Question

A solid sphere of mass $m$ rolls down an inclined plane without slipping, starting from rest at the top of an inclined plane. The linear speed of the sphere at the bottom of the inclined plane is $v$. The kinetic energy of the sphere at the bottom is

• A. $\frac{1}{2}m{v}^2$
• B. $\frac{5}{3}m{v}^2$
• C. $\frac{2}{5}m{v}^2$
• D. $\frac{7}{10}m{v}^2$

Answer 1

The correct option is D $\frac{7}{10}m{v}^2$

$\begin{array}{c}Thekineticenergyofthesolidsphereatthebottomis\\ K=K_r+K_R=\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2\\ =\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{2}{5}m{R}^2\right)w^2\left[forsolidsphere,I=\frac{2}{5}m{R}^2\right]\\ =\frac{1}{2}m{v}^2+\frac{1}{5}m{R}^2{w}^2\\ =\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2\left[Asv=Rw\right]\\ =m{v}^2\left[\frac{1}{2}+\frac{1}{5}\right]=\frac{7}{10}m{v}^2\end{array}$

Hence,

option $\left(D\right)$ is correct answer.