Question:
A pendulum bob has a speed of $3 \mathrm{~m} / \mathrm{s}$ at its lowest position. The pendulum is $50 \mathrm{~cm}$ long. The speed of bob, when the length makes an angle of $60^{\circ}$ to the vertical will be $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$_________ $\mathrm{m} / \mathrm{s}$.
Solution:
Applying work energy theorem :
$\mathrm{w}_{\mathrm{g}}+\mathrm{w}_{\mathrm{T}}=\Delta \mathrm{K}$
$-\mathrm{mgl}\left(1-\cos 60^{\circ}\right)=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}$
$\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gl}\left(1-\cos 60^{\circ}\right)$
$\mathrm{v}^{2}=9-2 \times 10 \times 0.5\left(\frac{1}{2}\right)$
$\mathrm{v}^{2}=4$
$\mathrm{v}=2 \mathrm{~m} / \mathrm{s}$