QA 33

#$&*

course PHY 121

8/1 4

033. Rotational KE and angular momentum

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Question: `q001. Note that this assignment contains 12 questions.

A rotating object has kinetic energy, since a rotating object has mass in motion.

However we cannot easily use 1/2 m v^2 to calculate this kinetic energy because

different parts of a typical object are rotating at different velocities.

For example a rigid uniform rod rotated about one of its ends is moving faster near its

far end than near its axis of rotation; it has a different speed at every distance from

its axis of rotation. However as long as the rod remains rigid the entire rod moves at

the same angular velocity.

It turns out that the kinetic energy of a rotating object can be found if instead of

1/2 m v^2 we replace m by the moment of inertia I and v by the angular velocity `omega.

Thus we have

KE = 1/2 I `omega^2.

What is the kinetic energy of a uniform sphere of radius 2.5 meters (that's a pretty big

sphere) and mass 40,000 kg when its angular velocity is 12 rad/sec (that's almost two

revolutions per second)?

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Your solution:

Moment of Inertia: mr^2

I = 40,000 kg (2.5 m)^2

I = 40,000 kg (6.25 m^2

I = 250,000 kg m^2

KE = .5 I `omega^2

KE = .5 (250,000 kg m^2)(12 rad/s)^2

KE - .5 (250,000 kg m^2)(144 rad^2/s^2)

KE = 18,000,000 Joules

confidence rating #$&*:

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Given Solution:

The KE is 1/2 I `omega^2. We first need to find I; then we can use the given angular

velocity to easily find the KE. For this sphere we have I = 2/5 M R^2 = 2/5 * 40,000

kg * (2.5 m)^2 = 100,000 kg m^2. The kinetic energy of the sphere is thus KE = 1/2 I

`omega^2 = 1/2 * 100,000 kg m^2 * (12 rad/s)^2 = 7.2 * 10^6 Joules.

STUDENT COMMENT: I don’t understand where the 2/5 came from.

INSTRUCTOR RESPONSE: Nearly all the mass of a sphere or a disk lies closer to its axis

of rotation than its rim. If all the mass of a sphere or a disk was concentrated at the

rim, its moment of inertia would be M R^2, but for the given reason the moment of

inertia is less than this.

It turns out that the moment of inertia of a disk is 1/2 M R^2.

A sphere is thicker near its axis of rotation, so the mass is concentrated even closer

to the axis than for a disk of the same mass and radius. The moment of inertia of a

sphere is 2/5 M R^2.

It requires calculus to understand where these actual formulas come from. For Phy 121

and Phy 201 students, the formulas should simply be remembered (it's easier to remember

if you understand that 2/5 is less than 1/2, and also understand why the moment of

inertia of the sphere is less than that of the disk).

Terminology is being used a little loosely here. Instead of just 'a disk' the formula

applies to a 'uniform disk rotating about a central axis perpendicular to its plane',

and instead of 'a sphere' we should say 'a uniform sphere rotating about an axis through

its center'.

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Self-critique (if necessary):

I forgot about the 2/5 part.

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Self-critique rating: 2

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Question: `q002. By carefully measuring the energy required to accelerate it from rest

to an angular velocity of 500 rad/s, we find that the KE of a certain uniform disk is

45,000 Joules. What is the moment of inertia of that disk?

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Your solution:

average velocity = (0 rad/s + 500 rad/s)/2 = 250 rad/s

KE = .5 I `omega^2

45,000 Joules = .5 I (250 rad/s)^2

45,000 Joules = .5 I (62,500 rad^2/s^2)

90,000 Joules = I (62,500 rad^2/s^2)

I = 1.44 kg m^2

confidence rating #$&*:

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Given Solution:

We know that KE = 1/2 I `omega^2, and we know the KE and we know `omega. Solving this

equation for I we obtain

I = 2 * KE / `omega^2.

So for this disk

I = 2 * (45,000 J) / (500 rad/s)^2

= 90,000 J / ( 250,000 rad^2 / s^2) = .36 kg m^2.

[ Note that if we know the mass or the radius of the disk we can find the other, since

I = 1/2 M R^2 = .36 kg m^2. ]

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Self-critique (if necessary):

I thought you would need the average velocity not the final. I have made changes in my

notes.

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Self-critique rating: 2

@&

KE is proportional to v^2, so it's not linear in v.

As a result, not to overstate thing but:

Any energy you get based on the average velocity is pretty much irrelevant to anything.

*@

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Question: `q003. A 3-kg mass is tied to a thin cord wound around the thin axle of a

disk of radius 20 cm and mass 60 kg. The weight descends 200 meters down a long

elevator shaft, turning the axle and accelerating the disk. If all the potential energy

lost by the weight is transferred into the KE of the disk, then what will be the angular

velocity of the disk at the end of the weight's descent?

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Your solution:

PE = F * `ds

F = m * a

F = 3 kg * 9.8 m/s^2

F = 29.4 N

PE = 29.4 N * 200 m

PE = 864.36 Joules

Moment of Inertia: mr^2

I = 60 kg (.2 m)^2

I = 60 kg (.04 m^2)

I = 2.4 kg m^2

KE = .5 I `omega

864.36 Joules = .5 (2.4 kg m^2) `omega

864.36 Joules = 1.2 kg m^2 `omega

angular velocity = 720.3 radians/second

confidence rating #$&*:

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Given Solution:

The 3-kg mass has a weight of 3 kg * 9.8 m/s^2 = 29.4 Newtons. As it descends 200

meters its PE decreases by `dPE = 29.4 N * 200 m = 5880 Joules.

The disk, by assumption, will gain this much KE (note that in reality the disk will not

gain quite this much KE because of frictional losses and also because the descending

weight will have some KE, as will the shaft of the disk; however the frictional loss

won't be much if the system has good bearings, the weight won't be traveling very fast

if the axle is indeed thin, and a thin axle won't have much moment of inertia, so we can

as a first approximation ignore these effects).

The KE of the disk is KE = 1/2 I `omega^2, so if we can find I our knowledge of KE will

permit us to find `omega = +-`sqrt( 2 KE / I ).

We know the radius and mass of the disk, so we easily find that I = 1/2 M R^2 = 1/2 *

60 kg * (.2 m)^2 = 1.2 kg m^2.

Thus the angular velocity will be +- `sqrt( 2 * 5880 J / (1.2 kg m^2) ) = +- 100 rad/s

(approx).

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Self-critique (if necessary):

It started off okay. Then I hit a bad button on the calculator or something because I

had the formula and numbers correct, but the answer was off. I forgot that the omega

was squared. I did not verify that the disk needs 1/2 mr^2 instead of plain mr^2.

Overall, I understand the errors. I do not promise to never make them again.

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Self-critique rating: 2

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Question: `q004. A rotating object also has angular momentum L = I * `omega. If two

rotating objects are brought together and by one means or another joined, they will

exert equal and opposite torques on one another and will therefore end up with an

angular momentum equal to the total of their angular momenta before collision.

What is the angular momentum of a disk whose moment of inertia is .002 kg m^2 rotating

on a turntable whose moment of inertia is .001 kg m^2 at 4 rad/s?

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Your solution:

.002 kg m^2 * `omega = .001 kg m^2 * 4 rs

(multiply .001 * 4 and divide the product by .002)

`omega = 2 rad/s

confidence rating #$&*:

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Given Solution:

The total moment of inertia of the system is .002 kg m^2 + .001 kg m^2 = .003 kg m^2.

The angular momentum of the system is therefore L = I * `omega = .003 kg m^2 * (4

rad/s) = .012 kg m^2 / s.

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Self-critique (if necessary):

I was confused on this one because I thought that I needed another velocity before I

could proceed. The disk and the turn table are rotating at the same velocity, which

with 20-20 hindsight makes sense. Is angular momentum just how fast they are going

around? And, where did the radians disappear to??

@&

Angular momentum, speaking very loosely, is just how strongly the object 'wants' to keep moving. It's the product of moment of inertia, which is analogous to mass and sort of acts like the mass of the rotating system in that it's the quantity that resists acceleration, multiplied by how quickly the object is rotating (i.e., its angular velocity).

More moment of inertia at a given angular velocity, more tendency to keep moving, harder to stop, that is, more angular momentum.

More angular velocity for a given moment of inertia, more tendency to keep moving, harder to stop, so more angular momentum.

More moment of inertia and more angular velocity, even harderer to stop, even morer angular momentum.

*@

@&

A meter of radius multiplied by a radian of angle gives you a meter of arc distance, so meter * radians can give you just meters.

A meter of arc distance divided by a radian of angle gives you a meter of radius, so a meter divided by a radian can give you a radian.

You probably want to leave it at that. Tracing whether different meters in a unit represent radii or arc distances tends to become ridiculously difficult.

*@

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Self-critique rating: 2

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Question: `q005. If a stick with mass .5 kg and length 30 cm is dropped on the disk of

the preceding example in such a way that its center coincides with the axis of

rotation, then what will be the angular velocity of the system after frictional torques

bring the stick and the disk to the same angular velocity?

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Your solution:

I = 1/12 * .5 kg * (.3 m)^2

I = 1/12 * .5 kg * .09 m^2

I = .0375 kg m^2

.003 kg m^2 + .00375 kg m^2 = .00675 kg m^2

.00675 kg m^2 (4 rad/s) = .027 kg m^2/s

confidence rating #$&*:

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Given Solution:

Since the stick and the disk exert equal and opposite torques on one another, the

angular momentum of the system will be conserved. Since we know enough to find the

moment of inertia of the new system, we will be able to easily find its angular

velocity.

The moment of inertia of the stick is 1/12 * .5 kg * (.3 m)^2 = .00375 kg m^2, so the

moment of inertia of the system after everything settles down will be the sum of the

original .003 kg m^2 and the stick's .00375 kg m^2, or .00675 kg m^2.

If we designate this moment of inertia by I ' = .00675 kg m^2 and the new angular

velocity by `omega ', we have L = I ' `omega ' so

`omega ' = L / I ',

where L is the .012 kg m^2 total angular momentum of the original system.

Thus the new angular velocity is

`omega ' = L / I ' = .012 kg m^2 / s / (.00675 kg m^2) = 1.8 rad/s, approx..

Thus when the stick was added, increasing the moment of inertia from .003 kg m^2 / s to

.00675 kg m^2 / s (slightly more than doubling I), the angular velocity decreased

proportionately from 4 rad/s to 1.8 rad/s (slightly less than half the original angular

velocity).

STUDENT QUESTION

Wouldn’t the radius be .15m since the whole length is .3m?

INSTRUCTOR RESPONSE

Good thinking, but it's not so in this case. The formula is 1/12 m L^2, and the fraction

1/12 takes account of the fact that it's rotating about its center. If a rod rotates

about its end, the formula is 1/3 m L^2 (note that 1/3 is 2^2 = 4 times as great as

1/12).

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Self-critique (if necessary):

I'm still working on the concepts of angular velocity, inertia, and angular momentum so

that I can understand enought to follow through all the way to the end. I just need

more practice.

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Self-critique rating: 2

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Question: `q006. An ice skater whose moment of inertia is approximately 1.2 kg m^2

holds two 5 kg weights at arm's length, a distance of 60 cm from the axis of rotation,

as she spins about a vertical axis at 6 rad/s (almost 1 revolution / sec ). What is her

total angular momentum and her total angular kinetic energy?

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Your solution:

moment of inertia for weights: mr^2

10 kg * (.6m)^2

10 kg (36 m^2)

3.6 kg m^2

Total inertia for system: 12. kg m^2 + 3.6 kg m^2 = 4.8 kg m^2

Total angular momentum:

L = I `omega

L = 4.8 kg m^2 * 6 rad/s = 28.8 kg m^2/s

KE = .5 I `omega^2

KE = .5 (4.8 kg m^2)(6 rad/s)^2

KE = .5 (4.8 kg m^2)(36 rad^2/s^2)

KE = 86.4 Joules

confidence rating #$&*:

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Given Solution:

The moment of inertia of each of the two weights is m r^2 = 5 kg * (.6 m)^2 = 1.8 kg

m^2, so the total moment of inertia of both weights is 3.6 kg m^2 and the moment of

inertia of the system consisting of the skater and the weights is 1.2 kg m^2 + 3.6 kg

m^2 = 4.8 kg m^2.

The angular momentum of the system is therefore 4.8 kg m^2 * 6 rad/s = 28.8 kg m^2 / s.

The total angular kinetic energy is KE = 1/2 I `omega^2 = 1/2 * 4.8 kg m^2 * (6 rad/s)^2

= 86.4 Joules.

STUDENT COMMENT: i wasn't sure what equation to use with the 5kg weights to get I, but

from here i used the correct equation to get L

INSTRUCTOR RESPONSE: Each 5 kg weight consists of particles that are all very nearly

the same distance from the axis. Thus r is about the same for all particles in each

weight, and you can therefore use I = m r^2.

STUDENT QUESTION

Do you automatically just know that the angular momentum is unchanged?

INSTRUCTOR RESPONSE

There isn't much torque acting on the skater--only the torqe produced by friction

between the ice and the point of her skate. Since the force on the point of the skate is

exerted very close to the axis of rotation, the torque is very small.

The skater pulls her arms in within a fraction of a second, so what little torque there

is has very little time to act. So the product of torque and time interval is short.

The impulse-momentum theorem for rotation says the the change in angular momentum I *

omega is equal to the product of torque and time interval. So there is very little

change in angular momentum.

Her moment of inertia decreases significantly, so her rate of spin increases

accordingly.

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Self-critique (if necessary): OK

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Self-critique rating: 2

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Question: `q007. The skater in the preceding example pulls the 5 kg weights close in

toward her stomach, decreasing the distance of each from the axis of rotation to 10 cm.

What now is her moment of inertia, angular velocity and angular KE?

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Your solution:

Moment of Inertia for weights: mr^2

10 kg * (.1 m)^2

10 kg * .01 m^2

.1 kg m^2

Total moment of inertia for system:

1.2 kg m^2 + .1 kg m^2 = 1.3 kg m^2

Total angular momentum: L = I * `omega

1.3 kg m^2 * 6 rad/s = 7.8 kg m^2/s

KE: .5 I `omega^2

.5 (1.3 kg m^2)(6 rad/s)^2

.5 (1.3 kg m^2)(36 rad^2/s^2)

23.4 Joules

confidence rating #$&*:

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Given Solution:

Her angular momentum must be conserved, so L = angular momentum remains at 28.8 kg m^2 /

s.

The moment of inertia for each of the two 5 kg masses is now only m r^2 = 5 kg * (.1 m)

^2 = .05 kg m^2 and her total moment of inertia is thus now 1.2 kg m^2 + 2 (.05 kg m^2)

= 1.3 kg m^2.

If we let I ' and `omega ' stand for the new moment of inertia and angular velocity, we

have L = I ' * `omega ', so `omega ' = L / I ' = 28.8 kg m^2 / s / ( 1.3 kg m^2) = 22

rad/s, approx..

Moment of inertia decreased from 4.8 kg m^2 to 1.3 kg m^2 so the angular velocity

increased by the same proportion from 6 rad/s to about 22 rad/s.

Her new kinetic energy is therefore KE ' = 1/2 I ' * ( `omega ' )^2 = 1/2 * 1.3 kg m^2 *

(22 rad/s)^2 = 315 Joules, approx..

[ Note that to increase KE a net force was required. This force was exerted by the

skater's arms as she pulled the weights inward against the centrifugal forces that tend

to pull the weights outward. ]

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Self-critique (if necessary):

I don't see why her angular momentum has to be conserved when the radius is changed for

the weights, which changes the total inertia, which was how we arrived at that angular

momentum to begin with. I understand the steps as they are given. I misread the

question about finding the new angular velocity rather than the new angular momentum.

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Self-critique rating: 2

@&

If there's no torque acting on her from outside, her angular momentum won't change. There's nothing to increase or decrease how hard it would be to stop her.

If her moment of inertia decreases, then, in order to still be as hard to stop, her rotation has to speed up.

*@

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Question: `q008. When a torque `tau acts through an angular displacement `d`theta, it

does work. Suppose that a net torque of 3 m N acts for 10 seconds on a disk, initially

at rest, whose moment of inertia is .05 kg m^2. What angular velocity will the disk

attain, through how many radians will it rotate during the 10 seconds, and what will be

its kinetic energy at the end of the 10 seconds?

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Your solution:

angular acceleration: Torque/Inertia

3 m N / .05 kg m^2 = 60 radians/s^2

60 radians/s^2 * 10 s = 600 radians/s - Change in velocity and final velocity because

initial velocity was 0 radians/s

(0 radians/s + 600 radians/s0/2 = 300 radians / second - average velocity

Radians in 10 seconds: 300 radians/s * 10 s = 3000 radians

KE = .5 I `omega^2

KE = .5 (.05 kg m^2)(300 rad/s)^2

KE = .5 (.05 kg m^2)(20,000 rad^2/s^2)

KE = 2250 Joules

confidence rating #$&*:

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Given Solution:

A net torque of 3 m N acting on the disk whose moment of inertia is.05 kg m^2 will

result in angular acceleration `alpha = `tau / I = 3 m N / (.05 kg m^2) = 60 rad/sec^2.

In 10 seconds this angular acceleration will result in a change in angular velocity

`d`omega = 60 rad/s^2 * 10 s = 600 rad/s. Since the torque and moment of inertia are

uniform the acceleration will be uniform and the average angular velocity will

therefore be `omegaAve = (0 + 600 rad/s) / 2 = 300 rad/s.

With this average angular velocity for 10 seconds the disk will rotate through angular

displacement `d`omega = 300 rad/s * 10 sec = 3000 rad.

Its kinetic energy at its final 600 rad/s angular velocity will be KE = 1/2 I `omega^2 =

1/2 * .05 kg m^2 * (600 rad/s)^2 = 9000 Joules.

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Self-critique (if necessary):

I used the average velocity instead of the final velocity for the kinetic energy.

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Self-critique rating: 2

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Question: `q010. Show that this 9000 Joule energy is equal to the product of the

torque and the angular displacement.

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Your solution: 3000 radians (angular displacement for the 10 seconds) * 3 m N (Torque) =

9000 Joules

confidence rating #$&*:

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Given Solution:

The angular displacement is 3000 rad and the torque is 3 m N. Their product is 9000 N m

= 9000 Joules. Note that the m N of torque is now expressed as the N m = Joules of

work. This is because a radian multiplied by a meter of radius gives a meter of

displacement, and work is equal to the product of Newtons and meters of displacement.

STUDENT COMMENT: good to see connection between the Nm ofa and multiplication of the

radians, always wwondered where they went or what was happening

INSTRUCTOR COMMENT: Right. Basically, any time a meter of radius is multiplied by a

radian of angle you get meters of arc. Any time you divide meters of arc by meters of

radius you get a radian. Sometimes it's not easy to see exactly where the radius and arc

come into a complicated calculation, so it's always worth thinking about.

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Self-critique (if necessary): OK

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Self-critique rating: 2

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Question: `q011. How does the previous example illustrate the fact that the work done

by a net torque is equal to the product of the torque and the angular displacement?

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Your solution:

The answer for kinetic energy .5 I `omega^2

was equal to

Torque * angular displacement.

(Linear work is Force times displacement, so this work was force/torque times

displacement/angular displacement.)

confidence rating #$&*:

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Given Solution:

From the net torque, moment of inertia and time interval we found that the KE increased

from 0 to 9000 Joules. We know that the KE increase of a system is equal to the net

work done on the system, so 9000 Joules of net work must have been done on the system.

Multiplying the angular displacement by the torque gave us 9000 Joules, equal to the KE

increase, so at least in this case the work done was the product of the angular

displacement and the net torque. It isn't difficult to prove that this is always the

case for any system, and that in general the work `dW done by a net torque `tauNet

acting through an angular displacement `d`theta is `dW = `tauNet * `d`theta.

UNIVERSITY STUDENT COMMENT (relevant only to students who know calculus): Speaking in

terms of calculus...

'dW=int('tau with respect to 'theta) from 'theta_1 to 'theta_2 = ('tau*'theta_2)-

('tau*'theta_1)='tau*(theta_2-'theta_1)='tau*'d'theta

Amazing!

INSTRUCTOR RESPONSE: Very good. That will of course work if tau is known as a function

of angular position theta (e.g., consider a cam accelerated by a falling mass, in the

same manner as the disk with bolts except that the rim of the cam is not at constant

distance from the axis of rotation). The shape of the cam may be described in terms of

polar coordinates, where the coordinate r is given in terms of the angle theta from the

polar axis of the cam.

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Self-critique (if necessary): OK

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Self-critique rating: 2

If you understand the assignment and were able to solve the previously given problems

from your worksheets, you should be able to complete most of the following problems

quickly and easily. If you experience difficulty with some of these problems, you will

be given notes and we will work to resolve difficulties.

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Question: `q012. To a uniform rod of length 30 cm and mass 80 grams, initially at rest

and constrained to rotate about an axis through its center, we add a 20-gram domino at

one end and a 50-gram magnet on the other side of the axis, at a distance of 6 cm from

the axis.

If the rod is subject to a torque of .03 m N for 5 seconds, what will be its angular

velocity?

What will be its kinetic energy?

Through what angular displacement will the object rotate?

How much work will therefore be done by the torque?

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Your Solution:

Moment of Inertia for Rod: 1/12 m L^2

1/12 (.08 kg)(.3 m)^2

1/12 (.08 kg)(.09 m^2)

.0006 kg m^2

Moment of Inertia for Magnet: mr^2

.05 kg (0.6 m)^2

.05 kg (.0036 m^2)

.00018 kg m^2

Moment of Inertia for Domino: mr^2

.02 kg (.15 m)^2

.02 kg (.0225 m^2)

.00045 kg m^2

Total Moment of Inertia:

.0006 kg m^2 + .00018 kg m^2 + .00045 kg m^2 = .00123 kg m^2

Angular Acceleration: Torque/Inertia

.03 m N / .00123 kg m^2

24.39 radians/s^2

Change in velocity:

24.39 radians/s^2 * 5 seconds = 121.95 radians/s

This is also the final velocity because initial velocity was ""at rest.""

Average velocity:

(0 rad/s + 121.95 rad/s)/2 = 60.975 radians/s

Displacement:

60.975 radians/s * 5 s = 304.875 radians

Work = Torque * angular displacement:

.03 m N * 304.875 radians

9.14625 N m, or Joules

KE = .5 I `omega^2

KE = .5 (.00123 kg m^2)(121.95 rad/s)^2

KE = .5 (.00123 kg m^2)(14871.8025 rad^2/s^2)

KE = 9.14615 Joules

@&

Excellent solution.

*@

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating:

&#This looks good. See my notes. Let me know if you have any questions. &#