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Question

(A) Let $v$ be the speed of the particle when it separates from the hemisphere.Since the surface is smooth,there is no friction,and mechanical energy

Vedclass · Accepted Answer

$\sqrt{2g(R-h)}$

Vedclass · Answer

(A) Let $v$ be the speed of the particle when it separates from the hemisphere.Since the surface is smooth,there is no friction,and mechanical energy is conserved.The particle starts from rest at the top of the hemisphere (height $R$ from the horizontal surface).The potential energy at the top is $U_i = mgR$.The potential energy at height $h$ is $U_f = mgh$.The kinetic energy at height $h$ is $K_f = \frac{1}{2}mv^2$.By the law of conservation of energy:$U_i = U_f + K_f$$mgR = mgh + \frac{1}{2}mv^2$$mg(R-h) = \frac{1}{2}mv^2$$v^2 = 2g(R-h)$$v = \sqrt{2g(R-h)}$Thus,the correct option is $A$.

$A$ particle slides from the top of a smooth hemispherical surface of radius $R$ which is fixed on a horizontal surface. If it separates from the hemisphere at a height $h$ from the horizontal surface,then the speed of the particle is

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