Question

A uniform solid sphere of radius $r=\frac{R}{5}$ is placed on the inside surface of a hemispherical bowl with radius $R(=5r)$. The sphere is released from rest at an angle $\theta =37^{\circ }$ to the vertical and rolls without slipping (Fig.). The angular speed of the sphere when it reaches the bottom of the bowl is

• A. $\sqrt{\frac{20}{7}Rg}$
• B. $\sqrt{\frac{40}{7}Rg}$
• C. $\sqrt{\frac{25}{7}Rg}$
• D. $\sqrt{\frac{10}{7}Rg}$

Answer 1

Answer: (B)

For the isolated sphere-Earth system, energy is conserved,
$\Delta U+\Delta K_{\text{rat }}+\Delta K_{\text{trans }}=0$

$mg(R-r)(\cos \theta -1)+\left[\frac{1}{2}m{v}^2-0\right]+\frac{1}{2}\left[\frac{2}{5}m{r}^2\right]{\omega }^2=0$
Substituting $v=r\omega$ we obtain
$mg(R-r)(\cos \theta -1)+\left[\frac{1}{2}m(r\omega {)}^2-0\right]$
$+\frac{1}{2}\left[\frac{2}{5}m{r}^2\right]{\omega }^2=0$
$mg(R-r)(\cos \theta -1)+\left[\frac{1}{2}+\frac{1}{5}\right]m{r}^2{\omega }^2=0$
$\omega =\sqrt{\left(\frac{10}{7}\right)\frac{(R-r)(1-\cos \theta )g}{r^2}}$ (i)
Here we have $r=\frac{R}{5}$ and $\theta =37^{\circ }$.
Substituting this values in equation (i), we get $\omega =\sqrt{\frac{40}{7}Rg}$