A copper wire of negligible mass, `1 m` length and cross-sectional area `10^(6) m^(2)` is kept on a smooth horizontal table with one end fixed. A ball of mass `1 kg` is attached to the other end. The wire and the ball are rotating with an angular velocity of `20 rad//s`. If the elongation in the wire is `10^(-3) m`.
a. Find the Young's modulus of the wire (in terms of `xx 10^(11) N//m^(2)`).
b. If for the same wire as stated above, the angular velocity is increased to `100 rad//s` and the wire breaks down, find the breaking stress (in terms of `xx 10^(10) N//m^(2)`).

Answer

`1`

Explanation

The restoring froce (`F`) produced in the wire provides the necessary centripetal force i.e.
`F=MLomega^(2)`…………i
(since radius of circular path `=` length of wire `=L`)
We have Young's modulus
`Y=(F//A)/(/_\L//L)` or `F=YA,(/_\L)/L`………….ii
comparing eqn i and we get
`YA(/_\L)/L=mLomega^(2)` or `Y=(mL^(2)omega^(2))/(A/_\L)`
Substituting given values we get
`Y=400xx10^(9)=4xx10^(11)N//m^(2)`
The breaking force
`F=mLomega_("max")^(2)=(1kg)(1m)(100rad//s)^(2)=10^(4)` newton
Therefore breaking stress`=` Breaking force)`/` Area
`=(10^(4))/(10^(-6))=10^(10)N//m^(2)`