Question

If $x$ is the ratio of rotational kinetic energy and translational kinetic energy of a rolling body and considering friction to be sufficient enough to prevent any slipping, which of the following statement is true

• A. $x=1$
• B. $x\le 1$
• C. $x\ge 1$
• D. $x=\frac{1}{2}$

Answer 1

The correct option is B $x\le 1$

Rotational kinetic energy $=\frac{1}{2}{I}_{CM}{\omega }^2$
Translational kinetic energy $=\frac{1}{2}m{v}^2$
condition of pure rolling $v=r\omega$
$\therefore x=\frac{\frac{1}{2}{I}_{\text{CM}}{\omega }^2}{\frac{1}{2}m{v}^2}=\frac{I_{\text{CM}}\times \frac{v^2}{r^2}}{m{v}^2}=\frac{I_{\text{CM}}}{m{r}^2}...\left(i\right)$
where,$I_{\text{CM}}=$moment of inertia of body about an axis passing through it's centre of mass.
Hence, from formula we know
$\frac{I_{\text{CM}}}{m{r}^2}=K$
Where $K$ is the radius of gyration and $K\le 1$
$\therefore x\le 1$