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. $\sqrt{\frac{10\text{g}\left(\text{R}+\text{r}\right)}{17}}$
• D. $\sqrt{\frac{10(\mathrm{R}+\mathrm{r})}{17{\mathrm{r}}^2}}$

Answer 1

Answer: (A)

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

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