A thin, uniform, 3.80 kg bar, 80.0 cm long, has very small 2.50 kg balls glued on at either end. It is supported horizontally by a thin, horizontal, frictionless axle passing through its center and perpendicular to the bar. Suddenly the right-hand ball becomes detached and falls off, but the other ball remains glued to the bar.
(a) Find the angular acceleration of the bar just after the ball falls off.
(b) Will the angular acceleration remain constant as the bar continues to swing?

Explanation:

Moment of inertia of bar and one ball attached at its one end

= 3.8 x .8² / 12 + 2.5 x .4²       ( MI of bar = m l² / 12 and sphere = m r² )  

= .20267 + .4

I = .60267 kg m²

Net force acting on the rod which will try to rotate will be weight of 2.5 kg ball . Because it was this force which was balancing the whole system. Torque due to this force = mg x r

= 2.5 x 9.8 x .4 Nm

= 9.8 Nm

angular acceleration = torque / moment of inertia

= 9.8 / .60267

= 16.26 rad / s²

b ) This angular acceleration will change because torque will change ( decrease in fact ) . It is to be noted that maximum value of torque acts when force is perpendicular . Since torque decreases , angular acceleration decreases.