Question

A solid sphere is rolling down an inclined plane. Then, the ratio of its translational kinetic energy to its rotational kinetic energy is

• A. 2.5
• B. 1.5
• C. 1
• D. 0.4

Answer 1

Answer: (A)

Translational kinetic energy,
$K_T=\frac{1}{2}m{v}^2$
Rotational kinetic energy $V_1$
$K_R=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times \frac{2}{5}m{R}^2\times \frac{v^2}{R^2}$
$=\frac{1}{5}m{v}^2$
$\frac{K_T}{K_R}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{5}m{v}^2}$
$=\frac{5}{2}=2.5$