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3-2.Motion in Plane
hard
A hemispherical bowl of radius $r$ is set rotating about its axis of symmetry in vertical. A small block kept in the bowl rotates with bowl without slipping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the block with the vertical is $\theta$, then find the angular speed at which the ball is rotating.
A$\omega=\sqrt{ rg \sin \theta}$
B$\omega=\sqrt{ g / rcos \theta}$
C$\omega=\sqrt{\frac{ gr }{\cos \theta}}$
D$\omega=\sqrt{\frac{ gr }{\tan \theta}}$
(AIIMS-2015)
Solution
(b)$Hint:$ Use the Newton's second law to balances the forces in $x$ and $y$ directions.
$Step\,1:$ Draw the free body diagram of the system.
Step 2:Balance the forces in horizontal and vertical direction. Balancing forces along vertical direction.
$N \cos \theta=m g$
Balancing forces along horizontal direction.
$\Sigma F _{ nes }=m\left( a _2\right)$
$N \sin \theta=m\left(\omega^2 r\right)$
Here, $r=r \sin \theta$.
$N$ is the normal reaction, $m$ is the mass of the block and $g$ is the acceleration due to gravity.
Step $3$: Solve the equations.
Dividing equation $1$ and $2$ and using equation $3$ , we get:
$\omega=\sqrt{\frac{g}{r \cos \theta}}$
Standard 11
Physics
