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)

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