Question

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity $\sqrt{10gl}$, where $l$ 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.

Answer 1

Equation of motion for the particle:

$T-mg\cos \theta =\frac{m{V}^2}{l},V_A=\sqrt{10gl}$

According to the law of conservation of energy:

$E_A=E_B$

$⟹\frac{1}{2}m{V}_A^2=\frac{1}{2}m{V}_B^2+mgl$

$⟹\frac{1}{2}\times 10gl=\frac{1}{2}{V}_B^2+gl$

$⟹V_B^2=2(5gl-gl)$

$⟹V_B^2=8gl$

$⟹V_B=\sqrt{8gl}$

$T_B=\frac{m{V}_B^2}{l}+mg\cos \left({90}^o\right)$

$⟹T_B=\frac{m\times 8gl}{l}$

$⟹T_B=8mg$

Again from the law of conservation of energy:

$E_A=E_C$

$⟹\frac{1}{2}m{V}_A^2=\frac{1}{2}m{V}_C^2+2mgl$

$⟹5gl=\frac{1}{2}{V}_C^2+2gl$

$⟹V_C=\sqrt{6gl}$

$T_C=\frac{m{V}_C^2}{l}+mg\cos \left({180}^o\right)$

$⟹T_C=\frac{m\times 6gl}{l}-mg$

$⟹T_C=6mg-mg$

$⟹T_C=5mg$

Again from the law of conservation of energy:

$E_A=E_D$

$⟹\frac{1}{2}m{V}_A^2=\frac{1}{2}m{V}_D^2+mg\times \frac{3l}{2}$

$⟹5gl=\frac{1}{2}{V}_D^2+\frac{3}{2}gl$

$⟹\frac{V_D^2}{2}=\frac{7}{2}gl$

$⟹V_D=\sqrt{7gl}$

$\therefore T_D=\frac{m{V}_D^2}{l}+mg\cos \left({120}^o\right)$

$⟹T_D=7mg-\frac{m}{2}g$

$⟹T_D=\frac{13}{2}mg$