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

Explanation

As the length of the wire is 5 m and diameter of sphere `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 increases 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)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 - 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, `(8piv^(2) //5.12) = (199.4pi - 8pi xx 9.8)`
i.e, `v^(2) = (121 xx 5.12//8)^(2) = 77.44`
So, v = 8.8 m/s