A particle mass of m is placed in equilibrium at the top of a fixed rough hemisphere of radius R

Question

A particle mass of m is placed in equilibrium at the top of a fixed rough hemisphere of radius R. Now the particle is given a gentle push so that it starts sliding on the surface of the hemisphere. It is found that the particle leaves the contact with the surface of the hemisphere at angular position $θ$ with the vertical where $\cos θ = \frac{3}{5}$ . If the work done against friction is $\frac{2 m g R}{10 x}$ , find $x$ .

Smart Achievers · Accepted Answer

(2)

$Angular position θ$

$m g \cos θ - N = \frac{m v^{2}}{R}$

$If it loses contact, N = 0$

$\Rightarrow v = \sqrt{g R \cos θ}$

$Now, \cos θ = \frac{3}{5} \Rightarrow v = \sqrt{\frac{3}{5} g R}$

$By work energy theorem, W_{m g} + W_{f} = \frac{1}{2} m v^{2}$

$m g R (1 - \cos θ) + W_{f} = \frac{1}{2} m \times \frac{3}{5} g R$

$W_{f} = - \frac{1}{10} m g R$

$\Rightarrow x = 2$

Solution

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Solution

Ans. (2) Angular position θ mg cosθ-N=mv2R If it loses contact, N=0 ⇒v=gRcosθ Now, cosθ=35⇒v=35gR By work energy theorem, Wmg+Wf=12mv2 mgR(1-cosθ)+Wf=12m×35gR Wf=-110mgR ⇒x=2

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