A bob of a pendulum of length 0.5m has a speed of $6 m s^{- 1}$ at its lowest point. Find the speed of the bob when the string of the pendulum makes $60^{0}$ with the vertical, (take g = $10 m s^{- 2}$

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$26 m s^{- 1}$

$\sqrt{31} m s^{- 1}$

$13 m s^{- 1}$

$1.3 m s^{- 1}$

answer is B.

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Detailed Solution

Taking point at lowest position is zero potential level and velocity at that point as u= 6 ms^-1
$\Rightarrow \frac{1}{2} {mu}^{2} = \frac{1}{2} {mv}^{2} + (mgh); v is velovity at 60^{0}$

$\Rightarrow v = \sqrt{u^{2} - 2 gh}$

$v = \sqrt{(6)^{2} (m / s)^{2} - 2 \times 10 (m / s^{2}) \times \frac{0.5}{2} (m)} = \sqrt{31} {ms}^{- 1}$

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