Asst 32 query

I sent that last night (Monday night). I do know that it took over 12 hours for a message to get to one of my students yesterday (sent at 4:30 in the afternoon, ran into the student an hour later and she didn't have it, ran into her again this morning and she finally got it earlier in the morning). I'm not sure what the problem is.

Let me know if you don't have it by now.

You are doing pretty well with the questions, but be sure to see my notes. I'll be glad to clarify anything you ask.

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10:40:17 Query 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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RESPONSE --> The falling weight has a loss in potential energy and I think the bolts have an increase in kinetic energy.

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10:41:55 ** 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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RESPONSE --> Ok so I was wrong with the bolts having an increase in KE. I had trouble visualizing the setup and that gave me problems with figuring out what gains KE. I don't have internet access to see the video file at the moment.

The video files are on your CDs or DVDs.

In any case, you were right. The KE of the bolts does increase, as you said and as was said in the given solution.

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10:43:02 What part or parts of the system experience(s) an increase in angular kinetic energy?

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RESPONSE --> My guess is the wheel and the axel both increase in angular kinetic energy

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10:43:15 ** The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE. **

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RESPONSE --> ok

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10:45:28 What part or parts of the system experience(s) an increasing translational kinetic energy?

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RESPONSE --> I'm not sure... maybe the string and the weight... other items in the system that are not rotating.

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10:47:37 ** Only the descending mass experiences an increase in translational KE. **

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RESPONSE --> ok so only the mass, nothing else... but the previous answer said that the string gained KE also... so what kind of KE does it have if it isn't translational or rotational?

The string is generally considered to have negligible mass.

However this is not always the case.

The part of the string that's still around the wheel can be considered to have angular KE, while the part that's been unwound and is descending can be considered to have translational KE.

However the entire string is moving at one speed, which is the speed of the descending mass, and you could just use this speed along with the mass of the string to calculate its KE.

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10:50:29 Does any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain.

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RESPONSE --> I'm trying to get online to see the video clip... but if the wheel and the bolts are placed symmetrically, I don't see why one bolt would receive more than the others. If some of the bolts were closer to the mass, I think they'd gain more KE but I really don't know.

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10:51:08 ** 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 **

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RESPONSE --> Oh I had it backwards. That makes sense.

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11:05:39 What is the moment of inertia of the Styrofoam wheel and its bolts?

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RESPONSE --> I = .5MR^2 is the equation for a uniform disk

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11:05:55 ** 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. **

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RESPONSE --> ok

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11:07:28 How do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass?

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RESPONSE --> Angular KE = .5 I * 'omega^2. If we get the angular velocity of the mass and the inertia, then we can use those to get the angular kinetic energy.

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11:08:40 ** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: 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. However in this case what we are really interested in is the final velocity of the falling mass, which is equal to the velocity of the part of the wheel around which it is wound. If we divide the velocity of this part of the wheel by the its radius we get the angular velocity of the wheel. **

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RESPONSE --> The answer stops at how to get the angular velocity of the wheel... so I'm assuming that I had the right formula for how to get the angular KE...

you were correct

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11:22:21 Query 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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RESPONSE --> Moment of inertia for the earth rotating on its axis is : I = .5 MR^2 = .5 * (6*10^24 kg) * (6.4*10^6m)^2 = 1.08*10^38kgm^2 Angular KE for the Earth on its axis is therefore: Ang KE = .5 I*'omega^2 I don't know how to get the angular velocities in order to get the angular KE.

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11:26:04 ** 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 is I=2/5 M r^2=6*10^24kg * ( 6.4 * 10^6 m)^2 = 9.83*10^37kg 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. **

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RESPONSE --> Well I'm not surprised that I didn't know how to do this. I wasn't completely off with my method of finding the inertia (though I had the wrong equation) and then using the angular KE equation, I just didn't know how to get the information I needed for them to work.

Some of this information--length of day, length of year--is common knowledge. Information on the Earth and Sun is in the inside front cover of your book.

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11:33:52 Query Gen Phy 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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RESPONSE --> Angular momentum and angular velocity are different right? The book's going over angular momentum and it's really confusing me b/c I'm not sure if that's what I'm supposed to be finding here or not. All I know is that the moment of inertia is I = .5mr^2 = .5m (2.4rev/s)^2

rev / s is a measure of angular velocity, not of radius. The radius will be in meters.

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11:37:39 ** The moment of inertia of the disk is I = 2/5 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 2/5 M R^2 and 1/12 M R^2. The combined moment of inertia is therefore 2/5 M R^2 + 1/12 M R^2 = 29/60 M R^2, and the ratio of the combined moment of inertia to the moment of the disk is ratio = (29/60 M R^2) / (2/5 M R^2) = 29/60 / (2/5) = 29/60 * 5/2 = 145 / 120 = 29 / 24. 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 = 24 / 29 * initial angular velocity = 24 / 29 * 2.4 rev / sec = 2 rev / sec, approx.. *&*&

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RESPONSE --> I thought the moment of inertia for a uniform disk was I = .5 MR^2.. the uniform sphere is 2/5mr^2... Yeah... fairly lost here. I gotta review it a lot in hopes of understanding it. I get the equations but I just cant figure out where I always need to start.

You are right about the .5 M R^2. I think this text problem was adapted from an earlier version that used a sphere, and I neglected to modify the solution. However other than that, the process given here is correct.

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11:37:43 Univ. 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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RESPONSE --> ok

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11:37:46 ** 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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RESPONSE --> ok"