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  4. A pendulum of mass 1 kg and length l=1 m is released from rest at angle θ=60°. The power delivered by all the forces acting on the bob at angle θ=30° will be: (Take g=10 m / s 2 )

Q. A pendulum of mass $1\, kg$ and length $l=1 \,m$ is released from rest at angle $\theta=60^{\circ}$. The power delivered by all the forces acting on the bob at angle $\theta=30^{\circ}$ will be : (Take $g=10 \,m / s ^{2}$ )

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Solution:

Potential energy at $A$,
image
$U_{A} =m g l\left(1-\cos 60^{\circ}\right) $
$ U_{A} =\frac{m g l}{2} $
$ U_{A} =U_{B}+K_{B} $
$ \frac{m g l}{2} =m g l\left(1-\cos 30^{\circ}\right)+\frac{1}{2} m v^{2} $
$ 5 =(10 \times 0.13)+\frac{v^{2}}{2} $
$3.66 =\frac{v^{2}}{2} $
$ v^{2} =7.32 $
$v =2.7 \,m / s $
Power $=F . v $
$=m g \sin 30^{\circ} \times v $
$=1 \times 10 \times \frac{1}{2} \times 2.7 $
$=13.5\, W$

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