Question

A disc is rolling on a surface without slipping. What is the ratio of its translational to rotational kinetic energies?

• A. $5:2$
• B. $2:1$
• C. $3:2$
• D. $2:3$

Answer 1

The correct option is B $2:1$

We know,

$v=\omega R$ in rolling motion.

$K{E}_T=\frac{1}{2}m{v}^2$

$K{E}_R=\frac{1}{2}I{\omega }^2=\frac{1}{2}\frac{m{R}^2}{2}\frac{v^2}{R^2}$

$\frac{K{E}_T}{K{E}_R}=\frac{\frac{1}{2}}{\frac{1}{4}}=\frac{2}{1}$

Hence option $\text{B}$ is correct answer.