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

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(B) For the disc to leave the surface of the hemisphere immediately at the top,the normal reaction $N$ between the disc and the hemisphere must be zer

Vedclass · Accepted Answer

$v = \sqrt {gR} $

Vedclass · Answer

(B) For the disc to leave the surface of the hemisphere immediately at the top,the normal reaction $N$ between the disc and the hemisphere must be zero at the highest point.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 centripetal force required for circular motion is provided by the net radial force:$mg - N = \frac{mv^2}{R}$To leave the surface immediately,we set the normal reaction $N = 0$:$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 $R$. What is the smallest horizontal velocity $v$ that should be given to the disc for it to leave the hemisphere immediately at the top and not slide down it? [There is no friction]

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