AP Physics C Mechanics 6.5 Rolling Energy

[Maximum number: 6]

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 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}

(c)

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.

6.5.A—Describe the kinetic energy of a system that has translational and rotational motion question 1

Question (a)

Question (b)

Question (c)