where will the plane end up in 4 hours? 3. Solution Strategy: a. Make an image of the exercise to get an idea of the solution:

Page 16 of 59

Chapter 4 (Motion in Two Dimensions)

b. First – we figure out the vectors for all three: i. ii.

iii. iv.

2400mi = 600 mi/hr 4hrs ⎧vagy = 600Sin(320°) = −385.673⎫ vag = 600@40° → ⎨ ⎬ ⎩ vagx = 600Cos(40°) = 459.627 ⎭ ⎧ vwgy = 70Sin(100°) = 68.937 ⎫ vwg = 70 @100° → ⎨ ⎬ ⎩vwgx = 70Cos(100°) = −12.156 ⎭ vwg = ?

c. Next – we use the pictures we made to figure out the unknowns for question a: i. vag − vwg = vaw ii. iii.

⎧ vagy − vwgy = vawy ⎫ ⎧ 600Sin(320°) − 70Sin(100°) = vawy ⎫ ⎧−454.61⎫ ⎬= ⎨ ⎬ ⎨ ⎬= ⎨ ⎩vagx − vwgx = vawx ⎭ ⎩600Cos(320°) − 70Cos(100°) = vawx ⎭ ⎩ 471.78 ⎭ ⎛ −454.61 ⎞ (−454.612 ) + (471.78 2 ) = 655.17 ArcTan ⎜ = −43.9° = 316.1° ⎝ 471.78 ⎟⎠

d. After that, we solve question b in the same way (but with its own equation of course) i. vag + vwg = vaw ii. iii. iv.

V.

⎧ vagy + vwgy = vawy ⎫ ⎧ 600Sin(320°) + 70Sin(100°) = vawy ⎫ ⎧−316.74 ⎫ ⎬= ⎨ ⎬ ⎨ ⎬= ⎨ ⎩vagx + vwgx = vawx ⎭ ⎩600Cos(320°) + 70Cos(100°) = vawx ⎭ ⎩ 447.47 ⎭ ⎛ −316.74 ⎞ (−316.74 2 ) + (447.47 2 ) = 548.22 ArcTan ⎜ = −35.3° = 324.7° ⎝ 447.47 ⎟⎠ dtotal = 548.22 × 4 = 2192.91 miles

Chapter 5: The Laws of Motion A. Background:

(return to top)

1. How is Kinematics different from Dynamics: In Kinematics, you’re asked to analyze a motion (i.e. circular, left/right, up/down, projectile). In Dynamics, you are asked to examine the forces which produce the motions you analyze in kinematics (gravity, normal, centripetal).

2. Fundamental Dynamic Equation:

∑ F = ma

B. Example 1:

1. Problem - You have a block with mass 2kg, which is being moved with an initial velocity of 5m/s for 15m. The force moving it is 15N.

2. Questions: Page 17 of 59

Chapter 5 (The Laws of Motion) a. What is the final velocity? b. Assume the same block but now it weighs 2lb, its initial velocity is 5ft/s, the force is 6lbs, and the distance is 15ft. what is the final velocity now?

3. Solution Strategy: a. Draw a diagram of the system

b. Set up the dynamic equation to analyze the forces: i. F = ma → 6 = (2)a

∑

a = 3m / s 2

c. Use kinematic equations along with the acceleration figure to calculate the final velocity: i. v f 2 = vi 2 + 2a(x f − xi ) v f 2 = 5 2 + 2(3)(15) v = 10.73m/s d. For question b, we simply switch values, but keep in mind the weight is not the mass, we need to convert it to mass to get an accurate value! 1slug 32 lb

i.

2lb ×

=

ii.

∑ F = ma

1

16

slug 6lb = ( 116 slug)(a)

a = 96ft/s 2

C. Example 2:

1. Problem – You have a block with mass 2kg, which is being moved vertically with an initial velocity of 5m/s for 15m. the force moving it is 15N

2. Questions: a. What is the final velocity? b. Assume the same block but now it weighs 2lb, its initial velocity is 5ft/s, the force is 6lbs, and the distance is 15ft. what is the final velocity now?

3. Solution Strategy: a. In this instance, keep in mind that gravity is playing a part, so include it in our initial dynamic calculations.

i.

∑ F = ma

ii.

v f = 5 + 2(−6.8)(15) 2

(6 − 19.6) = (2)a

2

a = −6.8

v f = 13.3i

since we get an imaginary number, this means with the given forces, the block will never make it to 15m. b. for part b, we again remember to convert the lbs to slugs before figuring out the acceleration.

iii.

i.

∑ F = ma

ii.

v f 2 = 5 2 + 2(64)(15)

(6lb − 2lb) =

1

16

a

a = 64ft/s2

v f = 44ft/s

D. Weight, Mass, and the Normal force

1. Although the normal force is sometimes hard to figure out, an intuitive way of thinking about it is think of it as your “apparent weight” because it is the normal force that gives us our feeling of weight. For example, a person in a rapidly accelerating and rising elevator will feel lighter because the normal force is higher (their weight will be the same). A person in the same elevator, which is rapidly accelerating and going down, will fell very little “apparent weight”, and as a matter of fact, if the acceleration of the elevator is the same as gravity, the elevator passenger will feel “weightless.”

E. Example 3 (friction): Page 18 of 59

Comment [as8]: Notes for Monday, July 3, 2006 begin here

Chapter 5 (The Laws of Motion) 1. Problem – You have a 3kg block being pulled with a force F. the coefficients of friction are μs= 0.6 and μk= 0.4

2. Questions: a. If the force applied is 5N, will the block move? If not, what is the magnitude of the force of static friction?

b. What is the minimum force needed to move the block? c. Once the block is in motion, what is the minimum force needed to keep it in motion without accelerating it?

d. If the applied force is 1N greater than the answer in “b,” what will the block’s acceleration be? e. If the applied force is 1N greater than the answer in “c”,” what will the block’s acceleration be? 3. Solution Strategy: a. See question b first, it will answer both a and b together. i. (Based on “b”) No. ii. fs = Fapplied = 5N b. We first need to use the equation of the force of static friction to figure out the maximum static friction force before the block slips.

i. (Fmin = fs max ) ≥ μ s n → fs max ≥ (.6)(29.4) = 17.64N c. Now we need to use the equation in part “b” but with μk instead: i. (Fmin = fk ) ≥ μ k n → fk ≥ (.4)(29.4) = 11.76N d. We figure out the acceleration (if any) by figuring out its net force: i. F = ma → F − μ s n = ma

∑

18.64 − (17.64) = 3a →

ii.

a = 2.29 3m/s 2

e. We do the same here except with the kinetic force: i. F = ma → F − μ k n = ma

∑

12.76 − (11.76) = 3a

ii.

→

a = .3m/s 2

F. Example 4:

1. Problem – the problem is the same as in example 3, but with the external force pulling at an angle of 50° above the horizontal

2. Questions: a. Same as in example 3. b. Same as in example 3. c. Same as in example 3. d. Same as in example 3. e. Same as in example 3. f. What is the minimum force, which will lift the block? What will its acceleration be then? 3. Solution Strategy: a. See question b first, it will answer both of these questions i. (Based on “b”) No. ii. fs = 5Cos50° = 3.21 b. Like before, we need to figure out the minimum force which can move the block, but this time, we need to keep in mind it is made up of two components, both of which affect the equation, so we will need to solve for two equations at once: i. Since the external force is reducing the normal force, we need to take it into account when calculating the vertical forces: a)

Page 19 of 59

∑F

y

= mg

→

n + FSin50° = 29.4

→

n = 29.4 − FSin50°

Chapter 5 (The Laws of Motion) ii.

When calculating the horizontal forces, we again need to calculate the horizontal component of the external force, this will combine both equations and give us the answer to “b”:

∑F

a)

x

= ma

→

FCos50° − μ s n = 3a

FCos50° − (.6)(29.4 − FSin50°) = 0 → 1.103F = 17.64 → F = 16.002

b) c)

.6428F − 17.64 + .4596F = 0

c. This is the same as the previous equation but the coefficient of static friction is replaced by the coefficient of kinetic friction:

FCos50° − (.4)(29.4 − FSin50°) = 0 .9492F = 11.76 → F = 12.389

i. ii.

→

.6428F − 11.76 + .3064F = 0

d. For this question, we simply increase the force by 1N and solve it as a dynamic equation: i. F = 17 (17)Cos50° − (.6)[29.4 − (17)Sin50°] = 3a ii. 10.9274 − 17.64 + 7.8137 = 3a a = .367m/s 2 e. We do the same, except we use the kinetic values instead: i. F = 13.389 (13.389)Cos50° − (.4)[29.4 − (13.389)Sin50°] = 3a

8.6063 − 11.76 + 4.1026 = 3a

ii. f.

a = .3163m/s 2

In order to solve this, we set the normal force to zero and calculate the vertical acceleration. After this, just plug it into the horizontal equation:

N + FSin50° = 29.4 N=0 FSin50° = 29.4 ∑ Fy = ma

i. ii.

∑F

iii.

x

= ma

38.378Cos50° − (0) = 3a

F = 38.379N a = 8.223m/s 2

G. Example 5:

1. Problem – the problem is the same as in example 3, but with the external force pulling at an angle of 50° above the horizontal and the block is on a 20° incline.

2. Questions: a. Same as in example 3. b. Same as in example 3. c. Same as in example 3. d. Same as in example 3. e. Same as in example 3. 3. Solution Strategy: (here one important tip is to accurately record the component vectors for all relevant forces. A good idea is to rotate the inclined plane so that the incline is the x-axis. All angles would need to be readjusted, but this often makes it easier)

a. Like the previous two, we need to go to “b” to figure out “a”: i. fs (5) = μ s n fs (5) = (.6)(27.6270 − (5)Sin30°) = 15.07N a)

ii.

Page 20 of 59

Since the force of static friction at 5N is less than the minimum force needed to move the block up the incline (fsmax), the block will not move up.

Since the block is on an incline and we know it will not move up, we now need to know if the block will move down. If not, then we need to find the force that keeps it from going down.

Chapter 5 (The Laws of Motion) We need to figure out the force that stops the block from moving down. Since the pull is down and friction acts against this force too, the friction force (which pulls up) plus the external force pulling the block up should equal the force pulling the block down.

a)

fs(up) + F(5) = FDownRamp

1)

3)

fs ( up ) + (5 )Cos30° = 29.4Sin20° fs(up) = 10.0554 − 4.3301

4)

fs ( up ) = 5.7253

2)

b)

So the force of static friction up the block (against gravity’s pull) fs(up) is the force keeping the block from sliding down the ramp.

b. As in example 4, we must remember the force is at an angle, and will affect both the x and y components of the equation, so we need to keep this in mind.

i. ii. iii.

∑F ∑F

y

= ma

x

= ma

n + FSin30° = mgCos20°

n = 27.6270 − 0.5F

FCos30° − (mgSin20° + μ s n) = 0

0.866F − [10.055 + (.6)(27.6270 − 0.5F)] = 0 F = fsmax = 22.8388

iv. 0.866F − 26.631 + 0.3F = 0 1.16F = 26.631 c. As in example 4, but we change our equation for the kinetic force: i. Fx = ma FCos30° − (mgSin20° + μ s n) = 0

∑

ii. iii.

0.866F − [10.055 + (.4)(27.6270 − 0.5F)] = 0 0.866F − 21.106 + 0.2F = 0 1.066F = 21.106

F = 19.799

d. Here we increase the minimum static force by 1 and calculate any acceleration. Keep in mind that since we are exceeding the static force, we use the kinetic coefficient

i. ii.

F = 23.8388 y = ma

iii.

∑F ∑F

iv.

4.1098 = 3a

x

n = 27.620 − 0.5(23.8388)

= ma

n = 16.2006

(23.8388)Cos30° − [10.055 + (.4)(16.2006)] = 3a a = 1.3700m/s 2

e. Here again, we increase the minimum kinetic force by 1 and calculate any acceleration i. F = 20.799 ii. Fy = ma n = 27.6270 − 0.5(20.799) n = 17.2205 iii.

∑ ∑F

iv.

1.0693 = 3a

x

= ma

(20.799)Cos30° − [10.055 + (.4)(17.2205)] = 3a a = 0.3564m/s 2

VI. Chapter 6: Circular Motion and Other Applications of Newton’s Laws (return to top)

A. Background:

1. Objects traveling in a circular path can often be analyzed from a Newtonian perspective if we keep in mind that Newton’s 2nd law still applies. In this chapter, we will look at forces in horizontal and vertical uniform circular motion. We will also analyze through a viscous medium, which could be considered an extension of the principles of friction presented in the previous chapter.

B. Horizontal Circular Motion

1. When calculating Horizontal Circular Motion, we will usually be dealing with a y-component that is in equilibrium (thus giving us a reference point from where to solve our problems). It is usually best to solve the y-component first for this reason.

Page 21 of 59

Comment [as9]: Notes for Wednesday, July 5, 2006 begin here

Chapter 6 (Circular Motion and Other Applications of Newton’s Laws)

C. Example 1:

1. Problem – A 6lb ball is being rotated about a rod. The ball is being held by two lines, the higher one at 40° to the horizontal and the lower line at 30° to the horizontal. The lower line’s length is 4ft. The highest tension in either string is 40 lbs before snapping. 2. Questions: a. What is the shortest period that this ball can have? 3. Solution Strategy: a. First we keep in mind this is an extension Newton’s 2nd law in that both the horizontal and vertical components should be set against each other to determine the forces acting on them. We use the fact the y-components are in equilibrium, and the fact one of the two strings will reach 40lbs first (since angles and lengths are unequal between T1 and T2). Our guess as to which one reaches first should be a matter of instinct (in this case, T1 because T2 and the ball are pulling down on it), but failing that, we can still figure it out. b. First we set the y-components equal to each other and set one to 40lbs.

i. ii.

∑F

=0 (T1 )(Sin40°) = (T2 )(Sin30°) + 6 (40)(Sin40°) − 6 T2 = T2 = 39.4230 Sin30°

(set T1 = 40)

y

iii. This proves our guess was right, when T1 hits 40, T2 will be close, at 39.42lbs c. Now that we know T1 and T2, we can figure out the shortest period they can have by doing a few algebraic manipulations on our centripetal acceleration equation keeping in mind that its velocity is the same as its angular frequency (ω) times its radius, and the angular frequency is a function of its period (T):

i. ii. iii.

mv 2 and v = rω and r 2 m(rω )2 4š 2mr ⎛ 2π ⎞ 2 F = = mr ω = mr ∑ x ⎜⎝ ⎟⎠ = r T T2 4π 2 (6 / 32)(4Cos30°) (T1 )(Cos40°) + (T2 )(Cos30°) = T2

∑F

x

= ma =

ω=

2π T

T = 0.629s

D. Example 2:

1. Problem – You have a car driving around a circle of radius 120 ft. the coefficient of static friction is (.8)

2. Questions: a. What is the fastest the car can go without slipping?

3. Solution Strategy: a. Like all horizontal motion problems, we should start out by figuring the vertical forces to get friction.

i. n = mg b. For horizontal forces, we recall friction is acting as a centripetal force.

i.

Page 22 of 59

ac = fs = μ s n

ac =

mv 2 r

μsn =

mv 2 r

Chapter 6 (Circular Motion and Other Applications of Newton’s Laws) mv 2 ii. (.8)(mg ) = v = (.8)(120)g 120 3600(sec/hr) iii. 55.42ft/s × = 37.8mi/hr 5280(ft/mi)

v = 55.42ft/s

E. Example 3: (Review of Chapter 5)

1. Problem: (this review example will help us better understand how to solve example 4). You have a block of mass m on an inclined plane. The coefficient of static friction is 0.6 and the coefficient of kinetic friction is 0.4. 2. Questions: a. What is the maximum inclination angle the plane can have before the block slips down? b. If θ is 10° higher than in question b, what will the acceleration be? 3. Solution Strategy: a. First we solve the y-component to figure out the normal force: a)

ii.

y

=0

N = mgCos(θ )

Now we figure out the x-component to get θ: a)

iii.

∑F

∑F

x

= ma

mgSin(θ ) − μ s n = 0

Combine both equations: a) b)

mgSin(θ ) = μ s [mgCos(θ )] mgSin(θ ) = μs Tan(θ ) = .6 mgCos(θ )

θ = 30.96°

b. For this question, we simply plug in 40.96 or 41° (for simplicity’s sake) and get the answer. Don’t forget to replace the coefficients of friction!

i. n = mgCos(θ ) + mgSin(θ ) − μ k n = ma ii. 9.8Sin(41°) − (.4)[(9.8)Cos(41°)] = a gSin(θ ) − μ k [gCos(θ )] = a iii. a = 3.47 4. Lessons: a. The next problem will be a car navigating a banked turn. The problem will be a combination of exercise 2 (horizontal circular motion) and exercise 3 (a banked road). The main thing to figure out is the normal force.

F. Example 4:

1. Problem – You have the same car from ex.2 but now the road is banked at an angle of 20°. 2. Questions: a. What is the maximum velocity the car can achieve without slipping? b. What is the minimum velocity the car can achieve without slipping? 3. Solution Strategy:

a. In order to figure out the maximum, we need to consider the fast car scenario: i. Time to figure out the vertical components to get the normal force:

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Comment [as10]: This images on this page were done at 300dpi. The ones before it were only done at 72dpi which caused them to print poorly. The only thing is the file size is 4 times greater… I decded to keep the image at 200dpi because it was the best tradeoff between size and graphics. The only way to improve the image would be to up it to 600dpi, but the file will become 5-10 times bigger and even worse, it will take forever to scroll through the document.

Chapter 6 (Circular Motion and Other Applications of Newton’s Laws)

ii.

mg Cos(θ ) − ( μ s )Sin(θ )

Now we set up the horizontal components to figure out the centripetal acceleration and plug in the y-based normal force equation to solve it. We remember that the force of static friction pulls the car back into a circular path, so one of its velocity vector components is centripetal. We also notice the normal force has a centripetal vector component too, so we put them together and form our centripetal acceleration equation:

mv 2 = n[Sin(θ ) + ( μ s )Cos(θ )] r

mac = (n)Sin(θ ) + ( μ s )(n)Cos(θ )

a)

iii.

n=

(n)Cos(θ ) − ( μ s )(n)Sin(θ ) − mg = 0

a)

Now plug in the equation for the normal force and solve: a)

⎞ mg mv 2 ⎛ [Sin(θ ) + ( μ s )Cos(θ )] =⎜ r ⎝ Cos(θ ) − ( μ s )Sin(θ ) ⎟⎠

b)

v2 =

c)

v=

d)

r ⎛ mg[Sin(θ ) + ( μ s )Cos(θ )] ⎞ m ⎜⎝ Cos(θ ) − ( μ s )Sin(θ ) ⎟⎠

v=

(32)(120)[Sin(20°) + (.8)Cos(20°)] Cos(20°) − (.8)Sin(20°) 3600(sec/hr) 79.409ft/s × = 54.14 mi/hr 5280(ft/mi)

(g)(r)[Sin(θ ) + ( μ s )Cos(θ )] Cos(θ ) − ( μs )Sin(θ ) v = 79.409ft/s

b. The slow car scenario is actually quite simple; notice the only difference is the force of static friction is acting in the opposite direction, if you do the math, you will notice it means all you do is switch the signs in the final equation we have:

mg Cos(θ ) + ( μ s )Sin(θ )

i.

n=

ii.

⎞ mg mv 2 ⎛ [Sin(θ ) - ( μ s )Cos(θ )] =⎜ r ⎝ Cos(θ ) + ( μs )Sin(θ ) ⎟⎠

iii.

v=

(g)(r)[Sin(θ ) - ( μs )Cos(θ )] Cos(θ ) + ( μ s )Sin(θ )

iv.

v=

(32)(120)[Sin(20°) - (.8)Cos(20°)] Cos(20°) + (.8)Sin(20°)

a)

mac = (n)Sin(θ ) - ( μ s )(n)Cos(θ )

v = 36.01i

The imaginary result simply means there is NO minimum speed. You can stop on the ramp because the friction force alone is enough to keep the car in place.

G. Vertical Circular Motion:

1. Vertical Circular Motion, unlike horizontal circular motion has a constantly changing Normal force, which makes calculating much more complicated. Since the Normal force is gravitationally influenced, it is different at every point in the curve from its maximum at the bottom (where n and g vectors are at 180°) to its minimum at the top (where n and g are pointing in the same direction). When solving equations with vertical circular motion, it is important to remember the direction of the normal force in relation to gravity (which is always pointing down and always the same magnitude). 2. Roller Coaster analogy: a. Revolution Type: in this type, the roller coaster does a loop-the-loop on the inside of the track (if it does it on the outside, the normal force would point in the opposite direction) and the normal force points to the center of the circle. At the bottom, the normal force is greatest because it needs

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Chapter 6 (Circular Motion and Other Applications of Newton’s Laws) to counteract the force of gravity plus provide enough force result in centripetal acceleration. At the top, the normal force is least because both the normal force and gravity are pointing towards the center of the circle, thus, the normal force doesn’t need much more to provide centripetal acceleration.

b. Colossus Type: in this type, the train goes up and down hills. The normal force is also greater at the bottom and least at the top, but for slightly different reasons. At the bottom of the turn, the normal force acts just like it did in the revolution-type example, with both forces opposing each other and the normal force pointing to the center of the radius. At the top though, unlike the loop, the normal force points straight up (180° away from the center of the circle) so only gravity is providing centripetal acceleration if at all. If the velocity is high enough up the hill and the rider lacked a restraint, gravity would be unable to provide centripetal force (pushing the rider back into the curve), the normal force would be zero, and the rider would fly out and become a projectile. i. Nota Bene: don’t try to memorize the equations above, if you understand the forces acting on the car, you can easily figure them out anyways, and if you can’t, then memorizing isn’t going to help much at all either… trust me!

H. Example 5:

1. Problem – A designer is making a roller coaster similar to the one above with a revolutiontype loop and a few colossus-type hills and valleys.

2. Questions: a. How fast should the ride be so the rider feels half their weight at the top of the hill? b. How fast should the ride be so the rider feels weightless at the top of the loop? c. How fast should the ride be so the rider feels twice as heavy at the bottom of the loop? d. How do we calculate the normal force at any given position in the circle? 3. Solution Strategy: a. Since the weight a person feels is actually the normal force, we just set the normal force to ½ and solve for the equation:

i.

normally n=mg

ii.

mg − ntop =

iii.

1 v 2 = r( g) 2

mv r

2

n=

so now

→

v=

mg −

1

2

mg

mg mv 2 = r 2

→

g−

1 v2 g= r 2

rg = 4.9r 2

b. Here again, we just set the normal force to zero and solve the top-of-loop equation: i.

mv 2 mv 2 rmg → 0 + mg = → v2 = m r r v = rg = 9.8r Guess what, this is also the minimum speed to do the loop!

ntop + mg =

ii. c. Do you notice a pattern? i.

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nbot − mg =

mv 2 r

→

2mg − mg =

mv 2 r

→

v2 =

rmg m

Chapter 6 (Circular Motion and Other Applications of Newton’s Laws) v 2 = rg = 9.8r Hey, that’s the same value as at the top of the loop! ii. d. This is just a matter of remembering that the normal force points to the center and is equal to the force of gravity and the cosine of the angle it makes with the vertical (n=mgCosθ) because the sine would actually be the tangent of the radius. Since at the bottom, we already know the equation, we simply add the cosine to allow for the change in angle… actually; we could add it to the top or bottom, as long as the angle’s origin begins at the vertical down (e.g. the +x axis would be 90° and the –x axis would be -90°) and it will work fine.

n − mgCosθ =

i.

⎛ v2 ⎞ n = m ⎜ gCosθ − ⎟ r⎠ ⎝

mv 2 r

I. Motion in the Presence of Resistive Forces (air, liquids)

1. Similarly to friction, resistive forces (air or water drag) oppose the object’s direction, but unlike friction, resistive forces in liquids tend to grow dramatically faster (the coefficient of kinetic friction can also vary with speed, but the change isn’t notable until the object goes very fast). a. In the case of slow moving particles or particles in a viscous fluid, the equation is: i. R = − bv b. And in the case of fast moving objects (in air):

R=

i.

1 D ρ Av 2 2

D= drag coefficient (dimensionless) ρ (rho)= density of air

a) b)

2. Terminal Velocity: when an object is initially dropped, its acceleration is equal to gravity. Over time, the resistive force increases until it equals the downward force: this is terminal velocity. Terminal velocity’s dependence on mass is the reason why heavier objects have a higher terminal velocity than lighter objects and why they hit the ground faster: 3. Developing the Terminal Velocity Equation: though its not essential to know how the equation is made (it’s not going to be on any tests), understanding how we got the equation provides a nice chance to review basic calc skills and to see how we tie together both the terminal velocity and force equations a. First we set up the conditions…

when mg = bv →

vTerminal =

if ∑ Fy = mg − bv

then

mg b

ma = mg − bv

so

m

dv = mg − bv dt

b. Now a few simple algebraic manipulations to prepare our integrate for time

b dv = g− v m dt

dv = dt g − (b m )(v)

⎯becomes ⎯⎯→

Next, we integrate this combined terminal velocity/force equation v

c.

0

t

dv

∫ g−(

b

m

)(v)

= ∫ dt 0

Now, we set up the substitution necessary to integrate this…

d. let u=g − (b m )(v)

so

du =

−b v m

and

Now we put it all back together and manipulate the logs

Page 26 of 59

dv =

−m du b

Comment [as11]: Notes for Monday, July 10, 2006 begin here

Chapter 6 (Circular Motion and Other Applications of Newton’s Laws) v −m

e.

∫ 0

f.

u

b

du = t

(g) − ( b m )(v )

(− m b )ln u g

→

(− m b )[ln[g − (b m )(v)] − ln(g)] = t

→

⎛ g − (b m )(v )⎞ bt =− ln ⎜ ⎟ ⎝ g ⎠ m

Get rid of the natural log by raising it to the power of Euler’s number…

g.

g − (b m )(v) − bt = e( m ) g

→

1−

bv − bt = e( m ) mg

→ 1 − e(

− bt

m

)

=

bv mg

h. And voila! The velocity of an object under drag as a function of time!

mg ⎛ (−bt m)⎞ = v(t) ⎜⎝ 1− e ⎟⎠ b

i.

We should also note that:

if t = 0

then

v=0

if t → ∞

then

v → mg b

4. Air-resistance modified kinematic equations: (− bt ) a. a(t) = v'(t) = ge m The acceleration of an object under air drag b. y(t) =

mg ⎡ m (−bt m) m ⎤ − bt ⎧ mg ⎫ [1 − e( m )]⎬dt = t+ e − ⎥ The position of an object under drag b b⎦ b ⎢⎣ b ⎭

∫ ⎨⎩

J. Example 6:

1. Problem – The instructor dropped a paper with a mass of 4.7g from a height of 2m. The time it took for the paper to fall is 1.37s.

2. Questions: a. Calculate b (the coefficient of resistance) from this data b. What is the terminal velocity for this paper? c. What is the final velocity of the paper and did it reach terminal velocity? 3. Solution Strategy: a. Since we have the position (y), time (t), and mass (m), we can plug this into the position terminal velocity equation.

i.

y(t) =

ii.

2=

mg ⎡ m (−bt m ) m ⎤ t+ e − ⎥ b⎦ b ⎢⎣ b

)( b ) ⎤ (.0047)(9.8) (.0047) (.0047 ) e⎡⎣ (1.37 (.0047 ) ⎦ [(1.37) + − ] b b b

→

b = 0.02764

b. Just plug in all your known values into the terminal velocity equation… i. mg = bv (.0047)(9.8) = (.02764)v vt = 1.6688 c. Use a kinematic equation to find the final velocity (velocity at time t) and compare… i. ii.

(.0047)(9.8) ⎡ (− (.02764 )(1.37 ) ) 1 − e (.0047 ) ⎤ = 1.6659m/s ⎦⎥ (.02764) ⎣⎢ v f 1.6659 = = 99.82% of terminal velocity vt 1.6688

v(t) =

VII. Chapter 7: Energy and Energy Transfer

(return to top)

A. Background

1. Work: when a force is applied to an object and moves it, the resultant displacement of the object can be quantified as work. In this way, work is the product of a force and the distance

Page 27 of 59

Chapter 7 (Energy and Energy Transfer) it moves the object. There are many ways to calculate it, but all are based on this basic concept. This should make it obvious but its worth stating: a force that does not result in displacement r r does no work. a. W = F ⋅ r = F r Cosθ

2. Relationship between Work and Kinetic Energy: Kinetic Energy has the same units (Joules) as work and is can actually be derived from the basic work equation to prove its equivalency… if we assume the angle between the force and the system is zero, our equation works out like this…

∫

r

r r

r

r

∫ (ma)dr (1)

r r ⎛ dv dr ⎞ m ∫ ⎜⎝ dt dt ⎟⎠ dt =

r

∫ (ma )dr =

b. W =

r r

∫ (F)dr Cos (0°) =

a. W = F ⋅ dr =

vf

∫ (mv )dv =

vi

1 2 1 2 mv f - mvi 2 2

c. KE = 1 2 mv 2 3. Work-Energy Theorem: if work is done on a system by external forces and the only change in the system is the speed, then the work is just the change in energy of the system. a. K f = K i +

∑W

1

2

mv 2f =

1

2

mvi2 + ∑ W

b. This means that if a 1kg object is moving at 2m/s and its final velocity is 4m/s, its Work is the same as its change of energy, which is 6 Joules.

4. Power: is the amount of energy (work) transferred during an interval or instant in time. W dW W a. Paverage = P instantaneous = lim = Δt → 0 Δt Δt dt

B. Example 1: (comparison of dynamic versus work approaches) 1. Problem - A 2kg block with an initial velocity of 3m/s is moved 15 meters with a force of 10N.

2. Questions: a. What is the final velocity of the box? b. What is the average power? 3. Dynamic (chapter 5) Solution Strategy: a. Figure out the acceleration (dynamics) and then figure out the velocity (kinematics):

F = ma v 2f = vi2 + 2ad

i. ii.

10 = 2a v 2f = 32 + 2(5)(15)

a=5 v f = 159 = 12.61m/s

b. This question cannot be answered with Dynamic equations 4. Work (Chapter 7) Solution Strategy: a. To figure this out, we find the total work and then plug it into the work-energy theorem. i. Set up the basic work equation and fill in the variables: r a)

ii.

r W = F ⋅ r = (10) ⋅ (15) = (10 )⋅ (15 )Cos (0° ) = 150J

Now that you have work figured out, use the work-energy equation to get final velocity: a)

1

2

mv 2f =

1

2

mvi2 + ∑ W

1

2

(2)v 2f =

1

2

(2)(3)2 + 150

b. To find the power, find the time the work was done and then solve… i. 12.61 = 3 + 5t t = 1.925s v f = vi + at ii.

Paverage =

W 150J = 78.13 watt Δt 1.925s

C. Example 2: (how is work affected by a change in direction?) Page 28 of 59

v f = 12.61

Chapter 7 (Energy and Energy Transfer) 1. Problem – The same block as the one above is pushed with three different forces (everything else remains the same). The first force pushing at 40° to the horizontal, the second is pulling at 40° from the other side (140°) and the last one is pulling up (90°). 2. Questions: a. What is the work done with the force in the 40° direction? b. What is the work done with the force in the 90° direction? c. What is the work done with the force in the 140° direction? 3. Solution Strategy: a. Figure out the done at 40°: r work r i. W = F ⋅ r = (10)(15)Cos(40°) = 114.906 b. Figure out the work done at 90°: i. (10)(15)Cos(90°) = 0 a)

The force doesn’t move it, and so does no work

c. Figure out the work done at 140°: i. (10)(15)Cos(140°) = −114.9 (The force is losing) ii. Is the result reliable? a) b)

1 2 1 2 mv f − mvi 2 2 1 1 −114.9 = (2 )v 2f − (2)(3)2 = 10.29i 2 2

W =

We got an imaginary number, which means there is no solution. That means the block never makes it to +15m and probably moves backwards instead. If it never makes it to 15m, then where does it stop?

iii.

iv.

a) b)

W = K f − Ki (x) (−10 )Cos (40°) = 1 2 (2)(0)2 − 1 2 (2)(3)2 x = 1.17 Å this is where the block stops moving forward (and then goes backwards).

D. Zero and Negative Work:

1. When a force is perpendicular to the displacement, no work is done. a. The Normal force is always perpendicular to displacement, so it never does work. b. Any force acting as a centripetal force does no work. 2. When a force is greater than 90° to the displacement, it does negative work. a. Friction always does negative work.

E. Example 3: (work done when force varies) 1. Problem – same block as in the previous

example with the force at 40° to the horizontal and the force is 10x 2. Questions: a. What is the work done on this system? 3. Solution Strategy: a. Since the force varies as a function of displacement, we integrate it to get the work: i.

W = ∫ (F)dr Cos (θ ) =

15

∫ (10x )dx Cos(40°) = 861.799 0

4. Note: a. In this example, the force (10x) will actually end up lifting the block at the 3.04 m mark, thus causing work in the vertical direction too. Since this is an introductory example, we won’t cover the y-component, but in tests, if asked for total work, you must include the sum of both the horizontal and vertical work.

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Chapter 7 (Energy and Energy Transfer)

F. Example 4: (work with a non-conservative force)

1. Problem – the same block as before (2kg, vi=3m/s, xf=15m) but the force is now r F = 10xiˆ + 3x 2 ˆj (a non-conservative force) 2. Questions: a. What is the total work done in the x and y directions?

3. Solution Strategy: a. Solve by integrating…but wait! 15

i.

15

?

∫ (F iˆ + F ˆj )⋅ (dxiˆ + dyˆj )= ∫ (10x )dxCos(0°)iˆ + ∫ (3x )dyCos(90°) ˆj = 1125iˆ + 0 ˆj 2

x

y

0

0

?

4. Notes: a. The work in the y direction is zero because displacement is zero. b. Although in theory, the 3x² would result in a lifting of the block at 2.55 feet, the problem does not state

G. Example 5: (work with a variable force across a limited path)

1. Problem – This time the block from the previous examples is going in a straight line from the points (1, 4) to (10, 2). The force is varying in both the x and y directions, but the path is limited to the straight line between the previously mentioned points. 2. Questions: r a. Find the work done by a conservative force ( F = 10xiˆ + 3y 2 ˆj )

r

b. Find the work done by a non-conservative force ( F = 10xiˆ + 3x 2 ˆj ) 3. Solution Strategy:

a. Here we integrate each based on their paths: i.

W =

xf

yf

10

2

xi

yi

1

4

∫ f (x)dx + ∫ f (y)dy = ∫ (10x )dx + ∫ (3y )dy = 439J 2

b. The trick here is to solve the y parameter. Since it depends on the x parameter, the easiest thing would be to integrate it on the x-scale (you can integrate it to the y-scale, but it would be unnecessarily difficult and time-consuming, especially given the fact you already have enough to solve it on the x-scale). ?

i.

∫ (3x )dy (Since we need to integrate it to x, we need to find and equivalent equation) 2

?

ii.

y − y1 =

y2 − y1 (x − x1 ) x2 − x1

(This is about the most no nonsense way to get the slope, since its based on the points we’re already given… no need to find functions based on the forces because the path is limited by the straight line)

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Comment [as12]: a.I need to get back to this example to finish the vertical component if possible.

Chapter 7 (Energy and Energy Transfer) 2−4 iii. y − 4 = x −1 10 − 1

2 2 y=− x− +4 9 9

2 34 y=− x+ 9 9

iv.

Now the trick… We differentiate implicitly to get our slope: dy = − 2 9 dx Now we can replace the dy with the dx in our integral and solve

v.

W =

10

∫ (3x )(− 2

2

9

dx ) = −222

1

The work is negative since the positive force is working against negative displacement. Now we add the x work and the y work and we’re done…

Wtotal = Wx + Wy = 495 − 222 = 273J

vi.

c. Extra: we want to change the path from a line to a parabola and compare the work done by conservative and non-conservative forces on this path. i. The conservative force would do the same work as before (439J) ii. The non-conservative force (3x²) though, would require a whole recalculating of the path times the force as we did in the first half of this example. First we use the quadratic formula for our model of a parabola and meld it with our current xequation. Since our points are given, we can create two equations that we can add together to give us one variable, from which we will then get our second variable.

a)

y = ax 2 + bx

b)

c) At y=4, the equation is: 4 = a(1) d) At y=2 (the end), the equation is:

2 = a(10)2 + b(10)

iii.

+ b(1) 2 = 100a + 10b

e)

Set up one equation to cancel a variable in the second equation and you get:

f)

Subtract the second equation from the first one to get:

10(4 = a + b)

40 = 10a + 10b

(2 − 40) = (100a − 10a) + (10b − 10b) = (− 19 45 ) + b b = 199 45

g)

From which we get: 4

h)

Which gives us the equation of the parabola:

−38 = 90a

a = − 19 45

y = − 19 45 x 2 + 199 45 x

Now we differentiate (like last time) to get the equivalent x-equation that we can integrate:

y = − 19 45 x 2 + 199 45 x dy = − 38 45 xdx + 199 45 dx iv.

2

dy = −(2)(19 45 )(x)dx + 199 45 dx dy = (− 38 45 x + 199 45 )dx

Now we put it back together and integrate on the x… 10

Wtot = Wx + Wy = 495J + ∫ 3x 2 (− 38 45 x + 199 45 )dx = 495 − 1914.9 = −1419.9 1

v.

As is obvious, the conservative path is still the same because it only depends on the beginning and end points. The non-conservative path, though, is totally dependent on the path, so we need to integrate the force on the parabolic path. So we decide on a quadratic equation (we could chose any equation for the parabola, but a quadratic is easier because of its familiarity) and fit the y parameter to conform to the x parameter. Once we get the path, we differentiate it implicitly to get the equivalency between y and x slopes (remember, derivatives are slopes?). This, times our non-conservative force can now be integrated over the x to get the work based on the path and the force on every point of that path.

H. What makes a force conservative?

1. Conservative Forces – are forces that only depend on the initial and final position to determine work. The path is irrelevant. Conservative forces always have a potential energy (chapter 8) associated with them.

Page 31 of 59

Comment [as13]: Notes for Wednesday, July 12, 2006 begin here

Chapter 7 (Energy and Energy Transfer) 2. Non-conservative Forces – are dependent on the path taken and so any irregular path must be taken into account when calculating work done. They do NOT have a potential energy associated with them.

I. Work Done by Gravity Forces (conservative)

1. Gravity is a conservative force which we can prove by comparing three scenarios:

2. Work done in a freefall: r r a. Wg = ∫ F ⋅ dr = F d Cosθ = (mg)(H i − H f )Cos(0°) = mgΔH b. If it hits the ground…(to find the velocity upon impact) i. Wg = mgΔH = 1 2 mv 2 v f = 2gh

3. Work done (against Gravity) going back up: a. Wg = F d Cosθ = (mg)(H i − H f )Cos(180°) = − mg ΔH 4. Work done along a frictionless incline: a. Wg = F d Cosα = (mg)(d)Cos(α ) b. dCos(α ) = H So… c. Wg = (mg)H = mgH 5. Conclusion – we can see that work done by or against gravity is conservative because it doesn’t depend on the path (straight drop or a drop on an incline). The fact that raising the block performs negative work also is part of why gravity is able to store potential energy.

J. Work done by Spring Forces (conservative)

1. A Spring is another example of a conservative force, which we can prove by calculating the work done based on Hooke’s Law (F=-kx): 2. Work Done in compression/extension: r r xf a. Ws = ∫ F ⋅ dr = ∫ (−kx)dx = − 1 2 kx 2f 0

3. Work done by the spring returning to its origin: 0

a. Ws =

∫ (kx )dx =

1

2

kx 2f

xf

4. Conclusion: work done on a spring is the same as work done against gravity; it results in negative work (which is where a conservative force’s potential energy is stored. When the spring returns to its natural position, the work it does then is positive.

K. Work done by Friction Forces (non-conservative) 1. This work is non-conservative because it always

depends on the path taken. Since the energy used up by friction is only stored as heat, it cannot return to the system and is essentially wasted; so no conservation of forces occurs. 2. Calculation of friction work (flat surface):

Page 32 of 59

Chapter 7 (Energy and Energy Transfer) d r r d a. W fk = ∫ F ⋅ dr = ∫ fk dxCos(180°) = − μ k n ∫ dx = − μ k nd 0

0

b. Conclusion: as we can see, friction’s work is negative because it always in a direction opposite to the displacement. We also see that the path of friction directly affects the work output.

L. Calculation of velocity on rough slopes of different lengths (showing friction depends on path)

1. Work done on the block by the time it gets to the bottom of the ramp Wblock = 1 2 mv 2f − 1 2 mvi2 = 1 2 mv 2f − 0 = 1 2 mv 2f 2. Calculating the work done by gravity minus friction to get our final velocity: 1 mv 2 = mgH + (− μ nd) a. Wblock = Wg + W f 2 f k b.

1

c.

1

2

mv 2f = mgH − μk mgCosθ d

1

2

v 2f = g( H − μ k Cosθ d )

vf =

2

v 2f = gH − μk gCosθ d 2g(H - μk dCosθ )

3. Conclusion: we can see in our final equation that the work done by gravity is dependent on the path and so the greater path (the longer slope) will result in greater friction work, and as a result, end with a lower final velocity.

M.

Example 6: (work with a variable force, Pt II)

1. Problem – A block is being pulled with a variable force of 2x at an angle of 40° for a distance of 20m. The coefficient of kinetic friction is 0.1.

2. Questions: a. What are the work done by the force (F), by friction (fk) and the total work? b. Find the final velocity. c. How long can this block be pulled before it is lifted? 3. Solution Strategy: a. Finding the first part is a simple work integration, friction isn’t much more complicated: 20

∫ (2x )dxCos40° = 306.418J

i.

WF =

ii.

Calculate the normal force and use it to figure out friction (keeping in mind μk is constant):

0

n + 2xSinθ = mg

n = (3)(9.8) − 2xSin40°

n = 29.4 − 2xSin40°

20

W fk = − fk ⋅ d = ∫ − μ k n ⋅ d = − μ k ∫ (n )dxCos(0°) = 0

20

W fk = − (.1) ∫ (29.4 − 2 xSin 40° )dx = −33.0885 J 0

iii.

Put them all together:

WTotal = WF + W fk = (306.418) + (−33.0885) = 273.3295J

b. Remember the work-energy theorem i. Wtotal = ΔK 273.330 =

Page 33 of 59

1

2

(3)v 2f − 0

v f = 13.50 m s

Chapter 7 (Energy and Energy Transfer) c. Just set the normal force equal to gravity to find the position at liftoff. i. 2 xSin40 = 29.4 x = 22.869

VIII.

Chapter 8: Potential Energy A. Background:

(return to top)

1. Conservative Forces: as stated in the earlier section on conservative forces, a force that is path independent (gravity, spring) is conservative and by its nature, stores energy done against it as potential energy. a. Path Independent: This is the first condition of a conservative force. The work done does not depend on the path, only the displacement between initial and final position.

b. Work on a closed path: This is the second condition of conservative forces. When an object goes around a complete path under the influence of a conservative force, then the total work is zero

i.

r

r

∫ A ⋅ dr = 0

c. Component notation: for conservative forces, the component and its magnitude will be from the same direction. r

i.

F = 2x 2iˆ − (3y + 5) ˆj Is conservative. Notice the iˆ is dependent on the x² and the ˆj is dependent only on the -3y-5. If instead, the iˆ were dependent on the y and/or the ˆj were

dependent on the x in any way, the equation would become non-conservative. d. Relationship to Potential Energy: there is always a function associated with conservative forces such that:

i.

r dU dU ˆ dU ˆ dU ˆ Fc = −ΔU = − =− i− j− k= dr dx dy dz

e. Units of Energy: the typical units are Newton Meters (Nm) aka Joules (J)

2. Non-Conservative Forces: if a force is NOT path independent and/or if the work done by a particle traveling a closed path is NOT zero, then the force is Non-Conservative (in other words, if any of the two conditions for conservative forces is not met). And non-conservative forces have no potential energy function associated with them. a. Component Notation: non-conservative forces will have at least one component dependent on a magnitude from a different direction.

i.

r F = 2x 2 iˆ − 3xˆj The ˆj is dependent on the x instead of the y.

b. Work on a closed path: unlike conservative forces, work always adds up as a path is traversed, so work on a closed path will NEVER equal zero.

i.

r

∫F

NonConservative

r ⋅ dr ≠ 0

B. Example 1:

r 1. Problem - you have the following functions: F = 2x 2 iˆ − (3y + 5) ˆj , 2. Questions: a. Find the associated Potential Energy function of this force.

3. Solution Strategy: a. We go back to the relationship between a force and potential energy: i. ii. iii.

Page 34 of 59

r dU ˆ Fc = −ΔU = − i− dx dU − = 2x2 and dx 2x3 then −U x = 3

dU ˆ j dy dU − = (−3y − 5) dy −3y 2 and −U y = − 5y 2

Chapter 8 (Potential Energy) iv.

⎛ 2x 3 ⎞ ⎛ 3y 2 ⎞ 2x 3 3y 2 −⎜ U(x, y) = − ⎜ − 5y ⎟ + C = − + + 5y + C Joules ⎟ 3 2 ⎝ 3 ⎠ ⎝ 2 ⎠

so

C. Example 2: (Comparison of Methods of Calculation; Ch5, Ch7, & Ch8)

1. Problem – You have a block which is dropped from an initial height of yi, 2. Questions: a. Calculate its final velocity using Newton’s Laws (Chapter 5) b. Calculate its final velocity using the Work-Kinetic Energy Theorem (Chapter 7) c. Calculate its final velocity using the Potential Energy Method (Chapter 8) 3. Solution Strategy: a. Actually, our solution harkens back to Chapter 2 (Kinematic Equations) i. v 2f = vi2 + 2 a( x f − xi ) = 0 − 2 g(0 − yi ) = 2 gy v f = 2gy

b. Here we use the fact that work is a change in kinetic energy i. Wg = ΔK mgh = 1 2 mv 2f − 1 2 mvi2 = 1 2 mv 2f − 0 ii.

gh =

1

2

v 2f

v=

2gh

c. Even though we saw this in Chapter 7, including the U makes it more Chapter 8-like i. WNonCon + WExtForces = ΔK + ΔU ii. since no non-conservative or external forces are acting on the block…

0 = (K f − K i ) + (U f − U i ) = (K f − 0) + (0 − U i ) = K f − U i

0=

1

2

mv 2f − mgh

1

2

v 2f = gh

vf =

2gh

D. Example 3: putting it all together

1. Problem – you have a block resting on top of an incline of 80ft at an angle of 25° with a coefficient of kinetic friction of 0.2. The block will slide down the ramp and travel through a loop with a 10m radius and a coefficient of static friction of 0.4. After passing the loop, it will compress a spring with a stiffness constant of 250 lb/ft. In this example, ignore the constant change of normal force in the loop and assume it is 10.

2. Questions: a. What is the final velocity at the top of the circle? Will it complete the circle? b. How much would the spring be compressed by if the height were 100ft? 3. Solution Strategy: a. We calculate total work with the Work-Kinetic-Potential energy equation: i. WNonCon + WExtForces = ΔK + ΔU ii. Time to figure out the variables involved in the incline’s friction calculation:

μ k1 = (0.2)

Page 35 of 59

n1 = 6Cos25°

d1 =

100 Sin25°

Chapter 8 (Potential Energy) iii.

Now we figure out the variables in the circle’s friction calculation: (keep in mind the distance traveled is really ½ the circumference at this point)

μ k 2 = (0.4) iv.

n1 = (10)

d1 =

(2)π (10) = 10π 2

There is only final kinetic energy, which is where we will get our velocity from

ΔK = K f − K i =

1

2

(6 32 )v2f − 0 = 3 32 v2f

v.

Total potential energy is the sum of the initial height minus the loop’s height.

vi.

Now we set up the equation with the known variables:

ΔU = U f − U i = mgh f − mghi = mg2r − mghi = (6 )(2 )(10 ) − (6 )(80 ) = −360

) ( ) ) − (0.4 )(10 )(10π ) = ( v )+ (−360 )

0 + − (μ k1n1d1 ) − (μ k 2 n2 d2 ) =

− (0.2 )(6Cos25° )(80 Sin25°

(

1

2

mv 2f − 1 2 mvi2 + U f − U i 3

2 f

32

vii. And the solution is: v f = 17.424 m/s , but does it make the loop? viii. Time to use the equation we developed back in chapter 6, example 5:

Vmin = rg = (10)(32) = 17.88 the block’s speed is a tad smaller than the minimum required to travel the loop. This means that at some point before it hits the top of the circle, the normal force will be less than the weight of the block and for a few moments, the block will behave like a projectile until it encounters the other side of the loop and returns on its path. Given the tiny difference, this is likely to be a tiny amount, but the math required to figure this out would be far greater than the math required to figure out the example, so we will ignore that for now and leave this problem at that. b. Now we reuse the equation we previously had, but keep in mind the height is now 100 feet and now the block is following the entire circle (not just half of it). Also, since the initial and final speeds of the block will be zero (its velocity is zero at maximum spring compression), we need only calculate the potential energy, and for final potential energy, we replace it with the spring potential energy equation we got from Chapter 7, section J.

(

)

i.

− (μ k1n1d1 ) − (μ k 2 n2 d2 ) = U f − U i

ii.

− (0.2 )(6Cos25° )(100 Sin25° ) − (0.4 )(10 )(2π 10 ) =

iii.

And we get x=0.8547ft.

(

1

2

)

250x 2 − (6 )(100 )

E. Equilibrium:

1. Equilibrium and Potential Energy: all potential energy functions have points of equilibrium, but the equilibrium can have three different types: a. Stable Equilibrium – when moved away from this equilibrium, the particle will return to it. Graphically, this looks like a U-shaped graph and the point of equilibrium is at the bottom. b. Neutral equilibrium – when moved away from this position, the particle will stay where it is so long as where it is moved is also a point of equilibrium. This is like a U shaped graph where the bottom of the U is extra long, looking like a floor. c. Unstable equilibrium – when moved away from this point of equilibrium, the particle will not return to it and instead will move away from it. Graphically, this looks like an upside down U-shaped graph with the particle at the top.

F. Example 4:

1. Problem: you have a the potential energy function for a spring which is -½ kx² 2. Questions: a. Find the associated force function b. Find the points of stable & unstable equilibrium

Page 36 of 59

Comment [as14]: Notes for Monday, July 17, 2006 begin here

Chapter 8 (Potential Energy) 3. Solution Strategy: a. When trading between energy and force, we can use this simple relationship (if you notice example 1, it follows this form): derive

i.

U

ii.

U(x) → F(x) − 1 2 kx 2 → F(x) dU 1 2 dU 1 2 − 2 kx = (−k) 2 x = −kx dx dx

integrate

F

iii. b. The points of stable & unstable equilibrium are the local max and min of U(x) or F’(x): i. U(x) = 0 @ x=0 There is only one point of equilibrium and it’s at zero and stable.

G. Example 5: 1. Problem: You have the Potential Energy function u ( x ) = −x 3 + 6 x 2 − 7 x + 12 J 2. Questions: a. Calculate the points of stable & unstable equilibrium. b. Calculate the associated force function. c. What is the U(x) for the points of equilibrium. d. What will happen if this particle is placed at x=0, x=-2, x=2 3. Solution Strategy: a. First, refer to step ‘b’ which gives us the force, from which we “solve” to

get the zeros of the force. Once we find those, we run the 2nd derivative test on the zero forces to see if they are local maxima (unstable) or minima (stable).

i. ii. iii.

F(x) = 0 −3x 2 + 12x − 7 = 0 F(x) = 0 at x = 0.709, 3.291 U ''( x ) = F '( x ) = −6 x + 12 Å use this for local max/min calculations

iv.

F '(0.709) = +7.7 positive 2nd derivative = concave up = local

min = stable

v.

F '(3.291) = −7.7 negative 2nd derivative = concave down =

local max = unstable

b. Here we just remember the relationship u(x) ⎯⎯⎯ → F(x) derive

i.

(

:

)

dU −x 3 + 6x 2 − 7x + 12 = −3x 2 + 12x − 7 = F(x) dx

c. Simple plug-n-chug i. U(.709 ) = 9.697 U(3.291) = 18.303 d. A view of the graph is the easiest thing to use to figure this out. Imagine the potential energy graph is a path and a marble was placed at any of the points in question. i. If its placed at x=0, it would oscillate from x=0 and x=1.59 (the two places where its y-value is 12) and if friction was a factor, it would eventually stop at 0.709. ii. If a marble was placed at x=2, it would oscillate between x=2 and x=-0.236 (the two places where its y-value is 14). iii. If the marble was placed at x=-2, it would roll towards the right, but since there is no other point on the graph on the right side that has a y-value of 58, it will just roll over the local max at x=3.291 and keep going right off the graph (theoretically into negative infinity).

IX. Chapter 9: Linear Momentum & Collisions A. Background: Page 37 of 59

(return to top)

Chapter 9 (Linear Momentum and Collisions) 1. Inertia: the principle that describes how an object will resist a change in its velocity. Since velocity is a vector and zero is a velocity, it encompasses everything from a body at rest will resist being put into motion to a body in motion will resist having its speed or its direction changed. Being a concept, it is not quantifiable; the quantification of inertia is momentum. r 2. Linear Momentum ( p )( kg ⋅ m s or slug ⋅ ft s ): is the inertia of a body. Most often it is used to refer to the inertia of a moving body because an object’s momentum is a product of its mass and its velocity. It is a vector quantity; thus it must be defined by magnitude & direction. There is also (angular momentum, but this will be covered in chapter 11) r r a. Formula: p = mv

r 3. Impulse: ( J )( N ⋅ s or lb ⋅ s ): the product of the force and the time during which it acts. Its units are the same as Linear momentum, but the units look different to distinguish each. a. Formula: I =

∫ Fdt

4. Center of Mass: The point in an object or group of objects where the object’s mass is evenly divided on all sides. If an object could be tossed with a slight rotation, it would rotate around its center of mass. If an object is hung by its center of mass, it would be perfectly balanced. In most equations, when considering the kinematic qualities (velocity, acceleration) of a particle, the center of mass can be used for modeling the particle’s motion because most objects obey kinematic physics at their center of mass. a. Formula: rcm =

∑m x

i i

rcm =

i

M

∫ rdm ∫ dm

5. Law of conservation of Linear Momentum: whenever two or more particles in an isolated system interact. The total momentum of the system will remain constant.

B. The relationship between Newton’s laws, Momentum & Impulse:

1. First: we start with Newton’s basic law and rewrite acceleration into its derivative form: r r dv r i. F = ma F=m dt b. Second: since mass is constant, we can include it in the derivative i.

F=

r r d(mv) dp Force is the time-derivative of momentum = dt dt

c. Next we find the total force for a certain time by integrating it by t. i.

r

r

∫ Fdt = ∫ dp

r r tfinal J = p tinitial

r J = (p f - pi )

We’re using shorthand here but the final and initial momentum are the momentum at the initial and final time. This final equation is known as the impulse approximation and simply states the impulse is the change in momentum of a system.

C. Example 1: (Comparison of Methods of Calculation; Ch5, Ch7, & Ch9)

1. Problem: you have a 5kg block that is being pushed a distance of 16m with an initial velocity of 2m/s and a force of 10N.

2. Questions: a. What is the final velocity using Newton’s Laws? b. What is the final velocity using the Energy Approach?

c. What is the final velocity using the ImpulseMomentum Approach?

d. What if the force was instead 10t, the distance were unknown, and time was 8s?

Page 38 of 59

Chapter 9 (Linear Momentum and Collisions) 3. Solution Strategy: a. This is a straightforward application of Newton’s laws and Kinematics i. F = ma 10 = 5a 2=a ii.

v 2f = vi2 + 2 ad

v f = 4 + 2(2)(16) = 8.25m/s

b. We keep in mind that work and kinetic energy are related

r

∫ F dx =

i.

W = ΔK

ii.

mad =

iii.

2(2)(16) = v 2f − 2 2

1

2

x

mv 2f − 1 2 mvi2

1

2

mv 2f − 1 2 mvi2

2ad = v 2f − vi2 v f = 4 + 64 = 8.25m/s

c. The limitation with this approach is it requires time. i. v f = vi + at 8.2462 = 2 + 2t t = 3.123 Of course, if we knew the final velocity, we wouldn’t need to figure it out using this approach, but there will be times when we will know the time and not the final velocity, so for now, lets assume we didn’t know how we got t=3.12 and use it to get the final velocity.

ii. iii.

r r J = Ft r r r J = p f − pi

r J = (10)(3.123) = 31.23

31.23 = 5v f − (5)(2)

v f = 8.25

d. Here we can use the more general impulse method, especially since everything is based on time instead of distance.

i. ii. iii.

r F = 10t r J = 320

r vi = 2 r 8 J = ∫ 10tdt 0 r r r J = p f − pi

d = 8s

320 = 5v f − 5(2)

v f = 66m/s

D. What is the advantage of using the momentum/impulse approach?

1. Distance is not needed: if you have at least 2 of time, velocity, and mass, you have enough to solve an equation.

2. Force as a function of time: would be impossible to solve any other way since the other ways only deal with force as a function of distance (work, for example).

E. Example 2: (varying mass and force)

1. Problem: you have a 5kg box that is open at the top and an initial velocity of 2m/s. the force acting on it is 10t+5 and the box travels an 8second distance and during the 8 seconds that it travels, rain falling into the box increases its mass by 200g per second. 2. Questions: a. What is the impulse imparted on the box? b. What is the final velocity? c. What is the work done on the block?

3. Solution Strategy: a. Impulse is the integral of force over a period of time i.

r J = ∫ Fdt =

∫

8

0

(10t + 5)dt = 360N ⋅ s

b. The velocity is the result of the change in momentum, but keep in mind the final mass has increased.

i.

Page 39 of 59

m f = mi +

dm t = 5 + (.2)8 = 6.6kg dt

Chapter 9 (Linear Momentum and Collisions) r ii. J = m f v f − mi vi 360 = 6.6v f − 5(2)

v f = 56.06m / s

c. Since you CAN’T calculate the integral of work without an x-distance, you must use the other definition of work:

i.

W=

1

2

mv 2f − 1 2 mvi2 =

1

2

(6.6)(56.06)2 − 1 2 (5)(2)2 = 10.361kJ

F. Collisions in one dimension

1. Relation to the impulse approximation: when multiple objects collide in the horizontal plane in one dimension, they transfer their momentum such that the momentum is conserved as long as no outside forces (impulses) change it. In reality, though, the change of momentum is NEVER instantaneous and occurs over a very small period of time. Luckily, most collisions (and in this chapter all of them) can be modeled by using the impulse approximation (that momentum is transferred instantaneously from one object to another). 2. Conservation of Momentum: always applies in collisions, regardless of type. This means that we need only know initial or final velocities and the mass to figure out the other velocities.

G. Types of collisions:

1. Perfectly Elastic (extremely rare): 99-100% Kinetic Energy is conserved. a. Example: pool balls colliding, momentum and kinetic energy are conserved almost 100%.

2. Semi-Elastic (most common): most kinetic energy is conserved a. Example: a rubber tennis ball, which keeps most of its kinetic energy after impact.

3. Inelastic: most kinetic energy is lost a. Example: two cars colliding.

H. Example 3: (1-dimensional perfectly elastic collision)

1. Problem: you have a 2kg block heading to the right at 3m/s and a 5kg block heading to the left at 1m/s. they have a perfectly elastic collision which lasts .8 seconds. 2. Questions: a. What are the final velocities of each block? b. What is the impulse on each block? c. What is the average force each block exerts on the other? 3. Solution Strategy: a. Since momentum AND kinetic energy are conserved, we use both equations together to find both unknown variables. We start with momentum and solve it for one variable. We then plug in the variable from momentum into the kinetic energy equation and that will give us our solution.

i.

m1v1i + m2 v2 i = m1v1 f + m2 v2 f

ii.

v1 f =

iii.

Don't let all the crazy subscripts confuse you, the subscripts are just for recordkeeping and don’t contribute mathematically to the equation.

iv.

1

v.

1

1

(2)(3) + (5)(−1) = (2)v1 f + (5)v2 f

5 2 − 2 v2 f

2

m1v12i + 1 2 m2 v22i =

2

(2)(3)2 + 1 2 (5)(1)2 =

1

2

m1v12i + 1 2 m2 v22i 1

2

(2)v12f + 1 2 (5)v22 f

23

2

= v12f + 5 2 v22 f

1 46 = 35 v 2 − 10 v = ( 1 2 − 5 2 v2 f )2 + 5 2 v22 f 4 2f 4 2f + 4 4 vii. v2 f = −1, 9 7 (or1.286) But -1 is our original velocity, so the only one left is 9/7, so: viii. v1 f = 1 2 − 5 2 (9 7 ) = − 19 7 or -2.714

vi.

23

2

b. We can use the impulse approximation.

Page 40 of 59

Comment [as15]: Notes for Wednesday, July 19, 2006 begin here

Chapter 9 (Linear Momentum and Collisions) r r r i. J on 5 kg = p f − pi = (5)(1.28) − (5)(−1) = 11.4Ns r r r ii. J on 2 kg = p f − pi = (2)(−2.714) − (2)(3) = −11.4Ns This shows us the impulse of one is the opposite of the other, as it should be.

c. Since the force is constant, there is no need to integrate i. J = Ft 11.4 = F (.8) F = 14.25

I. Example 4: (1-dimensional semi-elastic collision)

1. Problem: you have a 2kg block heading to the right at 3m/s and a 5kg block heading to the left at 1m/s. they have a semi-elastic collision which lasts .8 seconds, and the 5kg block rebounds with a velocity of ½ m/s. 2. Questions: a. What is the velocity of the 2kg block? b. What is the initial and final kinetic energy? c. What is the % kinetic energy loss? d. What is the impulse on each block? 3. Solution Strategy: a. We keep in mind momentum is retained, so this is simple algebra

i.

Pi(total ) = Pf (total )

1 = 2 v2 f + 5(.5)

v2 f = −.75

b. We already know the initial kinetic energy from Example 3, so we only need final kinetic energy. i. K f = 1 2 (5)(.5)2 + 1 2 (2)(.75)2 = 1.19J K i = 11.5J c. Simply figure out the difference as a percentage: i.

% Loss =

11.5 − 1.19 = 89.65% Loss 11.5

d. We canr again use the impulse approximation: J on 5 kg = (5)(.5) − (5)(−1) = 7.5 so i.

J on 2 kg = −7.5

J. Example 5: (1-dimensional inelastic collision)

1. Problem: you have a 2kg block heading to the right at 3m/s and a 5kg block heading to the left at 1m/s. they have an inelastic collision which lasts .8 seconds. 2. Questions: a. What is the final velocity of the system? b. What is the impulse on each block? c. What is the percentage loss of kinetic energy? d. Find the average force each block exerts on the other.

3. Solution Strategy: a. We know momentum is conserved, and since its inelastic, we will only have one final velocity, so we set up our equation with two initial velocities and one final velocity.

i. ii.

r r pi (total ) = p f (total ) 1 = 7v f

1 = (m1 + m2 )v f

vf =

1

7

b. The impulse approximation gives us the impulse

Page 41 of 59

Chapter 9 (Linear Momentum and Collisions) r i. J on 5 kg = (5)( 1 7 ) − (5)(−1) = 5.71

so

J on 2 kg = −5.71

c. We figure out the final kinetic energy since we got it in Example 3: K f = 1 2 (7)( 1 7 )2 = 114 or .0714 i.

11.5 − .0714 = 99.38% Loss 11.5

% Loss =

ii.

d. Since the force is constant, weruse the simpler impulse-force equation. r i. J = Ft 5.71 = F(.8) F = 7.14 N

K. Example 6: (2-dimensinoal elastic collision)

1. Problem: two balls on a frictionless plane are heading towards each other. Ball 1 is 2kg and is heading to the right at 3m/s and ball 2 is 5kg and is heading to the left at 1m/s. Both balls collide into each other, but one slightly above the other resulting in a change of ball 1’s trajectory from parallel to the x-plane to 70° above the x-plane. 2. Questions: a. Find ball 1’s final velocity b. Find ball 2’s final velocity c. Find ball 2’s angle of deflection 3. Solution Strategy: a. There are various parts to this first solution: breaking it into components, weeding out the trig functions, merging the momentum equations, merging with the kinetic equation, and solving. i. First break it down into its components: a) b)

ii.

iii.

r r p f ( total ) = pi ( total ) r r r r p fx = pix iˆ + p fy = piy ˆj

(

) (

)

(pr

fx

) (

r r r + p fy = pix + piy

Now solve the x and y equations. a)

2 v f 1Cos 70° + 5 v f 2 Cosβ = (2)(3) + (5)(−1) = 1

b)

2 v f 1Sin 70° − 5 v f 2 Sinβ = 0

Now we move the 70° to the left sides, square both equations, and add them to each other. Its critically important that you move the70° (known angle) to the side with only numbers, or else your equation will cancel itself out in the end and you will have lost time. a)

(5v

f2

Cosβ

) = (1 − 2v 2

Cos 70°

f1

)

2

25v 2f 2Cos 2 β = 1 − 4 v f 1Cos 70° + 4 v2f 1Cos 2 70°

iv.

Page 42 of 59

(x)

b)

25 v Sin β = 4 v Sin 70° (y)

c)

25v 2f 2Cos 2 β + 25v 2f 2 Sin 2 β = 1 − 4 v f 1Cos 70° + 4 v 2f 1Cos 2 70° + 4 v 2f 1Sin 2 70°

2

2 f2

2 f1

2

Now we extract the velocity and voila! Pythagorean identities! And when you’re done, you also have an equation with a variable on one side a)

v.

)

(

)

(

) = 1 − 4Cos70° (v )+ 4v

25 v 2f 2 Cos 2 β + Sin 2 β = 1 − 4 v f 1Cos 70° + 4 v 2f 1 Cos 2 70° + Sin 2 70° 2 f2

b)

25v

c)

v 2f 2 =

(1) = 1 − 4v f 1Cos70° + 4v (1) 1

25

( )

2 f1

25v

2 f2

f1

2 f1

− 4 25 Cos70° v f 1 + 4 25 v 2f 1

Now we set up the kinetic equation and solve for final velocity 2 squared so we can use it to solve our momentum equation.

Chapter 9 (Linear Momentum and Collisions) 2 2 a) 1 2 (2)v f 1 + 1 2 (5)v f 2 =

2

(2)(3)2 + 1 2 (5)(−1)2

(

)

v 2f 2 = (2 5 ) 11.5 − v 2f 1 = 23 5 − 2 5 v 2f 1

v 2f 1 + 5 2 v 2f 2 = 11.5 vi.

1

Now we substitute the kinetic variable into the momentum equation and solve

− 2 5 v 2f 1 =

a)

23

b)

0=

5

14

25

v − 2 f1

1

4

( )

25

− 4 25 Cos70° v f 1 + 4 25 v 2f 1

25

v f 1Cos 70° − 114 25

v f 1 = 2.903, −2.805

Negative values are invalid if we set up our equation right, because the only values we will get (whether angle or magnitude) will be positive, indicating the magnitude of either the angle or of the distance (yes, angles can be negative, but look at the equation, its set up so we can only get positive values, so in this case, only positive values are valid).

b. Now with the hard part done, this is easy; just solve the easiest (kinetic) equation i. v 2f 2 = 23 5 − 2 5 (2.903)2 = 1.205 = 1.0186 c. To solve for the angle, plug your velocities into either x or y equation: i. 2(2.2903)Cos70° + 5(1.1086)Cosβ = 1 β = ±100.24° a)

b)

Note 1: just like the previous double answer, this one’s negative value is invalid. Though instinctively, it seems that using -100.24 would be right (because it looks like the ball deflects clockwise 100 degrees, or down) and using 100.24 would be wrong (because it looks like the ball deflects counter-clockwise 100 degrees, or up), we are ignoring the fact that the answer only reflects the magnitude of the angle, not its absolute direction. Therefore, since 100.24 is the magnitude, 100.24° would be as meaningless as -100mph when discussing speed (as opposed to velocity). How then do we know what direction the 100 degrees is? You can do it the simple way (look at the picture, esp. the vectors) or add the original velocity vectors, which will only work if the 100° is clockwise (negative). Note 2: when solving for the angle, use the cosine formulas instead of the sine formulas because though the sine formulas can give you the answer, it could give you an answer that is off by 180° because of the inherent limitations of the sine function (remember the SAS calculations in trig class and how they might give you the wrong angle?, if not, then go look them up).

L. Finding the Center of Mass of an Object:

1. The Center of Mass: of an object is the most important point for most mechanics calculations because it is the point from where an object’s velocity is modeled. It is also the point around which it rotates if tossed in the air, and is the point on which it balances perfectly if suspended at one point. 2. The Center of Gravity: of an object is the same as the center of mass if gravity is uniform over the object (which on earth, it usually is unless the object is many miles long) 3. The Formulas: a.

M.

xcm =

∑ (m + x

FromOneEnd

M total

)

=

∫ xdm ∫ dm

ycm =

∑ (m + y

FromOneEnd

M total

)

=

∫ ydm ∫ dm

Example 7: (2 parts)

1. Problem: you have a rod with different masses fused into it. At one end, a sphere with 1m radius & 6kg. Next is 5m of rod with 10kg mass. Next is a sphere with 1.5m radius & mass 4kg. Finally, a rod 5m-long & 6kg mass. 2. Questions: a. Find the center of mass of this rod from the left center of the ball. b. Find the center of mass of this rod from the right 3. Solution Strategy: a. Calculate the distance of each object’s center from the center of the leftmost ball.

Page 43 of 59

Chapter 9 (Linear Momentum and Collisions) ∑ xi mi m x + m x + m x + m x 2 2 3 3 4 4 xcm = i = 1 1 i. m1 + m2 + m3 + m4 ∑ mi i

xcm

ii.

6(0) + 10(2.5 + 1) + 4(1.5 + 5 + 1) + 5(3 + 1.5 + 5 + 1) = = 4.7m 6 + 10 + 4 + 5

This is the distance of the center of mass from the center of the leftmost ball.

b. We do the same thing, and we should get the same position

5(1.5) + 4(1.5 + 3) + 10(2.5 + 3 + 3) + 6(1 + 5 + 3 + 3) = 7.3m 5 + 4 + 10 + 6

i.

xcm =

ii.

This is the distance from the edge of the rod to the center of mass. Check the answer

L = 13m

a)

xcm(right ) + xcm(left ) + 1mradius = 4.7 + 7.3 + 1 = 13

N. Example 8: 1. Problem:

a. Find the x and y center of mass of an object which has even density and fills a space bund by x³ from x=0 to x=3

2. Solution Strategy (even density): a. What we do is figure out the area to integrate, remember that in the area to integrate, there is a constant and a variable, so when we differentiate, we leave the constant alone and only differentiate the variable and its variable-dependent result i. Area of a vertical thin rectangle: da = y × dx . We use this when we want to calculate the x center of mass. Then, we can do the integral either with the x or y variable. ii. Area of a horizontal thin rectangle: da = (b − x)dy , where b is the final x value. We use this option when we want to calculate the y center of mass. Then, we can do the integral either with the x or y variable. b. Now we relate the area to density because our initial integration is based on density.

i.

surface mass density (σ ) =

mass d dm ⎯⎯ → → dm = σ da area da

c. Now we put our integrations together in the format: i.

rcm =

∫ rdm → ∫ rσ da = σ ∫ rda = ∫ rf (r )dr = The rest is just plug n’ chug ∫ dm ∫ σ da σ ∫ da ∫ f (r )dr 3

ii.

xcm =

∫ xdm ∫ dm

→

∫ xσ da ∫ σ da

=

σ ∫ xda σ ∫ da

=

∫ xydx ∫ ydx

∫ x( x )dx

=

∫ (x

3

)dx

∫ x dx 4

3

=

0 3

3 ∫ x dx

1

=

1

5

x5

4

x

3

0 4 3

= 2.4

0

0

27

iii.

ycm =

∫ ydm ∫ dm

→

∫ yσ da ∫ σ da

=

σ ∫ yda σ ∫ da

=

∫ y(3 − x )dy ∫ (3 − x )dy

=

∫ y(3 − y

3

)dy

= 7.71

0 27

∫ (3 − y 0

3. Part 2 (uneven density): Page 44 of 59

1

1

3

)dy Comment [as16]: Notes for Monday, July 24, 2006 begin here

Chapter 9 (Linear Momentum and Collisions) a. Let’s change the density from even, to density which changes as a function of x and y. b. Our areas to integrate remain the same c. The relation between area and density has been changed, but the only thing that happens is that now, sigma has a changing variable and so when we drop it in, we can’t just remove it like before, we need to express it as its new value and remove the constants and keep the variables.

i.

dm = σ da

σ=

and

kx 2 (as x↑, σ↑) (as y↑, σ↓) y

d. Now the integration is just as simple as before, as long as we’re systematic…

i.

xcm

ii.

⎛ kx 2 ⎞ ⎛ x2 ⎞ x x k da ∫ ⎜⎝ y ⎟⎠ ydx xσ da ∫ ⎜⎝ y ⎟⎠ xdm ∫ ∫ = → = = = 2 2 ∫ dm ∫ σ da ∫ ⎛⎜ kx ⎞⎟ da k ∫ ⎛⎜ x ⎞⎟ ydx ⎝ y ⎠ ⎝ y⎠

ycm

⎛ kx 2 ⎞ y⎜ (3 − x)dy ∫ ∫ ydm → ∫ yσ da = ⎝ y ⎟⎠ = = 2 ∫ dm ∫ σ da ∫ ⎛⎜ kx ⎞⎟ (3 − x)dy ⎝ y ⎠

27

3

2

1

3

0

27

1

∫⎜ 0

⎝

3

x dx ∫ x(x )dx = ∫ = 2.25 ∫ (x )dx ∫ x dx 3

0 3

2

2

0

27

∫x

2

(3 − y 3 )dy 1

0

27

⎛ x2 ⎞ 1 ∫0 ⎜⎝ y ⎟⎠ (3 − y 3 )dy

=

27

∫ (y ) (3 − y )dy ∫ (3y = ⎛ (y ) ⎞ ⎟ (3 − y )dy ∫ (3y ⎜ 1

3

2

2

3

− y)dy = 5.4

0 27

2

1

− 13

3

y ⎟ ⎠

− 1)dy

0

O. Rocket Propulsion

1. Difference from normal propulsion: unlike a car, a rocket propels itself by ejecting gasses.

The weight of these gases and their acceleration produce the force (by Newton’s 2nd law) that propels the rocket upwards. Add to this the fact that the mass is reducing by the fuel that is being spent, which increases the rocket’s acceleration further. The most important point though is: as per the law of conservation of momentum, momentum is conserved (e.g. though the rocket increases in speed, the momentum remains the same since all forces generated are internal)

a. Fengine =

dp d(mv) dm dm dv = = vexhaust + ma = vexhaust + m dt dt dt dt dt

2. How gas momentum and rocket momentum add up: d a. M rocket vrocket = vexhaust mexhaust ⎯⎯ → M r (dv) = ve dme b. dme = −dM r (the increase in exhaust mass corresponds to a decrease in rocket mass) vf

c.

Mf ⎛ dM ⎞ M r (dv) = −ve (dM r ) ⎯integrate ⎯⎯→ ∫ dv = ve ∫ ⎜ ⎟ Mi ⎝ M ⎠ vi

d. Thrust = ve

dM dt

3. How to work the “Rocket Equation” a. M f = 110 M i vi = 0 Page 45 of 59

ve = 500m/s

→

⎛M ⎞ v f - vi = ve ln ⎜ i ⎟ ⎝ Mf ⎠

Chapter 9 (Linear Momentum and Collisions) ⎛ ⎛ M1 ⎞ ⎞ b. v f = (500 )⎜ ln ⎜ ⎟ = (500 )(ln (10 )) = 1151m/s ⎝ ⎝ .1M ⎟⎠ ⎠ 1

if the rocket’s mass drops to 1/100th of its initial mass, the speed is even greater…

⎛

⎛ M1 ⎞ ⎞ = (500 )(ln (100 )) = 2303m/s ⎝ .01M 1 ⎟⎠ ⎟⎠

c. v f = (500 )⎜ ln ⎜

⎝

so dropping the mass really helps!

P. Example 9:

1. Problem: A rocket with an initial mass of 90,000kg and a velocity of 10m/s is losing mass at a rate of 50kg/s. the exhaust speed of the gas is 1000 m/s

2. Question: a. What is the final velocity after 10 minutes? b. What is the thrust force propelling the rocket forward? c. How many minutes will it take the rocket to achieve escape velocity? 3. Solution Strategy: a. First calculate the mass at 10 minutes, and then use the rocket equation. i. M f = M i − M lost (t) → 90, 000 − (50)(600) = 60, 000 ii.

⎛M ⎞ ⎛ 90000 ⎞ v f = vi + (ve )ln ⎜ i ⎟ = (10) + (1000)ln ⎜ = 415.47m / s ⎝ 60000 ⎟⎠ M ⎝ f⎠

b. Simple plug and chug…

Thrust = ve

i.

dM = (1000)(50) = 50kN dt

c. Set the rocket equation’s final velocity to escape velocity (11km/s), and then calculate time. i. ii.

X.

⎛ 90000 ⎞ → x = 1.518 (Almost nothing left!) 11000 = (10 ) + (1000 )ln ⎜ ⎝ x ⎟⎠ M f = M i − M lost (t ) → 1.518 = 90, 000 − (50)(60)(t ) = 29.999

Chapter 10: Rotation of a Rigid Object About a Fixed Axis (return to top) A. Background 1. Rotational Motion: a particle that is moving about a fixed axis (rotating) has

measurable vectors that quantify its position, velocity, and acceleration. The important distinction though, is how vector quantities (velocity, acceleration) are handled; in circular motion, the angle is the axis of rotation and the magnitude is the magnitude of the velocity or acceleration vector. a. Right-hand rule: If you imagine gripping the axis of rotation with your fingers pointing towards the direction of rotation, your outstretched thumb will point towards the angular velocity vector.

B. Similarities between Rotational & Linear Kinematics 1. Equivalent Counterparts: a. θ (rad) x(m) b. ω (rad/s) v(m/s) c. α (rad/s 2 ) a(m/s 2 ) 2. Kinematic Equations: Rotational Kinematics

Page 46 of 59

Linear Kinematics

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) ω f = ωi + αt v f = vi + at

θ f = θi + ω i t + 1 2 α t 2

x f = xi + vit + 1 2 at 2

ω 2f = ω i2 + 2α (θ f − θi )

v 2f = vi2 + 2 a( x f − xi )

θ f = θ i + 1 2 (ω i + ω f )t

(

)

x f = xi + 1 2 vi + v f t

3. Conversion Factors: a. s(Δx) = rθ b. vtangential = rω c. atangential = rα

C. Similarities between Rotational & Linear Dynamics 1. Dynamic Equations:

Rotational Dynamics

r

r

Linear Dynamics

ac = (rω ) r = rω 2 r r W = ∫ τ dθ

r r F = ma 2 ac = v r r r W = ∫ Fdr

W = ΔK = 1 2 Iω 2 r r P = ω t = τ ⋅ω r r L = Iω r r J = ∫ τ dt r r J = ΔL r r τ = dL dt

W = ΔK = 1 2 mv 2 r r P = Wt = F ⋅v r r p = mv r r J = ∫ Fdt r r J = Δp r dpr F = dt

τ = Iα

2

Chapter 5 6 7 7 7 9 9 9 9

2. New Definitions r a. Torque ( τ )( N ⋅ m or ft ⋅ lb ): can be thought of as a force that causes rotation. It is the product of the force acting on a lever by its distance from the fulcrum. On a disk, it's the same thing; the force acting to turn the disk times its distance from the axis of rotation.

r

r

r

τ =r×F

r

r

τ = Iα b. Moment of Inertia (MOI)( I )( kg ⋅ m 2 or ft ⋅ lb ⋅ s 2 ): this is like mass, but in a rotational context, i.

in other words, it describes the resistance the object will have to a change in the object’s rotation (whether at motion or at rest). An object’s moment of inertia will be lowest at its center of mass.

i.

I = ∫ r 2 dm

c. Angular r Momentum: r r i. L=r×p

D. Example 1:

1. Problem: A disk 1.2 m in diameter is rotating at an initial rate of 15rpm. It is made to accelerate at a rate of .8rad/s. we want to know some things after 10 seconds…

2. Questions: a. What is the final angular velocity? b. What is the final velocity at the rim of the disk, the centripetal acceleration, total acceleration, and angular acceleration?

c. How many turns has it made? 3. Solution Strategy:

Page 47 of 59

Comment [as17]: This is really a chapter 11 subject, but since the teacher included it in the formulas, I will include it here. I will flesh out the definition once I get to the chapter 11 notes for this definition.

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) a. Plug it in the related kinematic equation. But first translate rpm into angular velocity. i. ii.

15rev 2π rad 1min = 1.57rad/s 1min 1rev 60 s ω f = ω i + α t = 1.57 + (.8)(10) = 9.57 rad/s

b. Each velocity and acceleration has a simple formula: i. v f = rω f = (.6)(9.57) = 5.74 m/s ii.

ac = rω 2 = (.6)(9.57)2 = 54.95 m/s 2

iii.

atot = atan + ac = α r + 54.95 2 = atot = ArcTan

iv.

( ) atot

((.8)(.6))2 + 54.95 2

(

= 54.94 m/s 2

)

.48 ac = ArcTan 54.95 = 0.5° ahead of a c

c. Find the total radians and convert to turns. i. ii.

⎛ ωi + ω f ⎞ ⎛ 1.57 + 9.57 ⎞ t = 0+⎜ ⎟⎠ 10 = 55.7rad ⎟ ⎝ ⎝ 2 ⎠ 2 1turns 55.7rad = 8.86 turns 2π rad

θ f = θi + ⎜

E. Applying Tangential Velocity to the Rotation of the Earth

1. Tangential Velocity on Earth: is a result of the spin it has about its axis. Depending on where a person is, the spin can be more of a factor. At the equator, the spin is greatest, exerting the greatest tangential velocity. If this velocity matched the earth’s escape velocity, we would be flying off into space! a. vtangential = (Rearth )(Cosθ )(ω )

⎛ 1rev 2π rad 1hr ⎞ ⎝ 24hr 1rev 3600 s ⎟⎠

b. vtangential = (6370km )(Cosθ )⎜

c. vtangential = .463Cosθ km/s (Well below 11.2km/s) d. To find out how much faster the earth needs to spin to exceed escape velocity at the equator and at our latitude (33.56°):

i. ii. iii.

vtangential = vescape = 11.2km/s

11.2km/s = (.463x )Cos (0° ) km/s

11.2km/s = (.463x )Cos (33.56° ) km/s

x = 24.19 times

x = 29.02 times

So since we’re closer to the poles than the equator, the earth has to spin faster to exceed escape velocity at our latitude. A funny thing happens at the pole:

iv.

11.2km/s = (.463x )Cos (90° ) km/s

x=∞

At the pole (as at all locations on earth), your angular velocity is the same, but your tangential velocity is zero since your radius from the center of rotation is zero.

F. Moment of Inertia: definition

Comment [as18]: Notes for Wednesday, July 26, 2006 begin here

1. Relationship to Rotational Kinetic Energy: a. i = individual particles b. K Rot =

∑ K = ∑(

1

i

i

2

)

mi vi2 =

1

2

2

2 i

i i

but ω is the same everywhere, so

i

c.

Page 48 of 59

∑ (m r ω ) Krot =

1

2

⎛ 2⎞ 2 ⎜⎝ ∑ mi ri ⎟⎠ ω = i

1

2

Iω 2

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) 2. Dependence on mass and distance a. In this picture, we see three cars on the side of a mountain turning around the road. Given the same speed and road, the heaviest, widest car is most stable while the lighter car is less but the narrowest car is least stable (because distance is a greater factor than weight). The cars are numbered according to their stability. b. The distance is a greater factor in determining the MOI’s magnitude.

3. MOI and its relationship to Torque a. Force is to angular acceleration r related r i. F = matan since a tan = rα

r then F = mrα

b. Torque is related to force (when perpendicular) i.

r

r

r

r

90° τ = r F Sinθ ⎯Sin ⎯⎯ →F = τ r

c. And Torque’s relationship to MOI i.

r 2 r r =I mrα = τ r → τ = mr 2α ⎯mr ⎯⎯ → τ = Iα

G. Example 2:

1. Problem: two balls on a massless rod are at rest. A force F causes them to spin about a pivot point; one ball is a distance r from the pivot point while the second ball is half that distance from the pivot point. 2. Questions: a. What is the angular acceleration? b. What is the final angular velocity after 10 seconds?

c. What is the tangential acceleration of each mass? d. What is the tangential velocity of each mass? 3. Solution Strategy: a. Angular acceleration is a product of the MOI, which itself comes from all masses

r

r

(

( ) )r 2

τ → rF = I totalα = m1r 2 + m2 r 2 α r r r 4F 2 5 ii. Fr = 4 mr α → α = 5mr b. Angular velocity is just the product r of angular r acceleration and time (10s) 8F 4F i. ω f = ωi + αt = 0 + (10) = 5mr mr i.

c. Tangential Acceleration depends on its distance from the pivot point. i.

r r ⎛ 4F ⎞ 4F r r = at(edge) = rα = r ⎜ ⎝ 5mr ⎟⎠ 5m

at(mid)

r r ⎛ r ⎞ ⎛ 4 F ⎞ 2F = =⎜ ⎟⎜ ⎝ 2 ⎠ ⎝ 5mr ⎟⎠ 5m

d. Tangential Velocity is related to angular velocity by multiplying it by its radius of rotation.

vt(edge)

r r ⎛ 40 F ⎞ 8F = = rω = r ⎜ ⎝ 5mr ⎟⎠ m

vt(mid)

r r r r ⎛ 40 F ⎞ 4 F = = ω= ⎜ 2 2 ⎝ 5mr ⎟⎠ m

So we see that with the same force, tangential acceleration and velocity decrease as the radius decreases.

H. Example 3:

1. Problem: you have a rod with negligible mass attached to 4 spheres at different lengths. Sphere 1 is 5kg. Sphere 2 is 4kg and 2m left of Sphere 1 and 1m right from the pivot point. Sphere 3 is 6kg, 1m right from the pivot point and 2m left of Sphere 2. Sphere 4 is 3kg and is

Page 49 of 59

Comment [as19]: I just added the shadow to these spinny things, but I got a compatibility warning telling me that these shadows might not look too hot in Word for Windows 2003, so ill keep an eye on that for both these and won’t do much else (shadow-wise) till I can confirm the results.

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) 2m left of Sphere 3. The forces applied are: 30° (-50° absolute) on Sphere 1, 60° (-120° absolute) on Sphere 2, and 155° (25° absolute) on Sphere 4.

2. Questions: a. What is the MOI of the center of mass? b. What is the total torque of the system? c. What is the angular acceleration and velocity of the system after 10 seconds? d. What are the tangential acceleration and velocity of the system after 10 seconds? e. How many turns has the system made in that time? 3. Solution Strategy: a. Since we’re measuring individual particles, we only need to calculate the distance each item has from the center of mass.

∑x m = ∑m i

xcm

i.

i

i

i

=

5 (0 ) + 4 (2 ) + 6 (4 ) + 3(6 ) = 2.77 5+4+6+3

i

I CM = ∑ mi ri2 = 5 (2.78 ) + 4 (.78 ) + 6 (1.22 ) + 3(3.22 ) = 81.11kg ⋅ m 2 2

ii.

2

2

2

i

b. The torque is the sum r of all individual torques. r i. τ tot = rtot × Ftot = 3(12 )Sin50° + 1(10 )Sin120° + 3(8 )Sin155° = 155 N ⋅ m c. Angular acceleration is the ratio of torque to MOI. Angular velocity is the initial velocity plus the velocity due to angular acceleration.

ii. iii.

r

(

)

46.38 kgms −2 m 46.38Nm = = .5718s -2 (rad) 81.11kgm 2 I 81.11kgm 2 ω f = ω i + α t = 0 + .5718(10) = 5.718s -1 (rad)

α system =

i.

τ

=

Note: why is a radian a “unit less” measure? a) b)

The definition of a radian is a section of a circle that is equal to its radius. = 1m s = rθ ⎯rs⎯⎯ →θ = = 1m

1m →θ =1 1m

d. Tangential acceleration and velocity depends on the distance from the pivot point i. at ( L & R.End ) = rα = (3)(.5718) = 1.715m/s2 ii. iii. iv.

at ( InsideSpheres ) = rα = (1)(.5718) = 0.5718 m/s2

vt ( L & R.Ends ) = rω 10 s = (3)(.566 )(10 ) = 17.154 m/s

vt ( InsideSpheres) = rω 10 s = (1)(.5718 )(10 ) = 5.718 m/s

e. Just solve the rotational kinematic equation and convert radians to turns… 2 i. θ f = θ i + ω i t + 1 2 α t 2 = 0 + 0 + 1 2 (.5718 )(10 ) = 28.59rad ii.

Page 50 of 59

28.3rad

1turn = 4.55 turns 2π rad

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis)

I. How to use the Parallel Axis (PA) Theorem and the Integral Form to find the Moment of Inertia of an object. 1. Parallel Axis Theorem: When we know the MOI of an object through its center of mass ( I CM ) we can find the MOI through any other point which has a parallel axis to it by using: a.

2 I AnyAxis = I CM + mtotal dFromCM

b. The minimum MOI of any object is the MOI at the center of mass. This also means that the easiest way to rotate an object is through its center of mass. An object thrown in the air naturally rotates around its center of mass.

J. Example 4:

1. Problem: you have a thin rod of negligible mass with two spheres at each end. Both spheres have masses of 5kg and are 2m from each other.

2. Questions: a. Which pivot point has the lowest MOI by the direct method? b. Which pivot point has the lowest MOI by the PA method? c. What is the MOI at a point midway between PP2 and PP1? 3. Solution Strategy: a. We test each pivot point to see which gives us the lowest value: 2 2 i. I PP1 = mi ri2 = 5 (1) + 5 (1) = 10 kg ⋅ m 2

∑ i

ii.

I PP 2 = ∑ mi ri2 = 5 (0 ) + 5 (2 ) = 20kg ⋅ m 2

iii.

I PP 3 = ∑ mi ri2 = 5 (2 ) + 5 (0 ) = 20kg ⋅ m 2

2

2

i

2

2

i

b. We need to calculate the MOI of the center of mass before using the PA method. Since the object has evenly distributed mass, the center of mass is the midpoint between the spheres and the MOI at that point is what we got previously.

i.

I PP 2 = I CM + mtot r 2 = 10 + 10 (1) = 20 kg ⋅ m 2

ii.

I PP 3 = I CM + mtot r 2 = 10 + 10 (1) = 20kg ⋅ m 2

2

2

c. This is where the PA method is best; since we know the MOI of the center of mass, the rest is simple math.

i.

Direct: I ( pp 2 − pp1) =

ii.

PA:

∑m r

i i

2

= 5 (1.5 ) + 5 (.5 ) = 12.5 kg ⋅ m 2 2

2

i

I ( pp 2 − pp1) = I CM + mtot r 2 = 10 + 10 (0.5 ) = 12.5 kg ⋅ m 2 2

K. Example 5: (extension of Example 3)

1. Problem: we have the same rod as in example 3, but now we add pivot points at the center of mass and midway between the 6 and 3kg masses.

2. Questions: a. What is the MOI at the point between the 6 and 3kg masses by the direct method? b. What is the MOI at the point between the 6 and 3kg masses by the PA method? c. What is the highest MOI of this object? 3. Solution Strategy: a.

I 2 = 5 (5 ) + 4 (3) + 6 (1) + 3(1) = 170 2

2

2

2

( ) = 169.982 or 170

b. I AnyPoint = I CM + m tot r 2 = 81.111 + 18 2.2

2

c. the highest MOI is the point furthest from the center of mass.

Page 51 of 59

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) i. I REnd = 81.111 + 18(3.222)2 = 268 kg ⋅ m 2

L. Example 6: (Integrating to get the Moment of Inertia) 1. Problem: you have a rod of uniform mass of length L. 2. Questions:

a. Calculate the MOI passing through the center of mass by integration and substitution (Professor Kiledjian’s example)

b. Calculate the MOI passing through the center of mass by cull integration (Physics book example)

3. Solution Strategy: a. Here we figure out the MOI by finding the MOI of any axis and relating it to the PA theorem that will return the MOI of the center of mass. i. First we recall the relationship between mass and length and reduce it to a differential. Next we substitute which now allows us to integrate it on the x-axis L

a)

ii.

= λ dx I AnyAxis = ∫ x 2 dm ⎯dm ⎯⎯ → ∫ x 2 λ dx = λ ∫ x 2 dx =

λ L3

0

3

Now we need to figure out the value of λ, and we do that with another integration L

a)

iii.

0

Substitute the value of λ in the first equation for the second one a)

iv.

λ dx m = ∫ dm ⎯dm= ⎯⎯ → m = λ ∫ dx = λ L → λ = m L

I AnyAxis =

(m / L )L3 = mL2 3

3

Now use the PA theorem to get your answer:

mL2 mL2 = I CM + md 2 → − md 2 = I CM 3 3 2 1 mL2 ⎛ L⎞ = − m ⎜ ⎟ = mL2 ⎝ ⎠ 3 2 12

a)

I AnyAxis = I CM + md 2 →

b)

I CM

b. Here we simply recall that λ is the mass to length ratio (since mass is uniform) and integrate i. We do the same substitution as before, but with more baggage

( )

dm= λ dx = L dx I CM = ∫ r 2 dm ⎯⎯⎯⎯⎯ → ∫ x 2 m L dx m

a)

ii.

Now we recall that we want the MOI of the center of mass, which on a uniform rod is also at the center of the rod, so our lower and upper limits will reflect this. a)

I CM = m L ∫

L /2

− L /2

x 2 dx =

1 mL2 12

Very Powerful. Once we set up the integral, we could find the moment of inertia at any point on the rod by changing the limits

M.

Recap of Moments of Inertia:

1. Rolling down a ramp: if you put a hollow sphere, a solid sphere, a hollow ring, and a solid disk on a ramp. If they all had the same mass and radius, which one would make it down the ramp first? a. The rule here is based on the MOI of each item rolling down the ramp: The lower the MOI, the faster the object will roll down the ramp, so figuring out which one makes it down first is a matter of knowing each object’s MOI which will give us our answer…

i. ii.

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mr 2 (fastest) 2 Hollow Sphere = 3 mr Solid Sphere =

2

5

2

Comment [as20]: Notes for Monday, July 31, 2006 begin here

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) iii. Solid Disk = 1 2 mr 2 iv. Hollow Ring = mr 2 (slowest)

N. Example 7: 1. Problem: A rod has a linear density that increases as λ=kx³ from the left end of the rod. 2. Questions: a. Find the MOI for the left end, the center of mass, the minimum and maximum MOI. b. Answer part “a” if λ=kx³ was measured from the center of the rod. 3. Solution Strategy: a. Unlike a rod of uniform density (whose MOI at the ends is ⅓ML²), the left side of the rod will have a greater MOI on the left end than on the right end. The way to solve this is to use two integrals, one to get MOI, and the other to get the value of the constant (k) as it relates to mass. i. First we figure out the value of k, integrated into mass

( )

L

λ dx = kx dx mLEnd = ∫ dm ⎯dm= ⎯⎯⎯⎯ → ∫ kx 3 dx = k ∫ x 3dx → m = 3

ii.

0

kL4 4m →k= 4 4 L

Next we plug in our integrated k-mass value into our MOI equation and solve ⎛ 4 m⎞ dm = λ dx = ⎜ 4 ⎟ x 3 dx ⎝ L ⎠

⎛ 4 mx 3 ⎞ 4m L 2 I LEnd = ∫ x dm ⎯⎯⎯⎯⎯⎯→ ∫ x 2 ⎜ 4 ⎟ dx = 4 ∫ x 5 = mL2 L 0 3 ⎝ L ⎠ 2

iii.

Now we can figure out the center of mass’ MOI by getting its x-position and integrating.

∫ xdm = ∫ xλdx = ∫ x (kx )dx = ∫ = ∫ dm ∫ λdx ∫ (kx )dx ∫ 3

xCM

L

0 L

3

0

x 4 dx 3

=

x dx

L5

5 = 4L L 5 4 4

2

I any = I CM + md 2 →

2 2 2 ⎛4 ⎞ mL = I CM + m ⎜ L ⎟ → I CM = mL2 ⎝5 ⎠ 3 75

iv.

I max = I LEnd = 2 3 mL2

v.

I REnd = I CM + md 2 =

2 1 mL2 + m ⎡⎣(1 − 4 5 )L ⎤⎦ = mL2 75 15

b. This type of setup means the mass is symmetric down the middle, making it easier to figure out the MOI of the center of mass. (The 2 I plugged in the integral is because of the odd exponent. Since the range of integration is symmetric, integrating would give a zero, so instead we divide the range of integration in half and multiply the result by 2).

( )

= λ dx = kx dx mCM = ∫ dm ⎯dm ⎯⎯⎯⎯ → ∫ kx 3 dx = 2 k ∫ 3

i.

L /2

0

x 3dx → m =

kL4 32 m →k= 4 32 L

⎛ 32 mx 3 ⎞ dm = λ dx = kx 3 dx = ⎜ ⎟ dx ⎝ L4 ⎠

ii.

⎛ 32mx 3 ⎞ 64m L /2 1 I CM = ∫ x 2 dm ⎯⎯⎯⎯⎯⎯⎯⎯→ ∫ x 2 ⎜ dx = 4 ∫ x 5 dx = mL2 4 ⎟ 0 L 6 ⎝ L ⎠

iii.

I EitherEnd = I CM + md 2 =

1

mL2 + m (L 2 ) = 5 12 mL2 2

6

O. Solving Problems Involving Rotational Dynamics

1. Problems such as these are similar to linear dynamics as long as you recall the equations needed are very close to the ones we used previously. r r a. τ = rFSinθ , Iα

∫

b. I = r 2 dm

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Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) c. K = 1 2 Iω 2

P. Example 8:

1. Problem: You have two disks on top of each other, disk 1 (m = 10kg, r = 2m) and on top of it, the disk 2 (m = 5kg, r = 1m). Disk 1 has a force of 8N pulling at 90° from the horizontal, while disk 2 has a force of 6N pulling at 60° from the opposite horizontal axis. 2. Questions: a. Find the angular acceleration of the disks. b. What is the angular velocity after 12 seconds? 3. Solution Strategy: a. Remember torque = MOI times angular acceleration (MOI of a disk is ½ mr²)

r

r

r

i.

τ = Iα → (τ F1 − τ F 2 ) = (1 2 m1r1 + 1 2 m2 r2 )α

ii.

2 2 2 (8 ) − 1(6 )Sin 60° = ⎡⎣ 1 2 (10 )(2 ) + 1 2 (5 )(1) ⎤⎦ α → α = .48017

b. just plug it into the applicable rotational kinematics equation. i. ω f = ω i + α t = 0 + (.48 )(12 ) = 5.76 rad/s

Q. Example 9:

1. Problem: You have an apparatus where one block is suspended off a ledge. It is tied to another block and a hollow spherical pulley is transferring their tension. The suspended block is 2kg and is 4m from the floor. The block tied to the suspended block is 5kg and is resting on a flat surface with a coefficient of kinetic friction of 0.3. The hollow spherical pulley has a mass of 3kg and a radius of 0.8m. 2. Questions:

a. What is the acceleration of the suspended block? b. What is the suspended block’s final velocity as it hits the ground?

c. How much work is done by friction and gravity? d. Do part “a” and “b” by first doing part “c” and using the work-energy approach 3. Solution Strategy: a. We figure it out by making 3 related equations & adding them together i. 5k block = T1 − μ k n = ma → T1 − (.3)(49 ) = 5a → T1 - 14.7 = 5a ii.

2k block = 19.6 - T2 = 2a

iii.

Pulley =

τ = Iα → (T2 R − T1 R ) = (2 3 mR 2 )α r

α =a T2 - T1 = ⎡⎣ 2 3 (3)R ⎤⎦ α → T2 - T1 = (2 R )α ⎯R⎯⎯ → T2 - T1 = 2a

iv.

Add equations1 and 2 to get equation 4 and add it to equation 3 to solve: 1 2 4

T1 − 14.7 = 5a + 19.6 − T2 = 2a T1 - T2 + 4.9 = 7a

4 3

T1 - T2 + 4.9 = 7a T2 − T1 = 2a 4.9 = 9a → a = .54

b. A simple kinematic equation i.

v 2f = vi2 + 2ad → v f = 0 + 2 (.54 )(4 ) = 2.087

c. Work equations hearken back to chapter 7 and 8 i. Wg = mgh = (2 )(9.8 )(4 ) = 78.4

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Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) ii. W fk = − μ k nd = − (.3)(49 )(4 ) = -58.8 d. We set up a work/kinetic energy equation and find the velocity from which we will get our acceleration.

(

)

i.

Wtotal = ΔK total → Wg + W fk = (ΔK 5kB + ΔK 2kB + ΔK Pul )

ii.

(78.4 ) + (−58.88 ) = ⎡⎣ 2 (5 )v 2f + 1 2 (2 )v 2f + 1 2 Iω 2f ⎤⎦

iii.

19.6 = 7 2 v 2f + 1 2

iv.

1

19.6 =

7

2

(

2

3

)

=3 t = rω m pul R 2 ω 2f ⎯m⎯⎯ → 19.6 = 7 2 v 2f + R 2ω 2f ⎯v⎯⎯ → vt = v f

v 2f + vt2 ⎯ ⎯⎯ → 19.6 = 9 2 v 2f → v f = 2.087

R. Example 10: 1. Problem: A rod of length L with a density similar to example 7 (λ=kx³), except in this example, the rod is vertical and is falling down from the right end to the left end.

2. Questions: a. Calculate the tangential acceleration of the rod once its horizontal at the center of mass and at the end of the rod.

b. Calculate the tangential velocity of the rod once it is horizontal at the center of mass and at the end of the rod.

3. Solution Strategy: a. We will reuse a lot of example 7’s numbers for this example, which is lucky, if not, we’d have to reintegrate everything! i. We recall that the center of mass is 4/5th of the length.

6g 5L

= rF τ = Iα ⎯τ⎯⎯ → (4 5 L )(mg ) = (2 3 mL2 )α → α =

r

r

( )=

at = rα = (45L ) ii.

6g 5L

24 25

g Å Not quite as fast as gravity

We do the same thing as at the center of mass, but change the length to L

(L )(mg ) = (2 3 mL2 )α → α =

( )=

3g → at = rα = (L ) 2L

3g 2L

3 2

g Å more than gravity

b. We will use the work/potential energy equation to figure this out. i.

WCM = ΔK → mgh =

ii.

WLEnd = ΔK → mgh =

1

I ω 2 → mg (4 5 L ) =

2 CM 1

2

I LEndω 2 → mgL =

1

2

1

(

2

2

(

3

2

75

)

mL2 ω 2 → ω =

)

mL2 ω 2 → ω =

12g 5L

3g L

S. Chapter 10.9 (part of Test 5): Rolling Motion

1. Background: Rolling motion is the combination of translational and rotational motion. It occurs when an object encounters both types of motion. The tangential velocity at the TOP of the wheel and the center of mass velocity are the most important factors to consider when calculating rolling motion. 2. Examples of Rolling Motion:

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Comment [as21]: Notes for Wednesday, August 2, 2006 begin here.

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) a. Pure Rolling Motion: The tangential velocity at the top is equal to the velocity at the center of mass of the wheel. In pure rolling motion, the velocity at the contact point ( vtotal(bot) ) is zero.

i.

vtangential(top) = vCM

ii.

vtotal(top) = vtan + vCM = 2vCM

iii.

vtot ( front / back ) = 2

vtotal (bot ) = vCM − vtan = 0

iv. If a wheel is slipping, it will achieve rolling motion when atan = acm b. Overspin: when the tangential velocity is greater than the velocity at the center of mass of the wheel. If there is enough friction, it will slow down the tangential velocity until it matches the velocity of the center of mass. In real life, this is like when a car is taking off and burns rubber. c. Underspin – when the tangential velocity is less than the velocity at the center of mass of the wheel. In real life, this is like when a car slams the brakes and goes into a skid.

3. Kinetic Energy: an object undergoing rolling motion develops both angular and linear kinetic energy and both must be taken into account. 2 a. K = 1 2 I CM ω 2 + 1 2 mvCM 4. Rolling down a ramp: using the kinetic equation above, we can calculate the velocity of a ball rolling down a ramp: a. K =

1

2 I CM

( ) vCM

2

2 1 R + 2 MvCM → K =

b. K f + U f = K i + U i →

1

2

1

2

⎛ I CM ⎞ 2 ⎜⎝ 2 + M ⎟⎠ vCM R

⎛ I CM ⎞ 2 ⎜⎝ 2 + M ⎟⎠ vCM + 0 = 0 + mgh → vCM = R

2gh 1 + (Icm MR 2 )

T. Example 1: 1. Problem: A solid sphere of 6kg mass and 2m radius is given an initial push of 10 m/s on a surface whose coefficient of kinetic friction is 0.3

2. Questions: a. How long does it take it to reach pure rolling motion and how far has it traveled? b. What is its final tangential and center of mass velocity?

3. Solution Strategy: a. Figure out the dynamic aspect of it (Ch5), then the rotational aspect of it (Ch10): i. First, figure out the center of mass and tangential acceleration.

∑ F = ma → − μ n = ma → μ mg = ma → (.3)(9.8 ) = a → a = -2.94m/s r ∑ τ = Iα → FdSin90° = Iα → μ nR = ( mR )α k

ii.

k

2

k

5

2

α =a μ k mgR = (2 5 mR 2 )α → Rα = 5 2 μ k g ⎯R⎯⎯ → at = 5 2 (.3)(9.8 ) = 7.35m/s 2 t

iii.

Next, use the previous info to get the center of mass and tangential velocities.

v f (CM ) = vi(CM ) + at = 10 - 2.94t

v f (tan) = vi (tan) + atant = 7.35t

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2

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) iv.

Now put them together to get the time the tangential velocity will equal CM velocity.

v.

Finally, use that time to figure out how far it traveled…

v f = vi + at → 7.35t = 10 + 2.94t → t = .97s

x f = xi + vi t + 1 2 at 2 = 0 + 10 (.97 ) + 1 2 (−2.94 )(.97 ) = 8.33 m 2

b. Both are the same, so we solve for either. i. v f (cm ) = 10 − 2.94(.97) = 7.14 m/s

U. Example 2: 1. Problem:

a. You’ve got a solid disk of 2kg mass and 1.5m radius on a 40° incline which is 4m tall.

2. Questions: a. Find the tangential and center of mass acceleration b. Find the friction force c. Find the tangential, center of mass, and total top velocity at the bottom of the incline

d. Find the time it takes the disk to make it to the bottom of the incline

e. Solve part “a” by using the energy method. 3. Solution Strategy: a. Since we are told this is pure rolling motion, we figure out when both accelerations are the same. i. What torque rolls the ball? Friction of course!

∑τ = Iα → ( f )(R )(Sin90°) = ( r

1

s

( fs ) =

1

2

Rα = atan

mRα ⎯⎯⎯⎯ → fs =

1

2

2

)

mR 2 α → ( fs )(1) = (1 2 mR )α

mat

We know the moment our downward force equals our friction force, our ball will have achieved pure rolling motion. We can solve for acceleration at this point too.

ii.

∑ F = ma

f = 1 ma

CM

s t 2 → mgSinθ − fs = maCM ⎯⎯⎯⎯ → mgSinθ − 1 2 matan = maCM

tan = aCM gSinθ − 1 2 atan = aCM ⎯a⎯⎯⎯ → gSinθ =

3

2

a → a = 2 3 (9.8 )Sin (40° ) = 4.2m/s2

b. Given the acceleration, we use our previous friction/torque equation to find the friction force. i. fs = 1 2 ma = 1 2 (2 )(4.2 ) = 4.2N Yes, the friction force equals acceleration when your rolling item is a disk (I = ½ mr²) c. Simple kinematic equations now… i. ii. iii.

= 4 / Sin 40° v 2f ( CM ) = vi2(CM ) + 2ad ⎯d⎯⎯⎯ → v f ( CM ) =

(2 )(4.2 )(Sin440° ) = 7.23m/s

v f (tan) = v f ( CM ) = 7.23m/s v f (total @top) = v f (tan) + v f (CM ) = 14.46 m/s

d. Figure out the time now… i. v f = vi + at → 7.23 = 4.2t → t = 1.721s e. This is actually simpler in some respects, so long as you remember everything. i. WNonConservForces + WExtForces = K f − K i U f − U i

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(

ii.

0 + 0 = ⎡⎣

(

iii.

mgh =

mv 2f + 1 2

1

2

1

2

)(

)

)− (0 )⎤⎦ + ⎡⎣(0 ) − (mgh )⎤⎦ → mgh = ( mR )ω ⎯⎯⎯→ gh = v + v

mv + 1 2 Iω 2 f

1

2

2 f

2

2 f

Rω = v

1

2

2 f

1

4

2 f

1

2

mv 2f + 1 2 Iω 2f

Chapter 10 (Rotation of a Rigid Object About a Fixed Axis) iv.

gh =

3

4

v 2f → v f =

4

3

(9.8 )(4 ) = 7.23m/s

Notice how the speed at the bottom does not depend on the radius or mass of the disk

V. Example 3: 1. Problem: You have a system with a hollow sphere at the bottom of a ramp with a rope attached to its top (to pull it up while allowing it to roll upwards unhindered). At the top of the ramp, there is a hollow disk functioning as a pulley (mass 2kg and radius .5m). On the vertical side of the ramp, there is a block that weighs 20kg. 2. Questions: a. What is the final velocity of the block as it hits the ground? 3. Solution Strategy: a. The easiest way to solve this is the energy method, the setup is the hardest part, and you need to remember that the pulley and sphere have both angular and linear kinetic energy. You also have to remember the sphere has potential energy…

i. ii. iii.

(

( 0 = ⎡( ⎣

1

2

mv

0=⎡ ⎢⎣

(

1

2

{[mgh ]

)

2 f block

sphere

+

(

1

2

Iω

− (mgh )block

mv 2f

sphere

v.

) ( ) +(

) )

0 = K Block + K pulley + K sphere + U Sphere − U block

{[mgh ] iv.

)(

W NonConsForces + WExtForces = K f − K i U f − U i

)

block

2 f pulley

}

(

+ ⎡⎣ 1 2 mR 2

− (mgh )block

}

1

)( ) ⎤⎦ v

2

mv 2f

2

R

pulley

)

+

sphere

(

1

2

+

(

mv 2f

1

2

Iω 2

)

sphere

)

+

sphere

(( 1

2

⎤+ ⎦

2

3

mR 2

)( ) ) v

2

R

sphere

⎤+ ⎥⎦

⎡ ⎛ 20 ⎞ ⎛ (2 )(.5 )2 2 ⎞ ⎛ 4 2 ⎞ ⎛ (4 )(1)2 2 ⎞ ⎤ 0 = ⎢⎜ v2 ⎟ + ⎜ v ⎟ +⎜ 2 v ⎟ +⎜ 2 v ⎟⎥+ ⎠ ⎦⎥ ⎢⎣ ⎝ 2 ⎠ ⎝ (2 )(.5 ) ⎠ ⎝ 2 ⎠ ⎝ 3(1) ⎣⎡(4 )(9.8 )(2 Sin 40° ) − (20 )(9.8 )(4 )⎤⎦

(

) ( ) ( ) ( v )⎤⎦ + [50.395 − 784 ]

vi.

0 = ⎡⎣ 10 v 2 + v 2 + 2v 2 +

vii.

v=

4

3

2

(3 43)(733.605 ) = 7.154 m/s

XI. Chapter 11: Angular Momentum A. Background: B. Dd C. How the cross product determines the torque:

1. A block pulled at varying angles to the moment arm results in a torque (twisting force) which is the vector component of force at 90° to the moment arm. If we measure the angle of the force from the moment arm’s direction, this vector component is the sine of the force. One interesting point this illustrates is how the force is the same at two angles

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Chapter 10 (Rotation of a Rigid Object About a Fixed Axis)

D. Example 1:

r r 1. Problem: You have two vectors, A (2iˆ − 3 ˆj) and B (-iˆ + 5 ˆj) . 2. Questions: a. Find the r angle r θr between A & B b. Find C = A × B and its direction. 3. Solution Strategy: a. To find the angle, we add up the vectors and get the inverse trig function

Cosθ =

(A )(B ) = ⎡⎣(2 )(−1)iˆ ⎤⎦ + ⎡⎣(−3)(5 ) ˆj ⎤⎦ = A B

( 2 + 3 )( 1 + 5 ) 2

2

2

2

−17 → θ = 157.63° 18.384

b. To find the magnitude of C, do a little vector algebra, the direction is the sign…

ˆj kˆ ⎞ ⎛ iˆ r ⎜ ⎛ −3 0 ⎞ ˆ ⎛ 2 0 ⎞ ˆ ⎛ 2 −3⎞ ⎟ + k⎜ − j⎜ C = ⎜ 2 −3 0 ⎟ = iˆ ⎜ ⎝ −1 5 ⎟⎠ ⎝ −1 0 ⎟⎠ ⎝ 5 0 ⎟⎠ ⎜⎝ −1 5 0 ⎟⎠ kˆ ⎡⎣(2 )(5 ) − (−3)(−1)⎤⎦ = 7 kˆ Straight up

E. Angular Momentum and its relationship to linear momentum

1. The cross product: Angular momentum is the cross product of the moment arm and the linear momentum and the sine of the angle between them.

2. Different lengths mean different velocities: if a rod has two masses rotating about a fixed pivot, their angular momentum is the sum of the outer mass’ distance from the pivot and its momentum and the inner mass’ distance from the pivot and the momentum (which depends on its distance from the pivot). 3. Angular Momentum and linear motion: Any object with linear momentum also has angular momentum (in other words, it doesn’t need to be moving in a circular pattern). In order to determine its angular momentum though, you need to determine the reference point (pivot point) from which you calculate the angle between the pivot and the object. if there is a point of reference (the ground) from which to determine the angle made by the pivot and the object, then the angular momentum is the sine of this angle.

F. Example: 1.

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