A sphere of radius 0.1 m and mass 8pi kg is attachedto the lower end of a steel wire of length5.0 m and diameter 10^(-3)m. The wire is suspended from 5.22 m highceiling of a room. When the sphere is made to swing as a simplependulum, 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).

A sphere of radius 0.1 m and mass 8pi kg is attachedto the lower end o

1.	A sphere of radius 0.1 m and mass 8pi kg is attachedto the lower end of a steel wire of length5.0 m and diameter 10^(-3)m. The wire is suspended from 5.22 m highceiling of a room. When the sphere is made to swing as a simplependulum, 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).
Answer» Solution :As the length of the wire is 5m and diameter `2xx0.1=0.2m` andat lowestpoint 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 atlowest 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`.