Question

A hollow vertical cylinder of radius R and height h has smooth internal surface. A small particle is placed in contact with the inner side of the upper rim at a point P. It is given a horizontal speed v_o tangential to rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will

• A. $\frac{h}{2\pi R}$
• B. $\frac{v_0}{\sqrt{2gh}}$
• C. $\frac{2\pi R}{h}$
• D. $\frac{v_0}{2\pi R}\left(\sqrt{\frac{2h}{g}}\right)$

Answer 1

The correct option is D

$\frac{v_0}{2\pi R}\left(\sqrt{\frac{2h}{g}}\right)$

since the body has no initial velocity in the vertical direction.

$a_z$=-g, vertical displacement z=-h.
$\therefore z=at+\frac{1}{2}a{t}^2$
$\Rightarrow -h=0+\frac{1}{2}(-g)t^2$
$\Rightarrow T=\sqrt{\frac{2h}{g}}$ time taken to reach the bottom
let,t be the time taken to complete one revolution.
Then $t=\frac{2\pi R}{v_0}$
$\therefore$ number of revolution=$\frac{T}{t}=\frac{\sqrt{\frac{2h}{g}}}{\frac{2\pi R}{v_0}}=\frac{v_0}{2\pi R}\sqrt{\frac{2h}{g}}$