Query 32

#$&*

course PHY 121

7/30 2

032. `query 32

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Question: `qQuery experiment to be viewed. What part or parts of the system experiences

a potential energy decrease? What part or parts of the system experience(s) a kinetic

energy increase?

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Your solution:

At the beginning of the experiment, there is potential energy in the wheel as well as

the string/paper clips. As the paper clips decend, pulling the string, and turning the

wheel, the potential energy decreases as the kinetic energy of the system increases.

confidence rating #$&*:

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Given Solution:

`a** The mass on the string descends and loses PE.

The wheel and the descending mass both increase in KE, as do the other less massive

parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which

rotates at the same rate as the wheel but which due to its much smaller radius does not

move nearly as fast as most of the wheel). **

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Self-critique (if necessary): OK

There's no potential energy in the wheel?

@&

The wheel is slightly unbalanced, so if we move the unloaded wheel it to the right position and release it, it will probably rotate to a new position as its KE increases. However it's pretty well balanced, so any PE loss in such a process would be pretty much negligible.

If we move a balanced wheel from one angular position to another, part of the wheel goes up and part goes down. Gravity does positive work on the parts that go down and negative work on the parts that go up. The net work done by gravity is zero. So there is no change in gravitational PE as a well-balanced wheel moves from one position to another.

*@

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Self-critique Rating: 2

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Question: `qWhat part or parts of the system experience(s) an increase in rotational

kinetic energy?

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Your solution:

The wheel on the axel with its bolts increase in rotational kinetic energy.

confidence rating #$&*:

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Given Solution:

`a** The wheel, the bolts, the axle, and anything else that's rotating about an axis

experiences an increase in rotational KE. **

STUDENT QUESTION

I’m not quite sure what translation kinetic energy is, but it seems like it means that

kinetic energy is somehow moved from place to place??? I think maybe the string moves

the energy (if you can call it that) from the wheel to the paper clip.

INSTRUCTOR RESPONSE

Translational KE is 1/2 m v^2, where v is the velocity of the object itself as it moves

from one point to another (more technically, v is the velocity of the object's center of

mass).

This is contrasted with rotational KE, in which an object can stay in one place and

rotate about some axis.

An object can also move from point to point while rotating, so it can have both

rotational and translational KE.

For example the ball rolling down the grooved track moves from point to point, but it

also rotates as it moves.

In the current example, the wheel, axel and the bolts embedded in it rotate about the

axle, but the wheel doesn't go anywhere. So there is no translational KE in the wheel,

just rotational.

The only thing with translational KE is the descending mass.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `qWhat part or parts of the system experience(s) an increasing translational

kinetic energy?

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Your solution:

Translational kinetic energy is taking place with the string and the paper clips.

confidence rating #$&*:

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Given Solution:

`a** Only the descending mass experiences an increase in translational KE. **

STUDENT COMMENT: i must have overlooked the definition of translational KE, i didn't

know what it was

INSTRUCTOR RESPONSE: translational motion is motion from one point to another; in

rotational motion about an axis each point keeps following the same circle, repeating

the same points and otherwise never going anywhere, relative to that axis.

The axis of course could be moving from one place to another--i.e., undergoing

translational motion--so an object can undergo rotational motion as well as an

independent translational motion.

A simple example is a ball rolling down an incline. It moves from one end of the

incline to the other, but as it rolls it also rotates about an axis through its center.

The axis is horizontal and perpendicular to the incline. Since the center of the ball

is moving, the axis of rotation is also moving. The axis is translating from one point

to another as the ball rotates about it.

If you drive straight down the road, the back wheels of your car rotate about an axis

which runs straight through the center of your car's real axel. The axel has

translational motion (i.e., it moves down the road) and the wheels have the same

translational motion. Both the wheels and axel also rotate about a common central axis.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `qDoes any of the bolts attached to the Styrofoam wheel gain more kinetic

energy than some other bolt? Explain.

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Your solution:

I think the outside of the wheel has a greater velocity than the inside of the wheel

because of a larger circumference. This would mean that the bolts toward the outside

have more kinetic energy than the bolts toward the middle of the wheel.

confidence rating #$&*:

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Given Solution:

`a** The bolts toward the outside of the wheel are moving at a greater velocity relative

to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 **

STUDENT COMMENT: Oh... I need to think of it in terms of angular velocity

INSTRUCTOR RESPONSE: Think in terms of angular velocity as well as velocity. At any

instant all masses on the wheel have the same angular velocity, but the masses further

from the center have greater velocity (and therefore greater KE) than those closer to

the center.

STUDENT COMMENT: i had the right idea here, but had it backward, i thought the closer to

the fixed point the greater the velocity

INSTRUCTOR RESPONSE: That would be the case for a satellite orbiting a planet. However

in this case the entire wheel is rotating at a single angular velocity, so closer points

don't move as fast as distinct points.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `qWhat is the moment of inertia of the Styrofoam wheel and its bolts?

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Your solution:

The moment of inertia is calculated mass times the radius squared. You have to

calculate the moment of inertia for the wheel as well as for each of the bolts and put

them together for the total moment of inertia.

confidence rating #$&*:

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Given Solution:

`a** The moment of inertia for the center of its mass=its radias times angular velocity.

Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the

center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its

mass and R its radius. The wheel with its bolts has a moment of inertia which is equal

to the sum of all these components. **

STUDENT COMMENT:

I = .5mr^2 for the disk

I = .5mr^2 for each of the bolts

INSTRUCTOR RESPONSE:

The moment of inertia of a particle of mass m at distance r from the axis of rotation is

m r^2. A particle has all its mass concentrated at one specific location.

A hoop consists of a collection of particles, all at the same distance from the axis of

rotation. If we add up the m r^2 contributions from all the particles in the hoop, we

get M R^2, where M is the mass and R the radius of the hoop. Thus the moment of inertia

of the hoop is M R^2.

The disk consists of a collection of particle spread out at many different distances

from the axis. If we 'cut up' the disk into individual particles, we find that the sum

of the m r^2 contributions of the particles is 1/2 M R^2, where M is the mass of the

disk and R its radius. Thus the moment of inertia of the disk is 1/2 M R^2.

The mass of a bolt isn't all concentrated at a single distance from the axis, but all

the particles that make up the bolt are pretty close to the center of the bolt, so it

doesn't differ from a particle by much. Its moment of inertia is pretty close to m r^2,

where m is the mass of the bolt and r its distance from the axis.

You add the moment of inertia of the disk to the moments of inertia of the bolts, and

you end up with the moment of inertia of the system.

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Self-critique (if necessary):

I forgot to make the wheel .5 m r^2

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Self-critique Rating: 3

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Question: `qHow do we determine the angular kinetic energy of of wheel by measuring the

motion of the falling mass?

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Your solution:

You would see how far the mass drops. You would also need to look at how far the wheel

turned. Finally, you would need to time the drop. In this way, you can tell how many radians or degrees per second the wheel moved.

confidence rating #$&*:

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Given Solution:

`a** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration,

so the wheel also turns this fast.

INSTRUCTOR CRITIQUE: Acceleration isn't the rate at which something moves, or turns. It

is the rate at which the velocity (which is itself the rate at which the object moves,

or turns) changes.

We don't use the acceleration to find the angular KE, we use the velocity. The

acceleration, if known, can be used to find the velocity.

The question was how we use measurements of the motion of the descending mass to find

the angular KE:

By observing position vs. clock time we can estimate velocities, and determine the

velocity of the descending mass at any point.

The string is wound around the rim of the wheel. So the rim of the wheel moves at the

same speed as the string, which is descending at the same speed as the mass. So if our

measurements give us the speed of the descending mass, we know the speed of the wheel.

If we divide the velocity of the rim of the wheel by its radius we get the angular

velocity of the wheel. Recall that angular velocity is designated by the symbol omega.

Assuming we know the moment of inertia of the wheel, we find its KE, which is equal to

1/2 I omega^2. **

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Self-critique (if necessary):

I'm still getting this stuff mixed up.

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Self-critique Rating: 2

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Question:Principles of Physics and General College Physics problem 8.43: Energy to bring

centrifuge motor with moment of inertia 3.75 * 10^-2 kg m^2 to 8250 rpm from rest.

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Your solution:

KE = .5 I `omega

`omega is the angular velocity. The initial velocity was 0 rpm and the final velocity was 8250 rpm, so the average velocity would be 4125 rpm. To put that into seconds, you would need to divide by 60 seconds. This gives 68.75 revolutions per second. 2 `pi radians is one revolution, so 68.75 revolutions is 137.5 `pi radians per second could be your `omega.

Therefore you have:

KE = .5 (.0375 kg m^2)(137.5 `pi radians)

And now I'm stuck......

confidence rating #$&*:

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Given Solution: The KE of a rotating object is

KE = .5 I omega^2,

where I is the moment of inertia and omega the angular velocity.

Since I is given in standard units of kg m^2, the angular velocity should be expressed

in the standard units rad / sec. Since 8250 rpm = (8250 rpm) * (pi / 30 rad/sec) / rpm =

860 rad/sec, approx..

The initial KE is 0, and from the given information the final KE is

KE_f = .5 I omega_f ^ 2 = .5 * 3.75 * 10^-2 kg m^2 * (860 rad/sec)^2 = 250 pi^2 kg m^2 /

sec^2 = 14000 Joules.

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Self-critique (if necessary):

I'm not sure where the `pi/30 rad/sec came from or why the radians and pi's turned into Joules.

@&

One RPM is one revolution per minute, which is one revolution per 60 seconds. Thus

RPM = 1 rev / (60 sec).

1 revolution corresponds to 2 pi radians, so we get

RPM = (2 pi rad / (60 sec) ) = pi rad / (30 sec), or pi/30 radians / sec.

*@

@&

I'm at a loss to explain the 250 pi^2 or how it came from the preceding step, but .5 * 3.75 * 10^-2 kg m^2 * (860 rad/sec)^2 is about 14 000 Joules.

The units come out kg m^2 * rad^2 / sec^2. A meter of radius multiplied by a radian gives you a meter of distance along the arc, so m * rad is just m, and m^2 * rad^2 is just m^2.

So our units are kg m^2 / s^2, or Joules.

*@

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Self-critique Rating: 2

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Question: `qQuery gen problem 8.58 Estimate KE of Earth around Sun (6*10^24 kg, 6400 km

rad, 1.5 * 10^8 km orb rad) and about its axis.

What is the angular kinetic energy of the Erath due to its rotation about the Sun?

What is the angular kinetic energy of the Earth due to its rotation about its axis?

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Your solution:

confidence rating #$&*:

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Given Solution:

`a** The circumference of the orbit is 2pi*r = 9.42*10^8 km.

We divide the circumference by the time required to move through that distance to get

the speed of Earth in its orbit about the Sun:

9.42 * 10^8 km / (365days * 24 hrs / day * 3600 s / hr) =29.87 km/s or 29870 m/s.

Dividing the speed by the radius we obtain the angular velocity:

omega = (29.87 km/s)/ (1.5*10^8 km) = 1.99*10^-7 rad/s.

From this we get the angular KE:

KE = 1/2 mv^2 = 1/2 * 6*10^24 kg * (29870 m/s)^2 = 2.676*10^33 J.

Alternatively, and more elegantly, we can directly find the angular velocity, dividing

the 2 pi radian angular displacement of a complete orbit by the time required for the

orbit. We get

omega = 2 pi rad / (365days * 24 hrs / day * 3600 s / hr) = 1.99 * 10^-7 rad/s.

The moment of inertia of Earth in its orbit is M R^2 = 6 * 10^24 kg * (1.5 * 10^11 m)^2

= 1.35 * 10^47 kg m^2.

The angular KE of the orbit is therefore

KE = .5 * I * omega^2 = .5 * (1.35 * 10^47 kg m^2) * (1.99 * 10^-7 rad/s)^2 = 2.7 *

10^33 J.

The two solutions agree, up to roundoff errors.

The angular KE of earth about its axis is found from its angular velocity about its axis

and its moment of inertia about its axis.

The moment of inertia of the Earth as it spins on its axis is

I=2/5 M r^2= 2/5 * 6*10^24kg * ( 6.4 * 10^6 m)^2 = 9.83*10^37 kg m^2.

The angular velocity of the Earth about its axis is 1 revolution / 24 hr = 2 pi rad /

(24 hr * 3600 s / hr) = 7.2 * 10^-5 rad/s, very approximately.

So the angular KE of Earth about its axis is about

KE = .5 I omega^2 = .5 * 9.8 * 10^37 kg m^2 * (7.2 * 10^-5 rad/s)^2 = 2.5 * 10^29

Joules. **

INTERESTING STUDENT SOLUTION FOR ORBITAL VELOCITY AND INSTRUCTOR COMMENTARY

the moment of inertia of the earth in its orbit around the Sun is 1.35*10^47 kg m^2

the velocity is sqrt(6.67*10^-11*6*10^24/(6400000+1.5*10^10m)) = 163m/s

(instructor note: this method is good but the velocity is incorrect due to an

instructive oversite, as explained below)

INSTRUCTOR COMMENT: This is a very good way to find the result; however if the Earth is

considered to be a satellite around the Sun then its orbital velocity would be sqrt( G M

/ r), but M would be the mass of the Sun, not the Earth.

Alternatively you could divide the circumference of the Earth's orbit about the Sun by

the number of seconds in a year, as was done in the given solution.

The two methods should give you pretty much identical results.

Note that if you find the velocity of the Earth's orbit by the second method, you can

then use the result to find the mass of the Sun.

Also of interest: Once we obtained accurate results for G in physics laboratories, we

could then use the distance of the Moon (which was known long beforehand) along with the

orbital period of the Moon (known since we invented clocks) to determine the mass of the

Earth.

STUDENT QUESTION:

Why did you use the regular velocity instead of the angular velocity to find KE?

INSTRUCTOR RESPONSE:

If we use angular velocity to find KE, we use it with the moment of inertia to find KE =

1/2 I omega^2.

If we prefer to use the mass of the object, you use it with the velocity to find KE.=

1/2 m v^2

It could have been done either way. In the given solution I chose to use velocity and

mass to find the Earth's KE. Has we used the moment of inertia I = m r^2, with m the

mass and r the radius of the Earth, then we could have used it with the angular velocity

to calculate KE = 1/2 I omega^2, and would have obtained the same result.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qQuery problem 8.60 uniform disk at 2.4 rev/sec; nonrotating rod of equal

mass, length equal diameter, dropped concentric with disk. Resulting angular velocity?

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Your solution:

confidence rating #$&*:

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Given Solution:

`a** The moment of inertia of the disk is I = 1/2 M R^2; the moment of inertia of the

rod about its center is 1/12 M L^2. The axis of rotation of each is the center of the

disk so L = R. The masses are equal, so we find that the moments of inertia can be

expressed as 1/2 M R^2 and 1/12 M R^2.

The combined moment of inertia is therefore 1/2 M R^2 + 1/12 M R^2 = 7/12 M R^2, and the

ratio of the combined moment of inertia to the moment of the disk is

ratio = (7/12 M R^2) / (1/2 M R^2) = 7/12 / (1/2) =7/12 * 2 = 7/6.

Since angular momentum I * omega is conserved an increase in moment of inertia I results

in a proportional decrease in angular velocity omega so we end up with

final angular velocity = 6/7 * initial angular velocity = 6/7 * 2.4 rev / sec = 2.1

rev/sec, approximately.

STUDENT COMMENT:

I had no idea to do the ratio. I probably wouldn’t have ever thought of that either.

INSTRUCTOR RESPONSE:

You don't need to use the idea of ratio, it's simply convenient to do so.

You could equivalently obtain the expression for the angular momentum in terms of

initial angular momentum omega_0:

initial angular momentum of disk = I_disk * omega_0 = 1/2 M R^2 * omega_0

If omega_f is the angular momentum of the system after the rod is dropped then we have

final angular momentum of system = I_system * omega_f = 7/12 M R^2 * omega_f.

No external torque acts on the system so its angular momentum remains constant. Thus

initial angular momentum of disk = final angular momentum of system

1/2 M R^2 * omega_0 = 6/12 M R^2 * omega_f

We solve this to get

omega_f = (1/2 M R^2) / (7/6 M R^2) * omega_0 = 6/7 * 2.4 rev/s = 2.1 rev / s, approx..

CORRECTION BY INSTRUCTOR

The given solution incorrectly solve the problem for a sphere and a rod, not a disk and

a rod. It used the moment of inertia 2/5 M R^2 for the sphere where it should have use

1/2 M R^2, the moment of inertia of a disk.

The above solution can be easily revised to correct for this error.

STUDENT QUESTION

I am not sure how we know to use the values 1/2 or 1/12 in this situation???

I do see how we add the two values together though. I am still having a little trouble

understanding this. Is there another example I can look at.???

INSTRUCTOR RESPONSE

A uniform disk rotating about an axis through its center and perpendicular to its plane

has moment of inertia 1/2 M R^2.

A uniform sphere rotating about an axis through its center has moment of inertia 2/5 M

R^2.

A rod rotating about its center has moment of inertia 1/12 M L^2; rotating about its end

the moment of inertia is four times as great, 1/3 M L^2.

All students should know these formulas. Physics 231 students are expected to be able to

derive these formulas, and other, using calculus.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qUniv. 10.64 (10.56 10th edition). disks 2.5 cm and .8 kg, 5.0 cm and 1.6 kg,

welded, common central axis. String around smaller, 1.5 kg block suspended. Accel of

block? Then same bu wrapped around larger.

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Your solution:

confidence rating #$&*:

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Given Solution:

`a** The moment of inertia of each disk is .5 M R^2; the block lies at perpendicular

distance from the axis which is equal to the radius of the disk to which it is attached.

So the moment of inertia of the system, with block suspended from the smaller disk, is

I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + 1.5 kg * (.025 m)^2= .0032 kg

m^2 approx.

The 1.5 kg block suspended from the first disk results in torque

tau = F * x = .025 m * 1.5 kg * 9.8 m/s^2 = .37 m N approx.

The resulting angular acceleration is

alpha = tau / I = .37 m N / (.0032 kg m^2) = 115 rad/s^2 approx.

The acceleration of the block is the same as the acceleration of a point on the rim of

the wheel, which is

a = alpha * r = 115 rad/s^2 * .025 m = 2.9 m/s^2 approx.

The moment of inertia of the system, with block suspended from the larger disk, is

I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .05 m)^2= .006 kg m^2

approx.

The 1.5 kg block suspended from the first disk results in torque

tau = F * x = .05 m * 1.5 kg * 9.8 m/s^2 = .74 m N approx.

The resulting angular acceleration is

alpha = tau / I = .74 m N / (.006 kg m^2) = 120 rad/s^2 approx.

The acceleration of the block is the same as the acceleration of a point on the rim of

the wheel, which is

a = alpha * r = 120 rad/s^2 * .05 m = 6 m/s^2 approx. **

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Self-critique (if necessary):

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#*&!

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