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 $\text{P}$. It is given a horizontal speed $v_0$ tangential to rim. It leaves the lower rim at point $\text{Q}$, vertically below $\text{P}$. 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 $\frac{v_0}{2\pi R}\left(\sqrt{\frac{2h}{g}}\right)$


The vertical distance covered by a particle during travel from $\text{P}$ to $\text{Q}$ is $h$.

Let it takes time $t$ to come out from $\text{Q}$.

For vertical motion only, $u_y=0$

And from equation of motion,

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

$\therefore t=\sqrt{\frac{2h}{g}}$

Since the internal surface of cylinder is smooth, it will move with the same speed $v_0$ throughout its journey.

Let $n$ be the total number of revolutions made by the particle. So,

$n\left(2\pi R\right)=v_0\times t$

$\Rightarrow n=\frac{v_{0}t}{2\pi R}$

$\therefore n=\frac{v_0}{2\pi R}\sqrt{\frac{2h}{g}}$

Hence, option $\left(d\right)$ is correct.

Why this question ? Tip : During the entire journey total distance covered can be visualized as number of revolution times the circumference of circular path.