question_answer

If a sphere is rolling, then the ratio of its rotational kinetic energy to the total kinetic energy is

A)  1 :2         

B)  2:5

C)  2:7         

D)  5:7

Correct Answer: C

Solution :

 Let m be the mass, r the  radius of the sphere, let  v and co be linear and  angular velocities, in rolling down.     Total KE = Linear KE + Rotational KE Total KE\[KE=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}\] where, \[I\]is moment of inertia\[\left( I=\frac{2}{5}m{{r}^{2}} \right)\] Total \[KE=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}\left( \frac{2}{5}m{{r}^{2}} \right)\frac{{{v}^{2}}}{{{r}^{2}}}\] Total \[KE=\frac{1}{2}m{{v}^{2}}+\frac{1}{5}m{{v}^{2}}\] Total \[KE=\frac{7}{10}m{{v}^{2}}\] Hence ratio \[\frac{\text{Rotational}\,\text{KE}}{\text{Total}\,\text{KE}}=\frac{\frac{1}{5}m{{v}^{2}}}{\frac{7}{10}m{{v}^{2}}}=2:7\]