Question

For a rolling motion (without slipping) of a disc on a level surface is shown in figure, choose the correct statements.

I. $v_A=0$
II. $v_B=v_D>v_C$
III. $v_C=2r\omega$
IV. Velocity of centre of mass of body in pure rolling is zero.

• A. Both I and II
• B. Both II and III
• C. I, II and IV
• D. Both I and III

Answer 1

Answer: (D)

According to figure, the disc rotates with angular velocity $\omega$ about its symmetry axis passing through $O$.

At bottom point $A$, the linear velocity due to rotation is exactly opposite to the translational velocity $v_{CM}$,
i.e. $v_{\text{rot }}=r\omega =v_{CM}$
The point $A$ will be instantaneously at rest, if $v_{CM}=r\omega$,
i.e. $v_A=0$ when $v_{CM}=r\omega$.
Hence, for the disc, the condition for rolling without slipping is
$v_{CM}=r\omega$.
At the top point $C$ of the disc, linear velocity due to rotational motion and the translational velocity $v_{CM}$ are in the same direction, parallel to level surface.
Therefore, $v_C=v_{\text{rot }}+v_{CM}=v_{CM}+v_{CM}=2v_{CM}=2r\omega$
At $B$ and $D$, linear velocity due to rotation and the translational velocity $v_{CM}$ are perpendicular to each other. So,
$v_B=v_D=\left|v_{rot}+v_{CM}\right|=\sqrt{v_{CM}^2+v_{CM}^2+v_{CM}^2\cos 90^{\circ }}$
$=\sqrt{2}{v}_{CM}=\sqrt{2}r\omega$
Therefore, $v_B=v_D<v_C$
So, statements I and III are correct but II and IV are incorrect.