Question

A uniform solid cylinder rolls without slipping on a rough horizontal floor, its centre of mass moving with a speed $v$. It makes an elastic collision with a smooth vertical wall. After impact:

• A. its centre of mass will move with a speed $v$ initially
• B. its motion will be rolling without slipping immediately
• C. its motion will be rolling with slipping initially and its rotational motion will stop momentarily at some instant
• D. its motion will be rolling without slipping only after some time.

Answer 1

Answer: (A), (C), (D)

The correct options are
A its centre of mass will move with a speed $v$ initially
C its motion will be rolling with slipping initially and its rotational motion will stop momentarily at some instant
D its motion will be rolling without slipping only after some time.


During collision, all forces passes through the COM. Hence,
${\tau }_{com}=0,$
$\Rightarrow \Delta L=0$
$\Rightarrow I\omega =$constant,
$\therefore \omega$ will remain same.

Collision is elastic, thus after the collision, linear velocity will be same in magnitude and opposite in direction (So option A is correct).

Initially, cylinder rolls without slipping.
Hence, $v=\omega R$ (pure rolling)
After collision
${\overrightarrow{v}}_{P,g}\to$ velocity of $P$ w.r.t. ground.
${\overrightarrow{v}}_{P,g}={\overrightarrow{v}}_{P,com}+{\overrightarrow{v}}_{com,g}$
where ${\overrightarrow{v}}_{P,com}\to$ velocity of $P$ w.r.t COM $=-\omega R\hat{i}$
${\overrightarrow{v}}_{com.g}\to$ velocity of com with respect to ground $=-v\hat{i}$
$\Rightarrow {\overrightarrow{v}}_{P,g}=-\omega R\hat{i}-v\hat{i}$
$\Rightarrow {\overrightarrow{v}}_{P,g}=-2\omega R\hat{i}$ [From Eq. (i)]

Hence just after impact, motion of cylinder will be rolling with slipping. This friction will act rightward, which oppose the motion and reduce angular velocity.



Due to $\alpha$ (angular acceleration due to friction), $\omega$ decreases and becomes $0$ . Due to $\alpha$ after some time, angular velocity will be in anticlockwise direction and it will be equal to $v^{\prime }={\omega }^{\prime }R$ (it starts rotating without slipping) as shown. Then, there will be no slipping between ground and cylinder and no friction will act.
So options A, C, D are correct.