AP Physics C: Mechanics 6 Energy and Momentum of Rotating Systems Questions

Connect rotational energy, torque work, angular momentum, rolling constraints, and gravitational orbit models throughout AP Physics C Mechanics Unit Six.

Exam points

calculate rotational kinetic energy and use energy conservation to determine angular speed
relate torque through an angular displacement to work and a change in rotational kinetic energy
calculate angular momentum for rigid bodies and particles and relate angular impulse to its change
define a system and test external torque before applying conservation of angular momentum
analyze rotating collisions and mass redistribution while separating energy loss from momentum conservation
combine translational and rotational energy with v_cm = R omega for rolling without slipping
use static or kinetic friction to determine rolling acceleration and transitions between slipping and no-slip
relate gravitational potential, kinetic and total energy to the radius of a circular orbit
apply work, energy and angular momentum conservation to orbit transfers, elliptical motion and escape

Covered syllabus points

6.1 Rotational Kinetic Energy 6.2 Torque and Work 6.3 Angular Momentum and Angular Impulse 6.4 Conservation of Angular Momentum 6.5 Rolling 6.6 Motion of Orbiting Satellites

[Maximum number: 4]

Object A is a long, thin, uniform rod of mass M and length 2 L that is free to rotate about a pivot of negligible friction at its left end, as shown above.

Derive an expression for the angular speed of object B when it is in the position shown above. Express your answer in terms of $M, L, I_{\mathrm{B}}$ , and physical constants, as appropriate.

Note: Figure not drawn to scale.

For using conservation of energy
$U_{g 1}=K_{2}$

1 point

For correctly relating the change in rotational kinetic energy to the change in gravitational potential energy
$m g h=\frac{1}{2} I \omega^{2}$

1 point

For correctly substituting for h into the equation above
$m g\left(\frac{L}{2}\right)=\frac{1}{2} I \omega^{2}$

1 point

For an expression for $\omega$ that uses only the allowed symbols and is algebraically consistent with the previous steps
$\begin{aligned} \omega=\sqrt{\frac{2 m g h}{I}}=\sqrt{\frac{2 M g(L / 2)}{I_{\mathrm{B}}}} \omega=\sqrt{\frac{M g L}{I_{\mathrm{B}}}} \end{aligned}$

1 point

Total for part (f)

for question 215 points

[Maximum number: 5]

Mech.3.
A uniform rod of length d has one end fixed to the central axis of a horizontal, frictionless circular platform of radius R=2 d. Fixed at the other end of the rod is an ideal spring of negligible mass to which a block is attached. The block is set in frictionless grooves so that it can only move along a radius of the platform, as shown in Figure 1 above. The equilibrium length of the spring is d / 2. Below is a table showing the mass of the block and the masses and rotational inertias of the rod and platform.

Table for Question 2 — AP Physics C: Mechanics

A motor begins to slowly rotate the platform counterclockwise as viewed from above until the platform reaches a constant angular speed $\omega$ . Under these conditions, the spring has stretched by an additional length d / 2, as shown in Figure 2.

Answer the following questions for the platform rotating at constant angular speed $\omega$ . Express all algebraic answers in terms of $m, d, \omega$ , and physical constants, as appropriate.

(a)

Determine an expression for the angular momentum of the entire system about the axis of the platform.

Figure 3

While the system continues to rotate, a small mechanism in the pivot moves the rod slowly until the center of the rod is positioned on the axis, as shown in Figure 3 above. The same constant angular speed $\omega$ is maintained by the motor driving the platform.

[ 1 ]

1 point
Using a correct expression of angular momentum

L=I \omega

For an answer consistent with the answer from part (b) ii.

L=15 m d^{2} \omega

(b)

Is the angular momentum of the entire system increasing, decreasing, or staying the same? Increasing Decreasing Staying the same
Justify your answer.

[ 2 ]

(c)

In order to keep the system rotating with constant angular speed $\omega$ , is the motor doing positive work, negative work, or no work on the rotating system? Positive Negative No work
Justify your answer.

[ 2 ]

[Maximum number: 5]

A system consists of a small sphere of mass m and radius R at rest on a horizontal surface and a uniform rod of mass M=2 m and length $\ell$ attached at one end to a pivot with negligible friction, where $R \ll \ell$ . There is negligible friction between the surface and the sphere to the right of Point A and nonnegligible friction to the left of Point A. The rod is held horizontally as shown in Figure 1, then is released from rest. The total rotational inertia of the rod about the pivot is $\frac{1}{3} M \ell^{2}$ and the rotational inertia of the sphere about its center is $\frac{2}{5} m R^{2}$ . After the rod is released, the rod swings down and strikes the sphere head-on. As a result of this collision, the rod is stopped, and the ball initially slides without rotating to the left across the horizontal surface.

(a)

Derive an expression for the linear speed $v_{0}$ of the sphere immediately after colliding with the rod in terms of the length $\ell$ and physical constants as appropriate.

After sliding a short distance, at time t=0 the sphere encounters a region of the horizontal surface with a coefficient of kinetic friction $\mu$ , beginning at Point A as indicated in Figure 1. The sphere begins rotating while sliding and eventually begins rolling without sliding at Point B, also as indicated.

[ 2 ]

For using a correct expression for conservation of angular momentum 1 point
Example Response

\begin{aligned} & L_{\mathrm{i}}=L_{\mathrm{f}} \\ & I \omega=m v r \end{aligned}

For correctly substituting the expression for $\omega_{f}$ from part (a) 1 point
Example Response

\frac{2}{3} m \ell^{2} \sqrt{\frac{3 g}{\ell}}=m v_{0} \ell

Example Solution

\begin{aligned} & L_{\mathrm{i}}=L_{\mathrm{f}} \\ & I \omega=m v r \\ & \frac{2}{3} m \ell^{2} \sqrt{\frac{3 g}{\ell}}=m v_{0} \ell \\ & \therefore v_{0}=\sqrt{\frac{4}{3} g \ell} \end{aligned}

Scoring Note: The last equation is not needed for scoring the item but is presented for clarity.

(b)

Derive an expression for the time it takes the sphere to travel from Point A to Point B in terms of $v_{0}, \mu$ , and physical constants as appropriate.

[ 2 ]

(i) For indicating the linear speed is equal to $R \omega$ when slipping stops at Point B 1 point
Example Response

v=R \omega

For correctly substituting v and $\omega$ from parts (d)(i) and (d)(ii) 1 point
Scoring Note: Substituting the acceleration from part (d)(i) into a valid kinematic equation that includes time can earn this point.
Example Response

v_{0}-\mu g t=R \frac{5 \mu g}{2 R} t

Example Solution

\begin{aligned} & v=R \omega \\ & v_{0}-\mu g t=R \frac{5 \mu g}{2 R} t \\ & \therefore t=\frac{2 v_{0}}{7 \mu g} \end{aligned}

(c)

Derive an expression for the linear velocity of the sphere upon reaching Point B in terms of $v_{0}$ .

[ 1 ]

(ii) For correctly substituting the expression for time from (e)(i) into the expression for velocity in (d)(i)

1 point

Example Solution

\begin{aligned} & v=v_{0}-\mu g t \\ & v=v_{0}-\mu g \frac{2 v_{0}}{7 \mu g} \\ & \therefore v=\frac{5}{7} v_{0} \end{aligned}

Scoring Note: The last equation is not needed for scoring the item but is presented for clarity.

Total for part (e)

for question 3

15 points

[Maximum number: 11]

A uniform solid cylinder of mass $M=0.50 \mathrm{~kg}$ and radius $R=0.10 \mathrm{~m}$ is released from rest, rolls without slipping down a 1.0 m long inclined plane, and is launched horizontally from a horizontal table of height 0.75 m. The inclined plane makes an angle of 30° with the horizontal. The cylinder lands on the floor a distance D away from the edge of the table, as shown in the figure above. There is a smooth transition from the inclined plane to the horizontal table, and the motion occurs with no frictional energy losses. The rotational inertia of a cylinder around its center is $M R^{2} / 2$ .

(a)

Calculate the total kinetic energy of the cylinder as it reaches the horizontal table.

[ 2 ]

2 points
For correctly applying conservation of energy to the cylinder rolling down the incline

\begin{aligned} & U_{g_{-} \text {top }}=K_{\text {table }} \\ & K_{\text {table }}=m g h=(0.50 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m})(\sin 30) \end{aligned}

For a correct answer with units

K_{\text {table }}=2.45 \mathrm{~J}\left(\text { or } 2.5 \mathrm{~J} \text { using } g=10 \mathrm{~m} / \mathrm{s}^{2}\right)

(b)

Calculate the angular velocity of the cylinder around its axis at the moment it reaches the floor.

[ 3 ]

3 points
For correctly setting the kinetic energy of the cylinder equal to the sum of both the linear and rotational kinetic energy

K=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}

For correctly substituting into the above equation for the linear velocity and moment of inertia of the cylinder

\begin{aligned} & K=\frac{1}{2} M R^{2} \omega^{2}+\frac{1}{2}\left(\frac{1}{2} M R^{2}\right) \omega^{2}=\frac{1}{2} M(R \omega)^{2}+\frac{1}{4} M(R \omega)^{2}=\frac{3}{4} M(R \omega)^{2} \\ & \omega=\sqrt{\frac{4 K}{3 M R^{2}}}=\sqrt{\frac{(4)(2.45 \mathrm{~J})}{(3)(0.50 \mathrm{~kg})(0.10 \mathrm{~m})^{2}}} \end{aligned}

For correct substitution into the equation above

\omega=25.6 \mathrm{rad} / \mathrm{s}\left(\text { or } 26.0 \mathrm{rad} / \mathrm{s} \text { using } g=10 \mathrm{~m} / \mathrm{s}^{2}\right)

(c)

Calculate the ratio of the rotational kinetic energy to the total kinetic energy for the cylinder at the moment it reaches the floor.

[ 2 ]

2 points
For using a correct expression for the ratio of the rotational kinetic energy to the total kinetic energy of the cylinder

\frac{K_{r o t}}{K_{t o t}}=\frac{\frac{1}{2} I \omega^{2}}{\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}+m g h}=\frac{\frac{1}{4} M(R \omega)^{2}}{\frac{3}{4} M(R \omega)^{2}+M g h_{\mathrm{table}}}=\frac{(R \omega)^{2}}{3(R \omega)^{2}+4 g h_{\mathrm{table}}}

For substituting into the above equation

\begin{aligned} \frac{K_{r o t}}{K_{t o t}} & =\frac{[(0.10 \mathrm{~m})(25.6 \mathrm{rad} / \mathrm{s})]^{2}}{(3)(0.10 \mathrm{~m})^{2}(25.6 \mathrm{rad} / \mathrm{s})^{2}+(4)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.75 \mathrm{~m})} \\ \frac{K_{r o t}}{K_{t o t}} & =0.133\left(\text { or } 0.135 \text { using } g=10 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned}

point
1 point

Question 3

Alternate Solution
For using a correct expression for the ratio of the rotational kinetic energy to the total potential energy of the cylinder

\frac{K_{r o t}}{K_{t o t}}=\frac{\frac{1}{2} I \omega^{2}}{U_{t o t a l}}=\frac{\frac{1}{4} M(R \omega)^{2}}{M g(h+y)}

For substituting into the above equation

\begin{aligned} \frac{K_{\text {rot }}}{U_{\text {total }}} & =\frac{\frac{1}{4}(R \omega)^{2}}{g(h+y)}=\frac{\frac{1}{4}[(0.10 \mathrm{~m})(25.6 \mathrm{rad} / \mathrm{s})]^{2}}{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left((1.0 \mathrm{~m}) \sin \left(30^{\circ}\right)+(0.75 \mathrm{~m})\right)} \\ \frac{K_{\text {rot }}}{U_{\text {total }}} & =0.13\left(\text { or } 0.135 \text { using } g=10 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned}

(d)

Is the rotational kinetic energy of the sphere at the moment it reaches the floor greater than, less than, or equal to the rotational kinetic energy of the cylinder at the moment it reaches the floor? Greater than Less than Equal to
Justify your answer.

[ 2 ]

2 points
For selecting "Less than" and attempting a relevant justification For a correct justification

Example: Because the rotational inertia of the sphere is less than the rotational inertia of the cylinder, the sphere will rotate faster and, because $v=r \omega$ , will move with a greater linear speed. Because the mass is the same and the linear speed is greater, the sphere will have a greater linear kinetic energy. Because the total kinetic energies of the sphere and cylinder are the same, the sphere must have less rotational kinetic energy.

(e)

Is the horizontal distance the sphere travels from the table to where it hits the floor greater than, less than, or equal to the horizontal distance the cylinder travels from the table to where it hits the floor? Greater than Less than Equal to
Justify your answer.

[ 2 ]

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Covered syllabus points

Question 1

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question (a)

Question (b)

Question (c)

Question 4

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)