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Q.

A pendulum bob has a speed of $3 m s^{- 1}$ at its lowest position. The pendulum is 0.5 m long. Find the speed of the bob in $m s^{- 1}$ , when the length makes an angle of $60^{\circ}$ to the vertical. $[g = 10 m s^{- 2}]$

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Detailed Solution

$\frac{1}{2} m υ_{1}^{2} + m g l (1 - \cos θ) = \frac{1}{2} m υ^{2} \Rightarrow υ_{1}^{2} + 2 g l (1 - \cos θ) = υ^{2} \begin{array}{l} \Rightarrow υ_{1}^{2} + 2 \times 10 \times 0.5 (1 - \frac{1}{2}) = 9 \\ \Rightarrow υ_{1}^{2} = 9 - 5 = 4 \\ \Rightarrow υ_{1} = 2 m s^{- 1} \end{array}$

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A pendulum bob has a speed of 3ms−1 at its lowest position. The pendulum is 0.5 m long. Find the speed of the bob in ms−1, when the length makes an angle of 60∘ to the vertical. [g=10ms−2]