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 can be found using energy conservation principles. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. The potential energy at the top is $mgh$, where $m = 2.0 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $h = 7.0 \, \text{m}$. The kinetic energy at the bottom is a sum of translational ($\frac{1}{2}mv^2$) and rotational ($\frac{1}{2}I\omega^2$) kinetic energies. For a solid sphere, $I = \frac{2}{5}mr^2$, and the condition for rolling without slipping gives $v = \omega r$. Setting the potential energy equal to the total kinetic energy and solving for $\omega$ gives $\omega = \sqrt{\frac{10gh}{7r^2}}$, which when calculated with the given values results in approximately $5.8 \, \text{rad/s}$.

A Solid,uniform Sphere of Mass 2