Question

A small solid sphere of radius $r$ is released coaxially from point $A$ inside the fixed cylindrical bowl of radius $R$ as shown in figure. If the friction between the small sphere and the larger cylinder is sufficient enough to prevent any slipping. Find the ratio of total kinetic energy and rotational kinetic energy, when solid sphere reaches the bottom of the larger cylinder.

• A. $\frac{3}{2}$
  
  
• B. $1$
  
• C. $\frac{5}{2}$
  
  
• D. $\frac{7}{2}$
  

Answer 1

The correct option is D $\frac{7}{2}$


Let the mass $\left(m\right)$ and velocity of the $\text{COM}$ of sphere at the bottom of the cylinder be $v$.
Translational kinetic energy of the sphere at the point $B$ ${KE}_{Trans}=\frac{1}{2}m{v}_{\text{CM}}^2=\frac{1}{2}m{v}^2$
Rotational kinetic energy of the sphere at the point $B$:
${KE}_{Rot}=\frac{1}{2}{I}_{CM}{\omega }^2$
$\because I_{CM}=\frac{2}{5}m{r}^2,\omega =\frac{v}{r}$
${KE}_{Rot}=\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)\left(\frac{v^2}{r^2}\right)$
${KE}_{Rot}=\frac{1}{5}m{v}^2\dots \dots \left(1\right)$
Total kinetic energy of the solid sphere at the point $B$ is $K{E}_{Total}={KE}_{Trans}+{KE}_{Rot}$
$\therefore K{E}_{Total}=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2\dots \dots \left(2\right)$
When sphere reaches the bottom of the cylinder,
$\frac{K{E}_{Total}}{{KE}_{Rot}}=\frac{\frac{7}{10}m{v}^2}{\frac{1}{5}m{v}^2}=\frac{7}{2}$