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Law of Conservation of Mechanical Energy

A simple pend...

Question

A simple pendulum consists of a light string from which a spherical bob of mass, $M$ , is suspended. The distance between the point of suspension and the center of bob is $L$ . The bob at rest position is given tangential velocity $\sqrt{5 g L}$ , then the K.E of the bob, when the string makes an angle of $120^{0}$ with the vertical is:

A

0

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B

\frac{M g L}{2}

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C

M g L

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D

\frac{5 M g L}{2}

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Solution

The correct option is C $M g L$
Applying Energy conservation

$K_{i} + U_{i} = K_{f} + U_{f}$

$\frac{1}{2} M u^{2} + M g h_{i} = K_{f} + M g h_{2}$ ( $h_{2} = L + L sin 30^{\circ}$ )

$\frac{5 M g L}{2} + 0 = K_{f} = M g (L + L / 2)$

$K_{f} = (2.5 - 1.5) m g L = M g L$

Hence option C is correct

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