A hemispherical bowl of radius R is rotated a | vedclass

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Question

(A) The forces acting on the block are the normal force $N$ and the gravitational force $mg$. The block moves in a horizontal circle of radius $x = R

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{g}{R\omega^2}\right)$

Vedclass · Answer

(A) The forces acting on the block are the normal force $N$ and the gravitational force $mg$. The block moves in a horizontal circle of radius $x = R \sin 	heta$.For the vertical direction,the forces are balanced:$N \cos 	heta = mg$  ......$(i)$For the horizontal direction,the normal force provides the necessary centripetal force:$N \sin 	heta = m x \omega^2 = m (R \sin 	heta) \omega^2$  ......(ii)Dividing equation (ii) by equation $(i)$:$\frac{N \sin 	heta}{N \cos 	heta} = \frac{m R \sin 	heta \omega^2}{mg}$$	an 	heta = \frac{R \sin 	heta \omega^2}{g}$Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we have:$\frac{\sin 	heta}{\cos 	heta} = \frac{R \sin 	heta \omega^2}{g}$$\frac{1}{\cos 	heta} = \frac{R \omega^2}{g}$$\cos 	heta = \frac{g}{R \omega^2}$$	heta = \cos^{-1}\left(\frac{g}{R \omega^2}ight)$

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