Question

If 'R' is the radius of gyration of a body of mass 'M' and radius 'r' then the ratio of its rotational to translational kinetic energy in the rolling condition is.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Moment of inertia of a body , I = M $r^2$ = M $K^2$

As K = R [K is radius of gyration]

$\Rightarrow$ I = m $R^2$

As the given body is in rolling condition , V = r $\omega$

$\therefore$ $\frac{K{E}_R}{K{E}_T}$ = $\frac{\frac{1}{2}I{\omega }^2}{\frac{1}{2}m{V}^2}$ = $\frac{M{R}^2{\omega }^2}{M{r}^2{\omega }^2}$ = $\frac{R^2}{r^2}$