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 forces acting on the

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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 forces acting on the body are the component of gravity $mg \cos 	heta$ and the normal force $N$. The centripetal force is provided by the radial component of gravity:$mg \cos 	heta - N = \frac{mv^2}{r}$Since $N = 0$ at the point of losing contact,we have:$mg \cos 	heta = \frac{mv^2}{r} \implies v^2 = rg \cos 	heta$Using the principle of conservation of mechanical energy between the top of the hemisphere and the point where it loses contact:$mg r = mg(r \cos 	heta) + \frac{1}{2}mv^2$$gr = gr \cos 	heta + \frac{1}{2}v^2$Substitute $v^2 = rg \cos 	heta$ into the energy equation:$gr = gr \cos 	heta + \frac{1}{2}(rg \cos 	heta)$$gr = gr \cos 	heta + \frac{1}{2}gr \cos 	heta$$1 = \cos 	heta + \frac{1}{2} \cos 	heta = \frac{3}{2} \cos 	heta$$\cos 	heta = \frac{2}{3}$The height $h$ from the base is given by $h = r \cos 	heta$.Therefore,$h = r \left(\frac{2}{3}ight) = \frac{2}{3}r$.

$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y + 5 = 0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x = 1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x + 8y = 27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x = 9$

$A$ 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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