A bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$. The angle $ heta$ at which the speed of the bob is half of that at $A$, satisfies [image:4_image5.emf]

Question

A bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$. The angle $	heta$ at which the speed of the bob is half of that at $A$, satisfies [image:4_image5.emf]

Dr. Abubacker · Accepted Answer

(d) $V^{2} = U^{2} - 2g(L - L\cos	heta)$
$\frac{5gL}{4} = 5gL - 2gL(1 - \cos	heta)$
[image:4_image91.emf]
$5 = 20 - 8 + 8\cos	heta$
$\cos	heta = - \frac{7}{8}$
$\frac{3\pi}{4} < 	heta < \pi$

Q. A bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$. The angle $\theta$ at which the speed of the bob is half of that at $A$, satisfies

Explanation

Reviewed by Dr. Abubacker

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