A uniform rod of mass m and length l is rotating with constant angular velocity `omega` about an axis which passes thorugh its one end and perpendicular to the length of rod. The area of cross section of the rod is A and its Young's modulus is y. Neglect gravity. The strain at the mid point of the rod is:

Options

A. `(mn^(2)l)/(8Ay)`
B. `(3momega^(2)l)/(8Ay)`
C. `(3m omega^(2)l)/(4Ay)`
D. `(m omega)/(4Ay)`

Explanation

`dT=dm(l-x)omega^(2)`
`dT=(m)/(l)dx(l-x)omega^(2)`
`rArrint_(0)^(T)dT=int_(0)^(l//2)(m omega^(2))/(l)(l0x)dx`

`(m omega^(2))/(l)[lx-(x^(2))/(2)]_(0)^(l//2)=(m omega^(2))/(l)[(l^(2))/(2)-(l^(2))/(g)]`
`therefore` Tension at mid point is:
`T=(3)/(8)mlomega^(2)rArr` stress `=(3mlomega^(2))/(8A)`
`rArr` strain `(3ml omega^(2))/(8AY)`
Alternatively Tension at mid point can be found by using

`F_("ext")=ma_(cm)`
`T=(m)/(2)(omega^(2)(3l)/(4))=(3)/(8)m omega^(2)l`