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

Hint: To solve this question, we need to use the formulae for the rotational and the translational kinetic energy of an object. Using the expression for the moment of inertia of a solid sphere about its natural axis, we can determine the rotational kinetic energy. On taking their ratio, we will get the final answer.
Formula used: The formulae used for solving this question are given by
${K_T} = \dfrac{1}{2}m{v^2}$, here ${K_T}$ is the translational kinetic energy of an object of mass $m$ which is moving with a speed of $v$.
${K_R} = \dfrac{1}{2}I{\omega ^2}$, here ${K_R}$ is the rotational kinetic energy of a body rotating with an angular velocity of $\omega $, about an axis about which it has a moment of inertia of $I$.

Complete step-by-step solution:
Let the speed of the ball at the bottom of the hemispherical bowl be $v$, and its angular velocity at the same point be $\omega $.
Since the ball has the shape of a sphere of radius $r$, so the linear and the angular speeds can be related by
$v = \omega r$ ……………….(1)
If the mass of the ball is equal to $m$, then its translational kinetic energy can be given by
${K_T} = \dfrac{1}{2}m{v^2}$
Substituting (1) in the above equation, we get
${K_T} = \dfrac{1}{2}m{\omega ^2}{r^2}$....................(2)
Also, we know that the moment of inertia of a solid sphere about its natural axis is given by
$I = \dfrac{2}{5}m{r^2}$ …………………….(3)
Now, we know that the rotational kinetic energy of a rotating body is given by
${K_R} = \dfrac{1}{2}I{\omega ^2}$
Substituting (3) in the above equation, we get
${K_R} = \dfrac{1}{2} \times \dfrac{2}{5}m{r^2}{\omega ^2}$
$ \Rightarrow {K_R} = \dfrac{1}{5}m{r^2}{\omega ^2}$.....................(4)
Dividing (3) by (4) we get
$\dfrac{{{K_T}}}{{{K_R}}} = \dfrac{{\dfrac{1}{2}m{\omega ^2}{r^2}}}{{\dfrac{1}{5}m{\omega ^2}{r^2}}}$
On simplifying we finally get
$\dfrac{{{K_T}}}{{{K_R}}} = \dfrac{5}{2}$
Thus, the ratio of the translational and the rotational kinetic energies of the block at the bottom of the hemispherical bowl is equal to $5:2$.

Hence, the correct answer is option D.

Note: The location of the ball inside the hemispherical bowl is immaterial to the solution of this problem. In fact, the whole of the hemispherical bowl is immaterial. This is because the ratio of the translational and the kinetic energies of the ball will be fixed for any location of the ball, as can be appreciated by looking at the above solution.