Question

The radius of a wheel is $R$ and its radius of gyration about its axis passing through its centre and perpendicular to its plane is $K$ . If the wheel is rolling without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

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

Answer 1

Answer: (A)

$\frac{\text{Rotational KE}}{\text{Translational KE}}=\frac{\frac{1}{2}\text{I}{\text{ω}}^2}{\frac{1}{2}\text{M}{\text{v}}^2}=\frac{{\text{K}}^2}{{\text{R}}^2}$
$\left(\because I=M{K}^2,v=R\omega \right)$