Question

In the figure shown, the plank is being pulled to the right with a constant speed $V$. If the cylinder does not slip then,

$\left(A\right)$ the speed of the centre of mass of the cylinder is $2V$
$\left(B\right)$ the speed of the centre of mass of the cyllinder is zero
$\left(C\right)$ the angular velocity of the cylinder is $V/R$
$\left(D\right)$ the angular velocity of the cylinder is zero

• A. $A$ and $B$ are correct
• B. $B$ and $C$ are correct
• C. $C$ and $D$ are correct
• D. All are wrong

Answer 1

The correct option is B $B$ and $C$ are correct


Let $V_1$ be the centre of mass velocity of the cylinder and $\omega$ be its angular velocity.

Since, there is no slipping of the point of contact, $V_{AB}=0$, where $A$ is the lowermost point of the cylinder and $B$ belongs to the block. Taking right as positive,
$V_{AB}=0V_A-V_B=0\Rightarrow -V_1+R\omega =V\Rightarrow V_1=R\omega -V..............\left(1\right)$

The string moves along with the block speed $V$. No slipping between string and cylinder ensures that
$V_{CD}=0\Rightarrow V_C=V_D-V=-V_1-R\omega \Rightarrow V_1=V-R\omega ..............\left(2\right)$

$\left(1\right)$ and $\left(2\right)$ are possible, only if
$V_1=0$ and $V=R\omega \Rightarrow \omega =\frac{V}{R}$