Question

(p) A small solid cylinder of radius r and mass M slides down a smooth hill of height h from rest and gets onto the plank of mass M lying on the smooth horizontal plane at the base of the hill as shown in the figure. Due to friction between the cylinder and plank, the cylinder slows down and starts rolling without friction over the plank. While the cylinder is rolling (without sliding) over the plank, the velocity of the plank is $v_p$, the velocity of centre of mass of the cylinder is $v_c$ and angular velocity of the cylinder is “$\omega$”. The coefficient of friction between the cylinder and plank is , while gravity (g) is uniform in the space.
[Assume that height of the plank is negligible.]

The velocity of centre of mass of the cylinder after it starts pure rolling is

• A. $\frac{\sqrt{2gh}}{4}$
• B. $\frac{3}{4}\sqrt{2gh}$
• C. $\frac{2gh}{2}$
• D. $Noneoftheabove$

Answer 1

The correct option is A $\frac{\sqrt{2gh}}{4}$

$v_c-r_w=v_p$
$Mu=M{v}_c+M{v}_p$
$V_p=\mu gtand\omega =\frac{2\mu g}{r}t$
$Whereu=\sqrt{2gh}$
$\Rightarrow v_c=\frac{3}{4}\sqrt{2gh};v_p=\frac{2gh}{4},\omega \frac{2gh}{2r}$