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° with the upward vertical.

Answer 1

(a) Let the velocity at B be $v_1$.
$$\begin{gathered} \frac{1}{2}m{\nu }^2=\frac{1}{2}m{v}_1^2+mgl \\ \Rightarrow \frac{1}{2}m\left(10gl\right)=\frac{1}{2}m{\nu }_1^2+mgl \\ {\nu }_1^2=8gl \end{gathered}$$



So, the tension in the string at the horizontal position,
$$\begin{gathered} T=\frac{m{\nu }^2}{\mathrm{R}}=m\frac{8gl}{l} \\ =8mg \end{gathered}$$

(b) Let the velocity at C be $v_2$.
$\frac{1}{2}m{\nu }^2=\frac{1}{2}m{\nu }_2^2+mg\left(2l\right)$
$$\begin{gathered} \Rightarrow \frac{1}{2}m10gl=\frac{1}{2}m{\nu }_2^2+2mgl \\ \Rightarrow {\nu }_2^2=6gl \end{gathered}$$
So, the tension in the string is given by
${\mathrm{T}}_{\mathrm{C}}=\frac{m{v}_2^2}{l}-mg=5mg$

(c) Let the velocity at point D be ${\nu }_4$.
Again, $$\begin{gathered} \frac{1}{2}m{\nu }^2=\frac{1}{2}m{\nu }_3^2+mgl\left(1+\cos 60°\right) \\ \Rightarrow {\nu }_3^2=7gl \end{gathered}$$
So, the tension in the string,
$$\begin{gathered} {\mathrm{T}}_{\mathrm{D}}=\frac{m{{\nu }_3}^2}{l}-mg\cos 60° \\ =m\frac{\left(7gl\right)}{l}-0.5mg \\ =7mg-0.5mg \\ =6.5mg \end{gathered}$$