Question

A uniform solid sphere of radius $r=\frac{R}{5}$ is placed on the inside surface of a hemispherical bowl with radius $R(=5 r)$. 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. $\sqrt{\frac{20}{7} R g}$
• B. $\sqrt{\frac{40}{7} R g}$
• C. $\sqrt{\frac{25}{7} R g}$
• D. $\sqrt{\frac{10}{7} R g}$

Answer 1

Answer: (B)

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

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