Question

A horizontal cylinder is fixed, its inner surface is smooth and its radius is $R$. A small block is initially at the lowest point. The minimum velocity that should be given to the block at the lowest point, so that it can just cross the point, $P$, is $u$. Choose the correct option(s):

• A. If the block moves anti-clockwise then $u=\sqrt{3.5gR}$.
• B. If the block moves anti-clockwise then $u=\sqrt{3gR}$.
• C. If the block moves clockwise then $u=\sqrt{3.5gR}$.
• D. If the block moves clockwise then $u=\sqrt{5gR}$.

Answer 1

Answer: (A), (D)

The correct options are
A If the block moves anti-clockwise then $u=\sqrt{3.5gR}$.
D If the block moves clockwise then $u=\sqrt{5gR}$.

To reach the P point anti-clockwise, we can apply conservation of energy equation with speed at P = Zero

$N=\frac{m{u}^2}{R}+mg(3\cos \theta -2)$, at $\theta ={120}^{\circ }$
$N=0\Rightarrow N=\frac{m{u}^2}{R}+mg(3\cos {120}^{\circ }-2)=0$
$\Rightarrow u=\sqrt{3.5gR}$
If the block is moving clockwise, then to cross the point $P$, the block has to cross the highest point, so to cross the highest point $u=\sqrt{5gR}$