Question

The pendulum bob has a speed of $3m{s}^{-1}$ at its lowest position and the length of pendulum is $0.5m$. The speed of the bob when the length of the pendulum makes an angle of $60^{\circ }$ with the vertical will be .

• A. $1.5m{s}^{-1}$
• B. $2m{s}^{-2}$
• C. $2.7m{s}^{-1}$
• D. $3.2m{s}^{-2}$

Answer 1

Answer: (B)

Given, $I=0.5m,u=3m{s}^{-1}$
The situation is a shown below

Applying energy conservation at points $A$ and $B$
$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+mgh$
$\Rightarrow u^2=v^2+2gh$
$v^2=u^2-2gh=u^2-2g\left(I-I\cos 60^{\circ }\right)$
$[\text{ since, }h=MA=OA-OM=I-OB\cos 60^{\circ }$
$=I-I\cos 60^{\circ }]$
$=(3{)}^2-2(9.8)\left(0.5-0.5\times \frac{1}{2}\right)$
$\Rightarrow v^2\approx 4$
$\therefore v=2m{s}^{-1}$