question_answer

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

A) \[3.7\,\,kg\]                      

B) \[4\,\,kg\]

C) \[4.5\,\,kg\]                      

D)  \[5\,\,kg\]

Correct Answer: C

Solution :

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