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A simple pend...

Question

A simple pendulum consists of a $50 c m$ long string connected to a $100 g$ 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.

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Solution

In $Δ A B C$ :
$cos 37^{o} = \frac{A C}{A B}$
$\frac{4}{5} = \frac{A C}{50}$
$A C = \frac{4}{5} \times 50$
$= 40 c m$
$h = A D - A C$
$= 50 - 40 = 10 c m = 0.1 m$

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

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