Answer

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**Hint:**This question is based on the rotational motion. The easiest way to solve this kind of question is to assume the rolling. In case of rolling there will be linear velocity for the center of mass and rotational velocity for the entire body. At the topmost point these velocities add up while at the bottom most point, they act opposite to each other.

**Formula used:**

$v = R\omega $

**Complete answer:**

Disc is performing the rolling motion on the floor. We are asked to find out the angular velocity of it.

We have the formula for the angular velocity$(\omega )$ as $v = R\omega $

Where $v$ is the linear velocity due to rotation and $\omega $ is the angular velocity of the body which will be constant for every point of the rolling body.

According to the question no where it is mentioned that the disc is in pure rolling. So we will assume just rolling conditions.

The given scenario is clearly described in the below diagram.

Velocity of top most point will be ${v_1} = {v_c} + R\omega $

Velocity of the bottom most point will be ${v_2} = {v_c} - R\omega $

Where ${v_c}$ is the velocity of center of mass

If we subtract them, we will get

$\eqalign{

& {v_1} - {v_2} = \left( {{v_c} + R\omega } \right) - \left( {{v_c} - R\omega } \right) \cr

& \Rightarrow {v_1} - {v_2} = 2R\omega \cr

& \therefore \omega = \dfrac{{{v_1} - {v_2}}}{{2R}} \cr} $

Hence angular velocity of the disc will be $\omega = \dfrac{{{v_1} - {v_2}}}{{2R}}$

**Note:**

In case of pure rolling the velocity of point of contact with respect to the ground will be zero. This is the reason why we can consider the rolling motion of an object as a pure rotatory motion with respect to the point of contact and it is also called as the instantaneous center of the rotation.

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