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 point B . The angle θ at which the speed of the bob is half

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 point B . The angle θ at which the speed of the bob is half of that at A satisfies

TestHub · Accepted Answer

Suppose at point P velocity becomes half. We also know minimum velocity required to complete the circle is

$v_{A} = \sqrt{5 g L}$

Using conservation of energy from point $A$ to $P$

$\frac{1}{2} m {v_{A}}^{2} = \frac{1}{2} m {v_{P}}^{2} + m g h$
$\frac{1}{2} m v^{2} = \frac{1}{2} m {(\frac{v_{}}{2})}^{2} + m g h$
$\frac{3}{8} m v^{2} = m g h$

$\frac{3}{8} \times 5 g L = g h h = \frac{15}{8} L$

$\cos (θ) = \frac{L - h}{L} = 1 - \frac{h}{L} = - \frac{7}{8}$

Thus option d is right answer.

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