Athin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (the moment of inertia of the rod about this axis is ml2/3.) the rod is released when it is being held up above the level of horizontal and when it makes an angle of 37° with the horizontal. what is the angular acceleration of the rod at the instant it is released? (in units of rad/s/s )

Explanation:

Torque, is given by

where F is force and r is perpendicular distance

where is the angle of inclination

Torque, can also be found by

where I is moment of inertia and a is angular acceleration

Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity

Making a the subject, and already I is given as  

hence  

Taking g as 9.81, is given as 37 and L is 1.2

a is the best answer your welcome