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Question

By the energy conservation ($ME$) between bottom point and point $Y$

$\frac{1}{2} mv _0^2= mgh +\frac{1}{2} mv _1^2$

$	herefore v _1^2= v _0^2-

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$A,D$

Vedclass · Answer

By the energy conservation ($ME$) between bottom point and point $Y$

$\frac{1}{2} mv _0^2= mgh +\frac{1}{2} mv _1^2$

$	herefore v _1^2= v _0^2-2 gh \quad \ldots 	ext { (i) }$

Now at point $Y$ the centripetal force provided by the component of $mg$

$	herefore mg \sin 30^{\circ}=\frac{ mv _1^2}{ R }$

$	herefore v _1^2=\frac{ gR }{2}$

$	herefore$ from $(i)$

$\frac{ gR }{2}=v_0^2-2 gh$

At point $x$ and $z$ of circular path, the points are at same height but less then $h$. So the velocity more than a point $y$.

So required centripetal $=\frac{ mv ^2}{ r }$ is more.

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