Rolling: A solid, uniform sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an inclined plane of height 7.0 m. What is the angular velocity of the sphere at the bottom of the inclined plane?
A) 5.8 rad/s
B) 9.9 rad/s
C) 11 rad/s
D) 7.0 rad/s

Question

Quizplus · Accepted Answer

The angular velocity ($\omega$) of the sphere at the bottom can be found using energy conservation. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. The potential energy at the top is given by $PE = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $h$ is the height. The kinetic energy at the bottom is the sum of translational kinetic energy ($KE_{trans} = \frac{1}{2}mv^2$) and rotational kinetic energy ($KE_{rot} = \frac{1}{2}I\omega^2$), where $I$ is the moment of inertia of a sphere ($\frac{2}{5}mr^2$) and $\omega$ is the angular velocity. Since the sphere rolls without slipping, $v = r\omega$.Setting the potential energy equal to the total kinetic energy gives:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$Substituting $I = \frac{2}{5}mr^2$ and $v = r\omega$, and solving for $\omega$ gives:$$gh = \frac{1}{2}r^2\omega^2 + \frac{1}{2}\cdot\frac{2}{5}r^2\omega^2$$$$gh = r^2\omega^2\left(\frac{1}{2} + \frac{1}{5}\right)$$$$gh = r^2\omega^2\cdot\frac{7}{10}$$$$\omega^2 = \frac{10gh}{7r^2}$$$$\omega = \sqrt{\frac{10 \cdot 9.8 \cdot 7.0}{7 \cdot (1.7)^2}}$$$$\omega \approx 5.8 \, \text{rad/s}$$

Rolling: a Solid, Uniform Sphere of Mass 2