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

Connect rotational kinetic energy, torque work, angular momentum, conservation laws, and rolling motion in rotating systems.

[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 6 Energy and Momentum of Rotating Systems 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

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Covered syllabus points

6 Energy and Momentum of Rotating Systems question 1

6 Energy and Momentum of Rotating Systems question 2

Question (a)

Question (b)

Question (c)

6 Energy and Momentum of Rotating Systems question 3

Question (a)

Question (b)

Question (c)