Question

A solid sphere and a hollow sphere, both of the same size and same mass roll down an inclined plane .Then

• A. solid sphere reaches the ground first
• B. hollow sphere reaches the ground first
• C. both spheres reach the ground at the same time
• D. the time at which the spheres reach the ground cannot be specified by the data given

Answer 1

Answer: (A)

solid sphere reaches the ground first

Answer 2

Potential energy of fall is converted into kinetic energy of rotation and translation
$=mgh=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2$
where$\omega=\frac{v}{r}$
=$mgh=\frac{1}{2}I\left(\frac{v}{r}\right)^2+\frac{1}{2}mv^2$
For sphere $I=\frac{2}{5}mr^2$
$mgh=\frac{1}{2}\left(\frac{2}{5}mr^2 \right).\frac{v^2}{r^2}+\frac{1}{2}mv^2$
$\Rightarrow v=\sqrt{\frac{10}{7}gh}$
For hollow sphere $I=\frac{2}{3}mr^2$
$\Rightarrow v=\sqrt{\frac{6}{5}gh}$
.'.Since, v of sphere > velocity of hollow sphere