A small disc is placed on the top of a smooth | vedclass

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Question

(C) At the top of the hemisphere,the forces acting on the disc are the gravitational force $mg$ acting downwards and the normal reaction $N$ acting up

Vedclass · Accepted Answer

$V=\sqrt{g R}$

Vedclass · Answer

(C) At the top of the hemisphere,the forces acting on the disc are the gravitational force $mg$ acting downwards and the normal reaction $N$ acting upwards.The net centripetal force required for circular motion is provided by the difference between the gravitational force and the normal reaction:$mg - N = \frac{mv^2}{R}$For the disc to leave the hemisphere surface immediately,the normal reaction $N$ must become zero at the top point.Setting $N = 0$ in the equation:$mg - 0 = \frac{mv^2}{R}$$mg = \frac{mv^2}{R}$$v^2 = gR$$v = \sqrt{gR}$Thus,the smallest horizontal velocity required is $\sqrt{gR}$.

$A$ small disc is placed on the top of a smooth hemisphere of radius $R$. The smallest horizontal velocity $V$ that should be imparted to the disc so that it leaves the hemisphere surface immediately is (there is no friction):

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