A particle is placed at rest inside a hollow | vedclass

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(B) Let the particle be at an angle $	heta$ with the vertical. The forces acting on the particle are its weight $mg$ downwards,the normal force $N$ r

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$\left(1-\frac{\sqrt{3}}{2}\right) R$

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(B) Let the particle be at an angle $	heta$ with the vertical. The forces acting on the particle are its weight $mg$ downwards,the normal force $N$ radially outwards,and the frictional force $f$ tangential to the surface.For the particle to remain stationary,the component of weight along the tangent must be balanced by the limiting friction: $mg \sin 	heta = f = \mu N$.The component of weight perpendicular to the surface is balanced by the normal force: $N = mg \cos 	heta$.Substituting $N$ into the friction equation: $mg \sin 	heta = \mu (mg \cos 	heta) \implies 	an 	heta = \mu = \frac{1}{\sqrt{3}}$.Thus,$	heta = 30^\circ$.The height $h$ of the particle from the bottom of the hemisphere is $h = R - R \cos 	heta = R(1 - \cos 30^\circ)$.$h = R(1 - \frac{\sqrt{3}}{2})$.

$A$ particle is placed at rest inside a hollow hemisphere of radius $R$. The coefficient of friction between the particle and the hemisphere is $\mu = \frac{1}{\sqrt{3}}$. The maximum height up to which the particle can remain stationary is

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