A cylinder of length 1.5 m and diameter 4 cm is fixed at one end. A tangential force of `4 xx 10^(5) N` is applied at the other end. If the rigidity modulus of the cylinder is `6 xx 10^(10) Nm^(-2)` then, calculate the twist produce in the cylinder.

Explanation

Length of a cylinder l = 1.5 m
Diameter of a cylinder `d=4xx10^(-2)m`
Tangential force `F_(t)=4xx10^(5)N`
Rigidity modulus `eta_(R)=6xx10^(10)Nm^(-2)`
Twist produced `theta=` ?
Rigidity modulus `(r)/(R)=(F_(t)//DeltaA)/(theta)`
`=(F_(t)//DeltaA)/(x//h)`
`=(F_t)/(DeltaAtheta)`
Area of the cylinder `A=(pid^2)/(4)`
`=(3.14xx(4xx10^(-2))^(2))/(4)`
`=(3.14xx4xx4xx10^(-4))/(4)`
`=12.56xx10^(-4)m^(2)`
`"Shear stress"=("Force")/("Area")`
`=(4xx10^(5))/(12.56xx10^(-14))`
`=(4xx10^(9))/(12.56)`
`{:("Rigidity"),("modulus"):}}eta=("shear stress")/("shear strain")`
`therefore` Shear strain `theta=("shear stress")/(eta)`
`theta=(4xx10^(9))/(12.56xx6xx10^(10))`
`=(4xx10^(-1))/(75.36)=0.005307`
`therefore` Twist `=0.005307`