# Solve the following :

Question:

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity . $V 10 \mathrm{gl}$, where 1 is the length of the pendulum. Find the tension in the string when (a) the string is horizontal, (b) the bob is at its highest point and (c) the string makes an angle of $60^{\circ}$ with the upward vertical.

Solution:

(a) By law of conservation of energy,

$\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{1}{2} \mathrm{mv} \mathrm{v}_{2}^{2}+\mathrm{mgl}$

$\frac{1}{2} \times m \times(\sqrt{g l})^{2}=\frac{1}{2} \mathrm{mv}^{2}+\mathrm{mgl}$

$v_{2}=(\sqrt{8 g})$

and

$T_{B}=\left(m v_{2}^{2}\right) / R=\left[m \times(\sqrt{8 g})^{2}\right] / l$

$\mathrm{T}_{\mathrm{B}}=8 \mathrm{mg}$

(b) By law of conservation at $\mathrm{A}$ and $\mathrm{C}$

$\frac{1}{2} \mathrm{mv}_{1}^{2}=\frac{1}{2} \mathrm{mv}_{3}^{2}+\mathrm{mgh}$

$\frac{1}{2} \times(\sqrt{\log } \mid)^{2}=\frac{\frac{1}{2}}{2} \times v_{3}^{2}+g \times 2 l$

$v_{3}=(\sqrt{6 g l})$\

and

Tension

$T_{c}=\left[\left(\mathrm{mv}_{3}^{2}\right) / \mathrm{l}\right]-\mathrm{mg}$

$T_{\mathrm{c}}=5 \mathrm{mg}$

(c) By law of conservation of energy at $A$ and $D$

$h^{2}=\frac{1}{2} \mathrm{mv}_{4}^{2}+\mathrm{mgh}$

$\frac{1}{2}(\sqrt{\log l})^{2}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{4}^{2}+\mathrm{gl}\left(1+\cos 60^{\circ}\right)$

$\mathrm{v}_{4}=\sqrt{7 \mathrm{gl}}$

and

Tension

$\mathrm{T}_{\mathrm{D}}=\left(\mathrm{mv}^{2}\right) / \mathrm{l}-\mathrm{mg} \cos 60^{\circ}$

by putting $v=7 \mathrm{gl}$ we get,

$T_{D}=6.5 \mathrm{mg}$