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5.Work, Energy, Power and Collision
normal
The bob of a pendulum of length $l$ is pulled aside from its equilibrium position through an angle $\theta $ and then released. The bob will then pass through its equilibrium position with speed $v$ , where $v$ equals
A
$\sqrt {2gl\,\sin \,\theta } $
B
$\sqrt {2gl\,\left( {1 - \sin \,\theta } \right)} $
C
$\sqrt {2gl\,\left( {1 - \cos \,\theta } \right)} $
D
$\sqrt {2gl\,\left( {1 + \sin \,\theta } \right)} $
Solution
The height of fall is $\ell(1-\cos \theta)$
$\operatorname{Mg}\ell (1-\cos \theta)=\frac{1}{2} \mathrm{Mv}^{2}$
or $\quad v=\sqrt{2 g\ell (1-\cos \theta)}$
Standard 11
Physics
