Question

A sphere of radius $10cm$ and mass $25kg$ is attached to the lower end of a steel wire of length $5m$ and diameter $4mm$ which is suspended from the ceiling of a room. The point of support is $521cm$ above the floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate the velocity of the ball at its lowest position in $m/s$. $(Y_{steel}=2\times {10}^{11}N/m^2)$

Answer 1

$\Delta l=521-500-20$
$=1cm=0.01m$
$T-mg=\frac{m{v}^2}{R}$
$\therefore T=m(g+\frac{v^2}{R})$
$T=m(g+\frac{v^2}{l})$
$\Delta l=\frac{Tl}{AY}=\frac{m(g+\frac{v^2}{l})l}{\left(\pi d^2/4\right)Y}$
$\Delta l=\frac{mgl+m{v}^2}{\left(\pi d^2/4\right)Y}$
$\therefore V=\sqrt{\frac{\pi d^2\Delta lY}{4m}-gl}$
$V=\sqrt{\frac{\left(3.14\right)(4\times {10}^{-3}{)}^2(0.01\left)\right(2\times {10}^{11})}{4\times 25}-9.8\times 5}$
$V=31m/s$