Question

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

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

Answer 1

Answer: (B)

Given, $I=0.5\, m , u=3 \,ms ^{-1}$
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+m g h $
$ \Rightarrow u^2=v^2+2 g h $
$ v^2=u^2-2 g h=u^2-2 g\left(I-I \cos 60^{\circ}\right) $
$ {\left[\text { since, } h=M A=O A-O M=I-O B \cos 60^{\circ}\right.} $
$ \left.=I-I \cos 60^{\circ}\right] $
$ =(3)^2-2(9.8)\left(0.5-0.5 \times \frac{1}{2}\right)$
$\Rightarrow v^2 \approx 4 $
$ \therefore v=2 \,m s^{-1}$