Question

A simple pendulum consists of a $50cm$ long string connected to a $100g$ ball. The ball is pulled aside so that the string makes an angle of ${37}^{\circ }$ with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.

Answer 1

In $\Delta ABC$:
$\cos {37}^o=\frac{AC}{AB}$
$\frac{4}{5}=\frac{AC}{50}$
$AC=\frac{4}{5}\times 50$
$=40cm$
$h=AD-AC$
$=50-40=10cm=0.1m$

PE at B $=mgh$

This $PE$ converted into $K.E$ at $D$.
So, $\frac{1}{2}m{v}^2=mgh$
$v^2=2gh=2\times 9.9\times 0.1$
$v^2=1.98$ ....(1)
Tension at lowest position
$T=mg+\frac{m{v}^2}{r}$
$=\frac{100}{1000}\times 9.8+\frac{100}{1000}\times \frac{1.98}{0.50}$
$=0.98+0.396$
$=1.376N$