Question

A small ball of radius $r$ rolls down without sliding in a big hemispherical bowl of radius $R$. What would be the ratio of the translational and rotational kinetic energies at the bottom of the bowl.

• A. $2:1$
• B. $3:2$
• C. $4:3$
• D. $5:2$

Answer 1

The correct option is D $5:2$

The transnational kinetic energy is given by

$K_t=\frac{1}{2}m{v}^2$. . . . .(1)

Rotational kinetic energy is given by

$K_r=\frac{1}{2}I{\omega }^2$.. . . . . . . .(2)

where, $I=\frac{2}{5}m{r}^2$ moment of inertia

$\omega =\frac{v}{r}$ angular velocity

Equation (1) becomes,

$K_r=\frac{1}{2}\times \frac{2}{5}m{r}^2\times \frac{v^2}{r^2}$

$K_f=\frac{1}{5}m{v}^2$. . . . .(3)

Ratio, $\frac{K_t}{K_r}=\frac{m{v}^2/2}{m{v}^2/5}=\frac{5}{2}$

$K_t:K_i=5:2$

The correct option is D.