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Unit 5 Rotational Motion Lab

One wheel, measured six ways.

AP Physics C

Unit 5, Torque and Rotational Dynamics

5.1Rotational Kinematics 5.2Connecting Linear and Rotational Motion 5.3Torque 5.4Rotational Inertia 5.5Rotational Equilibrium and Newton’s First Law 5.6Newton’s Second Law in Rotational Form LabUnit 5 Rotational Motion Lab

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What you should be able to do

Measure an angle, an angular velocity and an angular acceleration, and relate each one to the linear motion of a point on the system.

Find the torque a string exerts on a wheel, and explain why the tension is not the weight hanging from it.

Measure a rotational inertia from the slope of torque against angular acceleration, and predict how adding mass at the rim changes it.

Decide whether a system is in rotational equilibrium, and find the quantity that puts it there.

Predict an angular acceleration from the net torque and the rotational inertia, then test the prediction.

Every part of this lab uses the same apparatus: a uniform disk of mass \(2.0\) kg and radius \(0.15\) m, turning on a fixed axle, with a string wound around a hub at its center. A mass hangs from the string. The hub radius, the hanging mass, and in one part the masses bolted to the rim, are yours to set. Plan on about one hour.

Work the parts in order. Each one hands you the measurement the next one needs.

Part 1. The angle, and the string that comes off the hub

Nothing is released yet. Turn the wheel by hand and watch two numbers: the angle it has turned through, and the length of string that has come off the hub.

The angular velocity is the derivative of the angular position, and the angular acceleration is the derivative of that. The angle you build up here by hand is the function those two derivatives are taken of.

Drag the wheel to turn it. The angle counts in radians and keeps counting past one full turn, and the panel reports the arc that has unwound from the hub.

Set it upSet the hub radius to \(0.080\) m. Turn the wheel through about two full turns, one way, without letting go and going back.

Report what happenedWrite down the angle in radians and the length of string off the hub. Show that those two numbers agree with \(s = r\,\Delta\theta\), and say what the reading in turns has to do with the reading in radians.

Now without the simulationA hub of radius \(0.060\) m turns through \(3.0\) rad. The length of string that comes off is

A. 50 m B. 0.020 m C. 0.18 m D. 180 m

Questions

1 The angle reading passes \(6.28\) rad and keeps going. The reason it is allowed to is that

A. the simulation forgets to reset it after each turn B. an angular position records the whole turning, not where the wheel happens to be pointing C. radians, unlike degrees, have no largest value D. the wheel has turned more than once, so the angle must be doubled

2 While you turn the wheel by hand at a steady rate, \(d\theta/dt\) is constant. During that time the angular acceleration is

A. constant and not zero, because the wheel is being turned B. the angular velocity divided by the time so far C. zero, because the angular acceleration is the derivative of the angular velocity D. undefined, because the wheel is not free to move on its own

Part 2. The falling mass and the turning wheel are one motion

Release the mass. The string stays on the hub, so the wheel and the mass are locked to each other: whatever length of string comes off is exactly how far the mass has fallen.

Set the hub radius and the hanging mass, then press Release. The panel reports the arc off the hub, the distance fallen, the angular velocity of the wheel, and the speed of the mass, all at the same instant. Back to the start returns it.

Set it upHub radius \(0.050\) m, hanging mass \(0.20\) kg. Release it, and read all four of those numbers while it is running.

Report what happenedCompare the arc off the hub with the distance fallen, and compare the speed of the mass with the hub radius times the angular velocity. Then say which of the four readings would change if you moved to a larger hub, and which would not.

Now without the simulationA hub of radius \(0.040\) m is turning at \(15\) rad/s. The string leaving it is moving at

A. 0.60 m/s B. 375 m/s C. 0.0027 m/s D. 3.8 m/s

Questions

3 A point on the rim of the wheel and a point on the hub are both carried around by the same rotation. Compared with each other they have

A. the same angular velocity and different speeds B. different angular velocities and the same speed C. the same angular velocity and the same speed D. different angular velocities and different speeds

4 The hanging mass has \(a = r\alpha\) with \(r\) fixed. Integrating that acceleration once with respect to time, starting from rest, gives

A. \(r\omega\), the speed of the mass B. \(r\theta\), the distance it has fallen C. \(r\alpha\) again, since \(r\) and \(\alpha\) are both constant D. \(\tfrac{1}{2}r\alpha t^2\), its acceleration

Part 3. The torque, and the tension that is not the weight

The string is the only thing touching the wheel that can turn it. Its tension acts at the hub, at right angles to the hub radius, so the torque it exerts is the tension times that radius.

The same release, with the tension and the torque now reported. Watch what the tension does when you change the hanging mass, and what it does when you change the hub radius.

Set it upHanging mass \(0.50\) kg, hub radius \(0.050\) m. Read the tension, then move the hub radius to \(0.100\) m and read it again.

Report what happenedBoth tensions are below \(4.9\) N, which is the weight of that mass. Explain why. Then say what moving to the larger hub did to the tension and what it did to the torque, and why those two went opposite ways.

Now without the simulationThe tension in the string is less than the weight of the hanging mass because

A. some of the weight is used up turning the wheel B. the string is light compared with the mass C. the hub radius is smaller than the radius of the wheel D. the mass is accelerating downward, so the upward force on it must be smaller than the downward one

Questions

5 A force of \(12\) N is applied to a wheel at a distance of \(0.25\) m from the axle, at \(30\) degrees to the line joining the axle to that point. The torque about the axle is

A. \(3.0\) N m B. \(0\) C. \(2.6\) N m D. \(1.5\) N m

6 With the same hanging mass, the tension comes closest to the full weight of that mass when

A. the hub radius is as large as it can be B. the hub radius is as small as it can be C. the rim masses are taken off D. the wheel has been turning for a long time

Part 4. Measuring the rotational inertia from a slope

Each setting of the hanging mass and the hub radius gives one torque and one angular acceleration. Collect several of those pairs and they fall on a line through the origin. The slope of that line is a property of the wheel alone.

For a continuous body that property is the integral \(I = \int r^2\,dm\), which for a uniform disk comes out to \(\tfrac{1}{2}MR^2\). The slope you are about to measure is that integral, arrived at by experiment instead of by calculus.

Change the hanging mass or the hub radius, then press Record this run to put that pair of readings on the graph at the right. The rim masses add mass at \(0.15\) m from the axle. Clear the record empties the graph.

Set it upWith the rim masses at zero, record three runs with three different hanging masses. Then set the rim masses to \(0.25\) kg each and record two more.

Report what happenedGive the rotational inertia reading before and after you added the rim masses, and say what the graph did when you added them. Then show that the slope of the first three points, torque divided by angular acceleration, is the rotational inertia and not something else.

Now without the simulationOn a graph of torque against angular acceleration for one rigid wheel, the slope of the line is

A. the tension in the string B. the rotational inertia of the wheel C. the angular velocity it reaches D. the mass hanging from the string

Questions

7 The wheel is a uniform disk of mass \(2.0\) kg and radius \(0.15\) m, and two masses of \(0.25\) kg each are bolted to its rim. The rotational inertia of the whole system about the axle is

A. \(0.0113\) kg m\(^2\) B. \(0.0225\) kg m\(^2\) C. \(0.0338\) kg m\(^2\) D. \(0.0563\) kg m\(^2\)

8 The two rim masses are moved from the rim to halfway in, without changing their masses. The rotational inertia of the system becomes

A. smaller, because each one contributes its mass times the square of its own distance from the axle B. smaller, because the disk now carries more of the total mass C. unchanged, because the total mass is unchanged D. larger, because they are closer to the axle

Part 5. Holding it still

A second string is wound the other way around the hub, with its own mass and its own radius. Nothing turns while the two torques match, and the panel says so.

The first mass and hub radius are on the left of the controls, the second mass and its radius on the right. The panel reports each torque, the net torque, and a verdict.

Set it upHanging mass \(0.20\) kg at a hub radius of \(0.050\) m. Set the mass on the other side to \(0.40\) kg, then find the radius for it that reads balanced.

Report what happenedGive the radius you found and both torques. Then state, as a rule about torques and not about masses, what has to be true for the wheel to stay still, and say why the heavier mass needed the smaller radius.

Now without the simulationA wheel is in rotational equilibrium. This tells you that

A. the net torque on it is zero, whether or not it is turning B. it is not turning C. no torques act on it D. its rotational inertia is zero

Questions

9 A mass of \(0.60\) kg hangs from a string at a radius of \(0.040\) m on one side of a hub. On the other side, at a radius of \(0.080\) m, the mass that holds the wheel still is

A. \(1.2\) kg B. \(0.60\) kg C. \(0.075\) kg D. \(0.30\) kg

10 A wheel is turning at a steady \(8.0\) rad/s and neither speeding up nor slowing down. The net torque on it is

A. steady and not zero, because something has to keep it turning B. zero, because its angular velocity is not changing C. equal to its rotational inertia times \(8.0\) rad/s D. impossible to decide without knowing the rotational inertia

Part 6. Predict it before you measure it

You now have everything the second law in rotational form needs. The torque comes from the tension at the hub, the rotational inertia comes from Part 4, and the angular acceleration is the one quantity left to find. Work it out on paper first, type it in, and then release.

Type your predicted angular acceleration into the box, press Release, and the panel compares your number with the measured one.

Set it upHub radius \(0.080\) m, hanging mass \(0.40\) kg, no rim masses, so the rotational inertia is \(0.0225\) kg m\(^2\). Predict the angular acceleration on paper. Type it in, then release.

Report what happenedGive your prediction, the measured value, and the two equations you solved together to get there. If you first divided the weight times the radius by the rotational inertia, you got about \(13.9\) rad/s\(^2\). Say why the real answer is below that.

Now without the simulationA net torque of \(0.36\) N m acts on a wheel whose rotational inertia is \(0.045\) kg m\(^2\). Its angular acceleration is

A. \(0.016\) rad/s\(^2\) B. \(0.13\) rad/s\(^2\) C. \(0.125\) rad/s\(^2\) D. \(8.0\) rad/s\(^2\)

Questions

11 Two wheels receive the same net torque. The one that reaches the larger angular acceleration is the one with

A. the smaller rotational inertia B. the larger rotational inertia C. the larger radius, whatever its rotational inertia D. the larger angular velocity to begin with

12 Taking the torque on the wheel to be the weight of the hanging mass times the hub radius gives an angular acceleration that is too large. The step that was skipped is

A. converting the angle to radians B. including the rotational inertia of the hub C. writing the second law for the hanging mass as well, which makes the tension smaller than the weight D. adding the mass of the string to the hanging mass

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