A sphere of radius ` 0.1m` and mass `8 pi kg` is attached to the lower end of a steel wire of length `5.0 m` and diameter `10^(-3)`. 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 . Young's modulus of steel is `(1.994xx10^(11) N//m^(2))`.

Explanation

Let `Deltal` be the extension of wire when the sphere is at mean position . Then, we have
`l+Deltal +2r=5.22`
or `Deltal =5.22-l-2r`
`= 5.22-5-2xx0.1`
`=0.02m` Let T be the tension in the wire at mean position during oscillations, then
`Y= (T//A)/((Deltal)//(l))`
`:. T=(YADeltal)/(l)=(Ypir^(2)Deltal)/(l)`
Substituting the values, we have
`T=((1.994xx10^(11))xxpixx(0.5xx10^(-3))^(2)xx0.02)/5`
`= 626.43 N`
The equation of motion at mean position is,
`T-mg =(mV^(2))/(R)` .....(i)
Here, ` R=5.22-r = 5.22-0.1 =5.12 m`
and m `= 8pi kg = 25.13 kg`
Substituting the proper values in Eq.(i) , we have
`(626.43) - (25.13xx9.8) = ((25.13)v^(2))/(5.12)`
Solving this equation, we get
` v = 8.8 m//s`