Question

A hollow vertical drum of radius, $r$, and height, $H$, has a small particle in contact with smooth inner surface of the upper rim at point P. The particle is given a horizontal speed, $u$, tangential to the rim. It leaves the lower rim at Q vertically below P. Taking $n$ as an integer for number of revolutions, we get:

• A. $n=\frac{2\pi r}{H}$
• B. $\frac{2\pi r}{\sqrt{2H/g}}$
• C. $n=\frac{2\pi r}{u\sqrt{2H/g}}$
• D. $n=\frac{u}{2\pi r}\sqrt{2H/g}$

Answer 1

The correct option is D $n=\frac{u}{2\pi r}\sqrt{2H/g}$

For vertical motion, using $S=ut+\frac{1}{2}a{t}^2$, we get $H=\frac{1}{2}g{t}^2$
or $t=\sqrt{2H/g}$.
For horizontal motion, distance covered is given by, $2\pi rn=ut$
or $2\pi rn=u\left(\sqrt{2H/g}\right)$
or $n=\frac{u}{2\pi r}\sqrt{2H/g}$