Question

A uniform ball of radius $r$ rolls without slipping down from the top of a sphere of radius $R$ . The spin angular velocity of the ball when it breaks away from the sphere is (Assume initial velocity negligible)

• A. $\sqrt{\frac{10 \mathrm{~g}(\mathrm{R}+\mathrm{r})}{17 \mathrm{r}^2}}$
• B. $\sqrt{\frac{10 \mathrm{~g}(\mathrm{R}-\mathrm{r})}{17 \mathrm{r}^2}}$
• C. $\sqrt{\frac{1 0 \text{g} \left(\text{R} + \text{r}\right)}{1 7}}$
• D. $\sqrt{\frac{10(\mathrm{R}+\mathrm{r})}{17 \mathrm{r}^2}}$

Answer 1

Answer: (A)

$\frac{\mathrm{mv}^2}{(\mathrm{R}+\mathrm{r})}=\mathrm{mg} \cos \theta ; \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2$
$m g(R+r)(1-\cos \theta)=\frac{1}{2} m v^2+\frac{1}{5} m v^2=\frac{7}{10} m v^2$

$\frac{1 0}{7} \text{mg} \left(1 - cos \theta \right) = \text{mg} cos \theta $
$m v^2=\frac{10}{7} m g(R+r)(1-\cos \theta) \frac{10}{7}=\frac{17}{7} \cos \theta$
or $cos \theta = \frac{1 0}{1 7}$
$\text{v} = \sqrt{\text{g} \left(\text{R} + \text{r}\right) cos \theta } = \sqrt{\frac{1 0}{1 7} \text{g} \left(\text{R} + \text{r}\right)}$
and $\omega=\frac{\mathrm{V}}{\mathrm{r}}=\sqrt{\frac{10 \mathrm{~g}(\mathrm{R}+\mathrm{r})}{17 \mathrm{r}^2}}$