Question

A pendulum bob has a speed $3\text{m/s}$ while passing through its lowest position . The length of the pendulum is $0.5\text{m}$. The speed of the bob when it makes an angle of ${60}^{\circ }$ with the vertical is (Take $g=10\text{m}/s^2$)

• A. $1\text{m/s}$
• B. $2\text{m/s}$
• C. $3\text{m/s}$
• D. $4\text{m/s}$

Answer 1

The correct option is B $2\text{m/s}$

Take the bob+earth as the system. The only external (nonconservative) force acting on the system is that due to the string. But its work done is zero as it is always perpendicular to velocity.
Hence, the total mechanical energy will remain constant.
So here, loss in kinetic energy equals gain in potential energy.


Gain in potential energy $=mgl(1-\cos \theta )$ {$\cos \theta =\frac{1}{2}$ (Given)}
Hence, $(\frac{1}{2}m{V}^2-\frac{1}{2}m{V}_1^2)=mgl(1-cos\theta )$
$⟹V_1=\sqrt{V^2-2gl(1-cos\theta )}$
$⟹V_1=\sqrt{(3{)}^2-2\times (10)\times (0.5\left)\right(1-\frac{1}{2})}$
$⟹V_1=2\text{m/s}$

Hence, the speed of the bob when it makes an angle of ${60}^o$ with the vertical is $2\text{m/s}$ .