Answer:
[tex]3.6\times 10^{-5}[/tex]
Explanation:
m = Mass of stone = 10.2 kg
v = Tangential velocity = 11.6 m/s
l = Length of wire = 3.62 m
r = Radius of wire = [tex]4.1\times 10^{-3}\ \text{m}[/tex]
A = Area of wire = [tex]\pi r^2[/tex]
Y = Young's modulus of steel = [tex]2\times 10^{11}\ \text{N/m}^2[/tex]
[tex]\varepsillon[/tex] = Strain
The force acting on the stone will be centripetal
[tex]F=\dfrac{mv^2}{l}\\\Rightarrow F=\dfrac{10.2\times 11.6^2}{3.62}\\\Rightarrow F=379.15\ \text{N}[/tex]
Stress is given by
[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{379.15}{\pi (4.1\times 10^{-3})^2}\\\Rightarrow \sigma=7179488\ \text{N/m}^2[/tex]
Young's modulus is given by
[tex]Y=\dfrac{\sigma}{\varepsilon}\\\Rightarrow \varepsilon=\dfrac{\sigma}{Y}\\\Rightarrow \varepsilon=\dfrac{7179488}{2\times 10^{11}}\\\Rightarrow \varepsilon=3.6\times 10^{-5}[/tex]
Strain in the wire is [tex]3.6\times 10^{-5}[/tex].
