A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is ` (ml^ 2 ) / ( 3 )`, the initial angular acceleration of the rod will be

Options

A. ` (2g ) /( 3 l ) `
B. ` mg "" ( l ) / (2 ) `
C. ` (3 )/( 2 ) gl`
D. ` ( 3g ) /( 2l ) `

Answer

D

Explanation

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is
` l = ( ml ^ 2 ) /( 2 ) `
Where m is mass of rod and l is length.
Torque ` ( tau = I alpha ) ` acting on centre of gravity of rod is given by
` tau = mg ""( l ) /( 2 ) or I alpha = mg ""(l ) /( 2 ) `
or ` ( m l ^ 2 ) /( 3 ) alpha = mg ""( l ) /( 2 ) or `
` alpha = ( 3g )/( 2l ) `