Question

A solid sphere of mass $m$ and radius $r$ is released from rest from the given position. If it rolls without sliding on the circular track of radius $R$, its speed when it reaches its lowest position will be

• A. $\sqrt{\frac{8}{7} g(R-r)}$
• B. $\sqrt{\frac{10}{7} g(R-r)}$
• C. $\sqrt{\frac{10}{9} g(R-r)}$
• D. $\sqrt{\frac{5}{7} g(R-r)}$

Answer 1

Answer: (B)

When the sphere reaches its lowest position it loses gravitational potential energy by $\Delta U$ and gains a kinetic energy by $\Delta K$, where $\Delta U=-m g(R-r)$

and $\Delta K=\frac{1}{2} m v_{c}^{2}\left(1+\frac{K^{2}}{R^{2}}\right)=\frac{1}{2} m v^{2}\left(1+\frac{2}{5}\right)=\frac{7}{10} m v^{2}$
According to conservation of energy,
$\Delta U+\Delta K=0$
or $-m g(R-r)+\frac{7}{10} m v^{2}=0$
or $v=\sqrt{\frac{10}{7} g(R-r)}$