Question: A particle moves down the inclined surface and completes a vertical circle at the foot of the plane....

Question

A particle moves down the inclined surface and completes a vertical circle at the foot of the plane. Ratio of heights H and h is

Answer 1

Velocity at the bottom of the circle = $\sqrt{5gr}$ .

Velocity at the top of the circle = $\sqrt{gr}$ Using P.E. = K.E., we

get

mg H = $\frac{1}{2}m\left( \sqrt{5gr} \right)^{2} = \frac{1}{2}m(5gr)$

(i) also mgh = $\frac{1}{2}m\left( \sqrt{gr} \right) = \frac{1}{2}mgr$

(ii) Dividing (i) and (ii), we get

$\frac{H}{h} = 5$