A bob of mass M is suspended by a massless st | vedclass

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(D) At position $A$, the velocity $V$ is just sufficient to reach the highest point $B$. For a vertical circular motion, the minimum velocity at the b

Vedclass · Accepted Answer

$\frac{3 \pi}{4} < 	heta < \pi$

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(D) At position $A$, the velocity $V$ is just sufficient to reach the highest point $B$. For a vertical circular motion, the minimum velocity at the bottom to reach the top is $V = \sqrt{5gL}$.Let the speed at angle $	heta$ be $v_{	heta}$. According to the problem, $v_{	heta} = \frac{V}{2} = \frac{\sqrt{5gL}}{2}$.Applying the law of conservation of energy between point $A$ and the point at angle $	heta$:$\frac{1}{2} M V^2 = \frac{1}{2} M v_{	heta}^2 + M g L(1 - \cos 	heta)$Substituting $V^2 = 5gL$ and $v_{	heta}^2 = \frac{5gL}{4}$:$\frac{1}{2} M (5gL) = \frac{1}{2} M \left(\frac{5gL}{4}ight) + M g L(1 - \cos 	heta)$Dividing by $MgL$:$\frac{5}{2} = \frac{5}{8} + 1 - \cos 	heta$$\cos 	heta = \frac{5}{8} + 1 - \frac{5}{2} = \frac{5 + 8 - 20}{8} = -\frac{7}{8}$.Since $\cos 	heta = -0.875$, and we know that $\cos(3\pi/4) \approx -0.707$ and $\cos(\pi) = -1$, the angle $	heta$ must lie in the range $\frac{3\pi}{4} < 	heta < \pi$.

$A$ 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:

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