Question

A hollow sphere of mass $M$ and radius $R$ is rolling without slipping on a rough horizontal surface. Then the percentage of it's total $KE$ which is translational, will be

• A. $72\%$
• B. $28\%$
• C. $60\%$
• D. $40\%$

Answer 1

The correct option is C $60\%$

Moment of inertia of hollow sphere about the axis of rotation passing through it's centre of mass,
$I=\frac{2}{3}M{R}^2$
Condition for pure rolling,
$v=R\omega$
$v=$ Translational speed of centre of mass
$\Rightarrow K{E}_{Trans}=\frac{1}{2}M{v}^2$
$\Rightarrow K{E}_{Rot}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times \left(\frac{2}{3}M{R}^2\right)\times \frac{v^2}{R^2}$
$\therefore K{E}_{Rot}=\frac{1}{3}M{v}^2$
Total Kinetic Energy
$\Rightarrow K{E}_{total}=K{E}_{Trans}+K{E}_{Rot}=\frac{1}{2}M{v}^2+\frac{1}{3}M{v}^2=\frac{5}{6}M{v}^2$
$\%$ of Translational kinetic Energy $=\frac{K{E}_{Trans}}{K{E}_{total}}\times 100$
$=\frac{\left(\frac{1}{2}M{v}^2\right)}{\left(\frac{5}{6}M{v}^2\right)}\times 100$
$\therefore \%$ of Translational kinetic energy$=60\%$