A small body of mass m slides down from top of a hemisphere of radius r . surface of block and hemis

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3-2.Motion in Plane

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

A

$\frac{3}{2}r$

B

$\frac{2}{3}r$

C

$\frac{1}{2}g{t^2}$

D

$\frac{{{v^2}}}{{2g}}$

Solution

$m g \cos \theta=\frac{m v^{2}}{R}$

$m g(R-R \cos \theta)=\frac{1}{2} m v^{2}$

$\operatorname{mgcos} \theta=\frac{2 m g R(1-\cos \theta)}{R}$

$\cos \theta=2-2 \cos \theta \Rightarrow \quad \cos \theta=\frac{2}{3}$

$h=r \cos \theta=\frac{2}{3} r$

Standard 11

Physics

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