Question

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be

• A. $\frac{K^2+R^2}{R^2}$
• B. $\frac{K^2}{R^2}$
• C. $\frac{K^2}{K^2+R^2}$
• D. $\frac{R^2}{K^2+R^2}$

Answer 1

The correct option is C $\frac{K^2}{K^2+R^2}$

Let mass of the ball os $m$ and velocity is $v$.

Moment of Inertia (MOI) $=m{K}^2$

Ball is rolling without slipping, $\Rightarrow v=R\omega$

Translational Kinetic Energy (TKE) $=\frac{1}{2}m{v}^2$

Rotational Kinetic Energy (RKE) $=\frac{1}{2}I{\omega }^2=\frac{m{K}^2{v}^2}{2R^2}$

$\therefore$ Total Energy = TKE + RKE $=\frac{1}{2}m{v}^2+\frac{m{K}^2{v}^2}{2R^2}$

$\Rightarrow$ Fraction of RKE associated with total energy $=\frac{\frac{m{K}^2{v}^2}{2R^2}}{\frac{1}{2}m{v}^2+\frac{m{k}^2{v}^2}{2R^2}}=\frac{K^2}{K^2+R^2}$