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

The Correct Option is A

Translational kinetic energy,
$K_{T}=\frac{1}{2} m v^{2}$
Rotational kinetic energy $V_{1}$
$K_{R}=\frac{1}{2} I \omega^{2}=\frac{1}{2} \times \frac{2}{5} m R^{2} \times \frac{v^{2}}{R^{2}}$
$=\frac{1}{5} m v^{2}$
$\frac{K_{T}}{K_{R}}=\frac{\frac{1}{2} m v^{2}}{\frac{1}{5} m v^{2}}$
$=\frac{5}{2}=2.5$