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(B) For the bob of mass $m$ moving in a horizontal circle of radius $r = \ell \sin 	heta$:$1$. The horizontal component of tension provides the centr

Vedclass · Accepted Answer

$\frac{m 	an 	heta}{m + M}$

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(B) For the bob of mass $m$ moving in a horizontal circle of radius $r = \ell \sin 	heta$:$1$. The horizontal component of tension provides the centripetal force: $T \sin 	heta = m \omega^2 (\ell \sin 	heta) = m \omega^2 r$.$2$. The vertical component of tension balances the weight: $T \cos 	heta = mg$.$3$. Dividing these equations,we get $	an 	heta = \frac{\omega^2 r}{g}$,or $T \sin 	heta = m g 	an 	heta$.$4$. Now,consider the block of mass $M$ and the rod. The vertical force exerted by the rod on the block is $N' = Mg + T \cos 	heta = Mg + mg = (M + m)g$.$5$. The horizontal force exerted by the rod on the block is the horizontal component of tension,$T \sin 	heta = m g 	an 	heta$.$6$. For the block not to slip,the maximum static friction $f_{max} = \mu N'$ must be greater than or equal to the horizontal force: $\mu (M + m)g \geq m g 	an 	heta$.$7$. Therefore,$\mu \geq \frac{m 	an 	heta}{M + m}$.

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