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. $v_{C}=2 r \omega$
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, $v_{C}=v_{\text {rot }}+v_{ CM }=v_{ CM }+v_{ CM }=2 v_{ CM }=2 r \omega$
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.