A sphere of radius 0.1 m and mass `8pi` kg is attached to the lower end of a steel wire of length 5.0 m and diameter `10^(-3)`m. The wire is suspended from 5.22 m high ceiling of a room. When the sphere is made to swing as a simple pendulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position. Y for steel `=1.994xx10^(11)N//m^(2)`.

Explanation

As the length of the wire is 5m and diameter `2 xx 0.1 = 0.2` m and at lowest point it grazes the floor which is at a distance 5.22 m from the roof, the increase in length of the wire at lowest point
`Delta L = 5.22 - (5+0.2) = 0.02m`
So tension in the wire (due to elasticity)
`T = (YA)/(L) Delta L`
` = (1.994 xx 10^11 xx pi (5 xx 10^(-4))^2 xx 0.02)/(5) = 199.4pi N`
and as equation of circular motion of a mass .m. tied to a string in a vertical plane is `(mv^2 //r)=T-mg cos theta`
So at lowest point `(mv^2//r)=T-mg [as theta =0]`
But here r = 5 +0.02 +0.1 = 5.12 m
So `(8 pi v^2 // 5.12) = (1.99 . 4pi - 8pi xx 9.8)`
i.e `v^2 = (121 xx 5.12//8) = 77.44`
so `v = 8.8 m//s `