A thin uniform rod of mass M and length L is free to rotate in vertical plane about a horizontal axis passing through one of its ends. The rod is released from horizontal position shown in the figure. Calculate the shear stress developed at the centre of the rod immediately after it is released. Cross sectional area of the rod is A. [For calculation of moment of inertia you can treat it to very thin]

Options

A. `(Mg)/(4A)`
B. `(Mg)/(10A)`
C. `(Mg)/(12A)`
D. `(Mg)/(16A)`

Answer

D

Explanation

Immediately after release
`Ialpha = tau rArr (ML^(2))/(3).alpha = Mg"(L)/(2)rArr alpha = (3)/(2)(g)/(L)`
Consider the half rod `BC`. Its `COM` has a downward acceleration
`a_(BC) = (3L)/(3) .alpha = (9)/(8)g`
`F_(Sheer) =` Sheer force applied by part `AC` on the part `BC`.
`(Mg)/(2) + F_(shear) = (M)/(2) a_(BC)`

`:.` Shear stress `= (mg)/(16A)`