Question

A pendulum bob has a speed of $3m{s}^{-1}$ at its lowest position The pendulum is $0.5m$ long.The speed of the bob, when the length makes an angle of ${60}^{\circ }$ to the vertical, will be $\left(g=10m{s}^{-2}\right)$:

• A. $3m{s}^{-1}$
• B. $\frac{1}{3}m{s}^{-1}$
• C. $\frac{1}{2}m{s}^{-1}$
• D. $2m{s}^{-1}$

Answer 1

The correct option is D $2m{s}^{-1}$

Given,

Height $h=L(1+\cos {60}^o)=0.5(1-\cos {60}^o)=0.25m$

Total Energy =kinetic energy + potential energy

Total Energy at lowest point $=\frac{1}{2}M{V}^2+Mgh=\frac{1}{2}M{\left(3\right)}^2+Mg\left(0\right)=\frac{9M}{2}$

Total Energy (at ${60}^o$)$=\frac{1}{2}M{V_1}^2+Mgh$

From conservation of energy

$\frac{1}{2}M{V_1}^2+Mgh=\frac{9M}{2}$

${V_1}^2=2(\frac{9}{2}-10\times 0.25)=4$

$V_1=2m/s$

Velocity of bob at angle ${60}^o$ is $2m/s$