A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m.What is the angular speed of the sphere at the bottom of the inclined plane?
A)5.1 rad/s
B)8.7 rad/s
C)9.7 rad/s
D)6.1 rad/s

Question

Quizplus · Accepted Answer

The angular speed of the sphere at the bottom 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 given by $PE = mgh$, where $m = 120\,kg$, $g = 9.8\,m/s^2$, and $h = 5.3\,m$. The kinetic energy at the bottom is the 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 is $v = r\omega$.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 $v = r\omega$, we get:$$mgh = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\omega^2$$Simplifying and solving for $\omega$, we find:$$gh = \frac{7}{10}r^2\omega^2$$$$ \omega = \sqrt{\frac{10gh}{7r^2}}$$Plugging in the given values:$$ \omega = \sqrt{\frac{10 \times 9.8 \times 5.3}{7 \times (1.7)^2}}$$$$ \omega \approx 5.1\,rad/s$$

A Solid Uniform Sphere of Mass 120 Kg and Radius