A L shaped rod of mass M is free to rotate in a vertical plane about axis A A as shown in figure. Maximum angular acceleration of rod is

Options

A. `(3g)/sqrt(10 I)`
B. `(9g)/(10 l)`
C. `(9g)/(5l)`
D. `(6g)/(5sqrt(2I))`

Answer

A

Explanation

Moment of inertia about axis AA. is:

`I_(AA.) = {(M/2) I^(2)/12 + (M/2) (5I)^(2)/4 + (M/2)I^(2)/3} = (5MI^(2))/6`
Taking point O as origin and x and y-axis along the lengths of each of the rods, we obtain `x_(C ) = (3I)/4` and `y_( C) =1/4`, so the length OC is, `OC = sqrt((9l^(2))/16 + l^(2)/16) = (sqrt(10)I)/4 = sqrt((10)I)/4`
( point C is centre of mass of the system )
Torque of Mg will be maximum when OC is horizontal:
`therefore alpha = tau/(I_(A A.)) = ((Mgsqrt(10I))/4)/((5MI^(2))/6)`
`alpha = (3g)/(sqrt(10)I)`