Question

A uniform sphere of mass $ 500{\text{ }}g $ rolls without slipping on a plane surface so that its centre moves at a speed of $ 0.02{\text{ }}m/s $ . The total kinetic energy of rolling sphere would be (in J)
(A) $ 1.4 \times {10^{ - 4}}J $
(B) $ 0.75 \times {10^{ - 3}}J $
(C) $ 5.75 \times {10^{ - 3}}J $
(D) $ 4.9 \times {10^{ - 5}}J $

Answer 1

Hint
As given in the question the rolling sphere does translational motion that slides down and rotational motion that is the rolling of the sphere. Hence the total kinetic energy must be due to both motions.
 $ \Rightarrow K{E_{translational}} = \dfrac{1}{2}m{v^2} $
 $ \Rightarrow K{E_{rotational}} = \dfrac{1}{2}I{\omega ^2} $

Complete step by step answer
Let us consider v as the velocity of the center of the sphere.
In the case of rolling without slipping, the angular speed $ \omega $ of the center becomes
 $ \omega = \dfrac{v}{r} $ where r is the radius of sphere
Total kinetic energy (E) of sphere will be sum of translational kinetic energy ( $ K.E_1 $ ) and rotational kinetic energy ( $ K.E_2 $ )
 $ \Rightarrow E = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2} $ where m is the mass of sphere and I is the moment of inertia of sphere
We know that moment of inertia of sphere about the diameter $ I = \dfrac{2}{5}M{R^2} $ where M is the mass of sphere and R is its radius
On substituting the values of moment of inertia and angular velocity, E becomes
 $ \Rightarrow E = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}({\dfrac{{2MR}}{5}^2}){\omega ^2} = \dfrac{1}{2}m{v^2} + \dfrac{1}{5}m{v^2} = \dfrac{7}{{10}}m{v^2} $
Putting the value of mass and velocity in the above equation we get,
 $ \Rightarrow E = \dfrac{7}{{10}} \times \dfrac{1}{2} \times {0.02^2} = 1.4 \times {10^{ - 4}}J $
So, the correct option is (A).

Additional Information
During the roll of the sphere, it needs to be kept in mind that the frictional force acts in the direction of motion of the center of mass of the sphere. Also, it is to be noted that the instantaneous speed of the point of contact is zero but the instantaneous acceleration at the contact point is not zero. Pure rolling is observed when the point of contact of the body with ground is stationary.

Note
There is often confusion between slipping and sliding. In case of sliding, the body has no angular velocity but in case of slipping, angular velocity exists.