A bob of a pendulum of length $0.5 \mathrm{~m}$ has a speed of $6 \mathrm{~ms}^{-1}$ at its lowest point. Find the speed of the bob when the string of the pendulum makes $60^{\circ}$ with the vertical, (take $\left.g=10 \mathrm{~ms}^{-2}\right)$

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Aicharya · Accepted Answer

Given, length of pendulum, $l=0.5 \mathrm{~m}$

Speed at lowest point, $v_1=6 \mathrm{~m} / \mathrm{s}$ Angle made by string with vertical, $\theta=60^{\circ}$ In $\triangle O B C, \frac{O C}{O B}=\cos \theta=\cos 60^{\circ}$ $\frac{l-h}{l}=\cos 60^{\circ}$ $\Rightarrow \quad h=l\left(1-\cos 60^{\circ}\right)$ $h=0.5\left(1-\frac{1}{2}\right)=0.25 \mathrm{~m}$ Using law of conservation of mechanical energy, we get $\begin{aligned} & K_i+U_i=K_f+U_f \\ & \frac{1}{2} m v_1^2+0=\frac{1}{2} m v_2^2+m g h\end{aligned}$ [Taking initial potential energy at ground is equal to zero.] $\begin{aligned} \frac{1}{2} m v_2^2 & =\frac{1}{2} m v_1^2-m g h \\ v_2^2 & =v_1^2-2 g h\end{aligned}$ Substituting the values, we get $v_2=\sqrt{(6)^2-2 \times 10 \times 0.25}$ $=\sqrt{31} \mathrm{~m} / \mathrm{s}$ Hence, the final velocity at an angle $60^{\circ}$ is $\sqrt{31} \mathrm{~m} / \mathrm{s}$.

A bob of a pendulum of length $0.5 \mathrm{~m}$ has a speed of $6 \mathrm{~ms}^{-1}$ at its lowest point.…

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