Question
The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $$h$$ from rest without sliding is
A.
$$\sqrt {\frac{{10}}{7}gh} $$
B.
$$\sqrt {gh} $$
C.
$$\sqrt {\frac{6}{5}gh} $$
D.
$$\sqrt {\frac{4}{3}gh} $$
Answer :
$$\sqrt {\frac{{10}}{7}gh} $$
Solution :
When solid sphere rolls on inclined plane, then it has both rotational as well as translational kinetic energy
Total kinetic energy $$K = {K_{{\text{rot}}}} + {K_{{\text{trans}}}} = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2}$$
For sphere, moment of inertia about its diameter $$I = \frac{2}{5}m{r^2}$$
$$\eqalign{ & \therefore K = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{5}m{r^2}{\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{5}m{v^2} + \frac{1}{2}m{v^2}\,\,\left( {{\text{as}}\,v = r\omega } \right) \cr & = \frac{7}{{10}}m{v^2} \cr} $$
On reaching sphere at $$O,$$ it has only kinetic energy
$$\eqalign{ & \therefore PE = {\text{Total}}\,KE \cr & mgh = \frac{7}{{10}}m{v^2} \cr & \Rightarrow v = \sqrt {\frac{{10gh}}{7}} \cr} $$
When solid sphere rolls on inclined plane, then it has both rotational as well as translational kinetic energy
Total kinetic energy $$K = {K_{{\text{rot}}}} + {K_{{\text{trans}}}} = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2}$$
For sphere, moment of inertia about its diameter $$I = \frac{2}{5}m{r^2}$$
$$\eqalign{ & \therefore K = \frac{1}{2}\left( {\frac{2}{5}m{r^2}} \right){\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{5}m{r^2}{\omega ^2} + \frac{1}{2}m{v^2} \cr & = \frac{1}{5}m{v^2} + \frac{1}{2}m{v^2}\,\,\left( {{\text{as}}\,v = r\omega } \right) \cr & = \frac{7}{{10}}m{v^2} \cr} $$
On reaching sphere at $$O,$$ it has only kinetic energy
$$\eqalign{ & \therefore PE = {\text{Total}}\,KE \cr & mgh = \frac{7}{{10}}m{v^2} \cr & \Rightarrow v = \sqrt {\frac{{10gh}}{7}} \cr} $$
