A sphere of radius `10 cm` and mass `25 kg` is attached to the lower end of a steel wire which is suspended from the ceiling of a room. The point of support is `521 cm` above the floor. When the sphere is set swimming as a simple pendulum, its lowest point just grazes the floor. Calculate the velocity of the ball at its lowest position.

Explanation

Young's modulus of steel` =20xx10^(10)N//m^(2)`
Unstretched length of the wire `=500cm`
Radius of the steel wire `=0.05cm`
A wire can be considered as a spring of suitable force constant. For a wire
`Y=(F/A)/(/_\l//l)=(FL)/(A/_\l)impliesF=Y(Al)/l/_\l`
Now `F=k./_\l`
Therefore `k` (force cosnant `=F/(/_\l)=(YA)/l`
`implies k=(20xx10^(10)xxpi(0.05xx10^(-2))^(2))/5=pixx10N//m^(2)`
Now we may treat the problem as if the mass were suspended by a spring of force constant `k=10^(4)piN//m^(2)`. Considering the dynamics of circular motion at the lowest point
`T-mg=(mv^(2))/r`
Obviously, at the lowest point the wire elongates by `1 cm`
`:. T=k/_\l=pixx10^(4)xx0.01=pixx10^(2)N`
`:.pixx10^(2)-25xx9.8=(25xxv^(2))/5.21`
`v^(2)=((100pi-245)xx5.21)/25=14.42`
`impliesv=3.8ms^(-1)`