A hemispherical bowl of radius r is rotating | vedclass

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(B) Step $1$: Identify the forces acting on the block. The forces are the gravitational force $mg$ acting downwards and the normal reaction force $N$

Vedclass · Accepted Answer

$\omega=\sqrt{\frac{g}{r \cos 	heta}}$

Vedclass · Answer

(B) Step $1$: Identify the forces acting on the block. The forces are the gravitational force $mg$ acting downwards and the normal reaction force $N$ acting perpendicular to the surface of the bowl towards the center of the sphere.Step $2$: Resolve the forces into vertical and horizontal components. The angle between the normal force $N$ and the vertical axis is $	heta$.Vertical equilibrium: $N \cos 	heta = mg$  (Equation $1$)Horizontal component provides the centripetal force: $N \sin 	heta = m \omega^2 R$,where $R$ is the radius of the circular path of the block. From the geometry of the bowl,$R = r \sin 	heta$.So,$N \sin 	heta = m \omega^2 (r \sin 	heta)$ (Equation $2$)Step $3$: Solve the equations. Divide Equation $2$ by Equation $1$:$\frac{N \sin 	heta}{N \cos 	heta} = \frac{m \omega^2 r \sin 	heta}{mg}$$	an 	heta = \frac{\omega^2 r \sin 	heta}{g}$$\frac{\sin 	heta}{\cos 	heta} = \frac{\omega^2 r \sin 	heta}{g}$$\frac{1}{\cos 	heta} = \frac{\omega^2 r}{g}$$\omega^2 = \frac{g}{r \cos 	heta}$$\omega = \sqrt{\frac{g}{r \cos 	heta}}$

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