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3-2.Motion in Plane
medium
A particle describes a horizontal circle of radius $r$ on the smooth surface of an inverted cone as shown. The height of plane of circle above vertex is $h$. The speed of particle should be
A$\sqrt{r g}$
B$\sqrt{2 r g}$
C$\sqrt{g h}$
D$\sqrt{2 g h}$
Solution
(c)$N \cos \left(90^{\circ}-\theta\right)=\frac{m v^2}{r}$
$N \sin \left(90^{\circ}-\theta\right)=m g$
$N=\frac{m g}{\cos \theta} \ldots (i)$
and $N \sin \theta=\frac{m v^2}{r} \ldots (ii)$
dividing $(ii)$ and $(i)$
$m g \tan \theta=\frac{m v^2}{r}$
$g\left(\frac{h}{r}\right)=\frac{v^2}{r} \Rightarrow v=\sqrt{g h}$
Standard 11
Physics
