Question

A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is
A. $\dfrac{2}{5}$
B. $\dfrac{2}{7}$
C. $\dfrac{3}{5}$
D. $\dfrac{3}{7}$

Answer 1

We have to know what the friction is. This is the force which resists the relative motion of solid surface, fluid and material elements sliding against each other. This resistance force is called frictional force. There are two types of dry friction. One is static friction and the other one is kinetic friction. The coefficient of friction is equal to $\tan \theta$. $\theta$ is called angle.

Complete step by step answer:
${f_r} = \dfrac{{0.5I{\omega ^2}}}{{\dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}}}$
Here $I = \dfrac{2}{5}m{r^2}$
Kinetic energy for translation is equal to $\dfrac{1}{2}m{v^2}$
Kinetic energy for rotation is equal to $\dfrac{1}{2}I{\omega ^2}$
For rolling on a stationary surface $V = \omega R$
Therefore kinetic energy for rotation will be $\dfrac{{m{v^2}}}{5}$
Therefore the total kinetic energy would be,
\dfrac{{m{v^2}}}{5}$$$$ + \dfrac{1}{2}m{v^2} $= \dfrac{{7m{v^2}}}{{10}}$
Therefore the fraction of total energy associated with the rotation is $\therefore\dfrac{{\dfrac{{m{v^2}}}{5}}}{{\dfrac{{7m{v^2}}}{{10}}}} = \dfrac{2}{7}$

So the right answer will be option number B.

Note: We can get confused between kinetic energy of rotation and kinetic energy of translation. But both are not very similar to each other. We should properly know what those two kinetic energies mean. Kinetic energy for translation is $\dfrac{1}{2}m{v^2}$. Here $m$ is the mass of the body and v is the velocity of that body which is in motion. The kinetic energy for rotation is $\dfrac{1}{2}I{\omega ^2}$, here $I$ is equal to moment of inertia around the axis of rotation. And the omega is equal to the angular velocity of that rotating body. This is only applicable for a rotating body.