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(C) Given: Length of pendulum $l = 2 \ m$,angular displacement $	heta = 60^{\circ}$,mass $m = 200 \ g = 0.2 \ kg$,acceleration due to gravity $g = 10

Vedclass · Accepted Answer

$2 \sqrt{2.5} \ rad/s$

Vedclass · Answer

(C) Given: Length of pendulum $l = 2 \ m$,angular displacement $	heta = 60^{\circ}$,mass $m = 200 \ g = 0.2 \ kg$,acceleration due to gravity $g = 10 \ m/s^2$.When the pendulum bob moves in a horizontal circle,the radius of the circular path is $r = l \sin 	heta$.For the bob to move in a horizontal circle,the forces acting on it are tension $T$ and weight $mg$.The vertical component of tension balances the weight: $T \cos 	heta = mg$ ... $(i)$The horizontal component of tension provides the necessary centripetal force: $T \sin 	heta = m r \omega^2 = m (l \sin 	heta) \omega^2$ ... (ii)Dividing equation (ii) by equation $(i)$:$\frac{T \sin 	heta}{T \cos 	heta} = \frac{m l \sin 	heta \omega^2}{mg}$$	an 	heta = \frac{l \omega^2 \sin 	heta}{g}$Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we get:$\frac{1}{\cos 	heta} = \frac{l \omega^2}{g}$$\omega^2 = \frac{g}{l \cos 	heta}$Substituting the values: $\omega^2 = \frac{10}{2 	imes \cos 60^{\circ}} = \frac{10}{2 	imes 0.5} = \frac{10}{1} = 10$$\omega = \sqrt{10} = \sqrt{4 	imes 2.5} = 2 \sqrt{2.5} \ rad/s$.

$A$ simple pendulum of length $2 \ m$ is given a horizontal push through an angular displacement of $60^{\circ}$. If the mass of the bob is $200 \ g$,the angular velocity of the bob will be (Take acceleration due to gravity $g = 10 \ m/s^2$).

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