AP Physics C Mechanics 6.5 Rolling Motion Overview

[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 ]

[Maximum number: 3]

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 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}

(b)

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

Covered syllabus points

6.5 Rolling question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

6.5 Rolling question 2

Question (a)

Question (b)