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 $V_0$ tangential to rim. It leaves the lower rim at point $Q$, vertically below $P$. The number of revolutions made by the particle will be:

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

Given,

Vertical distance, $s_y=$Height of cylinder, $h$

Radius of cylinder $=R$

Initial velocity in horizontal direction $=v_o$ (which is tangential to cylinder)

Initial in vertical direction, $u_y=0$

Acceleration in vertical direction $a_y=$Acceleration of gravity, $g$

Apply kinematic equation

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$h=0\times t+\frac{1}{2}g{t}^2$

$t=\sqrt{\frac{2h}{g}}$ ……… (1)

Tangential velocity of circulation = Horizontal velocity

Total displacement in circulation is multiple of number of turn and Parameter of circle $s=n\times 2\pi R$

Displacement = tangential velocity X time

$s=v_o\times t$

put value of time from equation (1)

$s=v_o\sqrt{\frac{2h}{g}}$

$n=\frac{v_o}{2\pi R}\sqrt{\frac{2h}{g}}$

Number of turns from P to Q, $n=\frac{V_o}{2\pi R}\sqrt{\frac{2h}{g}}$