Question

A uniform ball of mass ‘$m$’ rolls without sliding on a fixed horizontal surface. The velocity of the lowest point of the ball with respect to the centre of the ball is $V$. Find out the total kinetic energy of the ball?

Answer 1

Hint: Given that the mass of the ball is fixed on a horizontal surface without sliding from that point the center of the ball with the lowest point of the ball, based on this information we have found the kinetic energy of the ball. For this calculation we are using the moment of inertia and angular speed of the ball.

Complete step by step answer:
Given data, the uniform ball mass is, $m$ and velocity of the lowest point from the center of the ball is $V$. For this we are using the moment of inertia and angular speed.
Moment of the inertia of rotating ball is, $I = \dfrac{2}{5}m{R^2}$
Angular speed of the rotating ball is, $\omega = \dfrac{v}{R}$
By using kinetic energy due to rotation is,
$K.{E_1} = \dfrac{1}{2}I{\omega ^2}$
By substituting the values of the moment of inertia and angular speed in kinetic energy is,
$K.{E_1} = \dfrac{1}{2} \times \dfrac{2}{5}m{R^2} \times {\left[ {\dfrac{v}{R}} \right]^2}$
$\Rightarrow K.{E_1} = \dfrac{1}{5}m{v^2}$

Now we are going to find the kinetic energy due to translational motion,
$K.{E_2} = \dfrac{1}{2}m{v^2}$
For finding the kinetic energy at the lowest point on the surface is,
$K.{E_t} = K.{E_1} + K.{E_2} \\
\Rightarrow K.{E_t} = \dfrac{1}{5}m{v^2} + \dfrac{1}{2}m{v^2} \\
\Rightarrow K.{E_t} = m{v^2}\left( {\dfrac{1}{5} + \dfrac{1}{2}} \right) \\
\therefore K.{E_t} = \dfrac{7}{{10}}m{v^2} \\ $
We have proved the total kinetic energy.

Note: By using kinetic energy in moments of inertia and angular speed we have proven the lowest point of the ball in the horizontal surface. The procedure we used in the above solution part is the kinetic energy formula due to rotation and translation, after we have added both the kinetic energies then we got the total kinetic energy.