A small body of mass m slides down from the t | vedclass

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(B) Let the body lose contact at an angle $	heta$ with the vertical. At this point,the normal force $N$ becomes zero.The radial component of the grav

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$\frac{2}{3}r$

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(B) Let the body lose contact at an angle $	heta$ with the vertical. At this point,the normal force $N$ becomes zero.The radial component of the gravitational force provides the necessary centripetal force:$mg \cos 	heta - N = \frac{mv^2}{r}$Since $N = 0$,we have $mg \cos 	heta = \frac{mv^2}{r}$,which implies $v^2 = rg \cos 	heta$.Using the principle of conservation of mechanical energy between the top of the hemisphere and the point of contact loss:$mg r = mg(r \cos 	heta) + \frac{1}{2}mv^2$$gr = gr \cos 	heta + \frac{1}{2}v^2$Substituting $v^2 = rg \cos 	heta$ into the energy equation:$gr = gr \cos 	heta + \frac{1}{2}(rg \cos 	heta)$$gr = \frac{3}{2}rg \cos 	heta$$\cos 	heta = \frac{2}{3}$The height $h$ from the base is given by $h = r \cos 	heta$.Therefore,$h = r \left(\frac{2}{3}ight) = \frac{2}{3}r$.

$A$ small body of mass $m$ slides down from the top of a hemisphere of radius $r$. The surfaces of the block and the hemisphere are frictionless. The height at which the body loses contact with the surface of the sphere is

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