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 `2xx0.1=0.2m` 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
`DeltaL=5.22-(5+0.2)=0.02m`
So tension in the wire (due to elasticity)
`T=(YA)/(L)DeltaL`
`=(1.994xx10^(11)xxpi (5xx10^(-4))^(2)xx0.02)/(5)=199.4piN`
and as equation of circular motion of a mass .m. tied to a string in a vertical plane is `(mv^(2)//r)=T-mgcos theta`
So at lowest point `(mv^(2)//r)=T-mg " "["as" theta =0]`
But her `r=5+0.02+0.1=5.12m`
So, `(8pi v^(2)//5.12)=(1.99.4pi-8pi xx9.8)`
i.e., `v^(2)=(121xx5.12//8)=77.44, " so v" =8.8m//s`.