Question 265: Work, Energy and Power & Circular Motion

Question

A simple pendulum of length $$2 \mathrm{~m}$$ is given a horizontal push through angular displacement of $$60^{\circ}$$. If the mass of bob is 200 gram, the angular velocity of the bob will be (Take Acceleration due to gravity $$=10 \mathrm{~m} / \mathrm{s}^2$$ ) $$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)$$

Accepted Answer

Given : $$l=2$$ m, $$	heta=60^\circ$$, m = 200 g = 2g
From the figure,
$$\mathrm{T}=\mathrm{mr} \omega^2$$ ..... (i)
also $$\mathrm{T} \cos 	heta=\mathrm{mg}$$ ..... (ii)
putting (i) into (ii)
$$m r \omega^2-\cos 	heta=m g$$ ..... (iii)
putting the given values into equation (iii)
$$\begin{aligned}
& 2 	imes 2 	imes \omega^2 \frac{1}{2}=2 	imes 10 \quad \ldots\left(\because \cos 60=\frac{1}{2}ight) \
& \omega^2=10 \\
& \Rightarrow \omega=\sqrt{10} \quad \ldots .(\because \sqrt{10}=\sqrt{2 	imes 2 	imes 2.5}) \\
& =2 \sqrt{2.5} ~\mathrm{rad} / \mathrm{s}
\end{aligned}$$

Answer

$$2 \sqrt{2} ~\mathrm{rad} / \mathrm{s}$$

Answer

$$3 \sqrt{2} ~\mathrm{rad} / \mathrm{s}$$

Answer

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

Answer

$$3 \sqrt{2.5} ~\mathrm{rad} / \mathrm{s}$$

Work, Energy and Power & Circular Motion

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