Consider a 1m long steel wire with the cross section of `0.05cm^(2)` and `Y_("steel")=2xx10^(11)Nm^(-2)`.Now a 15 kg mass is attached to the end of the wire and it is whirled in a vertical circle with an angular frequency of 2 rev/s at the bottom of the circle. When the mass is at the lowest point of its path, find the elongation of the wire in mm:
(Take `g=10ms^(-2)` & `pi^(2)=10`)

Answer

`02.55`

Explanation

When the mass is at the lowest point of the vertical circle, the stretching force is
`F\mg+mre^(2)=mg+mr(2pin)^(2)`
`=15xx10+15xx(1)xx(2pixx2)^(2)` `=2550N`
Elogation `l=(FL)/(AY)`
`l=(2550xx1)/(0.05xx10^(-4)xx2xx10^(11))`
`=2.55xx10^(-3)m`
`:.I=2.55mm`