Question

When a hollow sphere is rolling without slipping on a rough horizontal surface then the percentage of its total $K.E$ which is Translational is

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

Answer 1

The correct option is C $60\%$

Let sphere of mass is $M$ and radius is $R$.

Moment of inertia(MOI), $I=\frac{2}{3}M{R}^2$

Hollow sphere is rolling without slipping. $\Rightarrow v=R\omega$, Where$v$ is speed of centre of mass and $\omega$ is angular speed.

$\Rightarrow$ Translational KInetic Energy$=\frac{1}{2}M{v}^2$

$\Rightarrow$ Rotational KInetic Energy$=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times \left(\frac{2}{3}M{R}^2\right)\times {\left(\frac{v}{R}\right)}^2=\frac{1}{3}M{v}^2$

$\Rightarrow$ Total Kinetic Energy$=\frac{1}{2}M{v}^2+\frac{1}{3}M{v}^2=\frac{5}{6}M{v}^2$

$\Rightarrow$ $\%$ of Translational Energy$=\frac{\frac{1}{2}M{v}^2}{\frac{5}{6}M{v}^2}\times 100=60\%$