Question 1831
(A) To find the inertia for any body is I=Icm+Mh^2. The center of mass in this case for the whole system is {{{Xcm=(m1X1+m2X2)/(m1+m2)}}}. Xcm=1.72 m from the hinge. The inertia of the center of mass is the inertia of the rod at the center of mass which is {{{(mr^2)/(12)}}}. The r is the distance we just got and m is the mass of the rod  itself. Icm=3.43 kgm^2. If M is the total mass of the system while h is the distance at the center of mass the total inertia you get is 3.43+(6+7.5+14)(1.72)^2=84.37 kgm^2.


(B) The net torque is T=rF. Since the force is applied tangentially you don't have to worry about the sin of the angle between the radius and the force. The Force is simply the total mass of the system multiplied by 9.8 with which we get (6+7.5+14)(9.8)=269.5 N. The radius is once again the distance from the center of mass which is 1.72m so the net torque is T=(1.72)(269.5)=463.54 N*m.


(C) Remember the other definition of torque is that it is equall to the inertia mutiplied by the angular acceleration. This means that the angular acceleration is equivalent to the net torque divided by the inertia of the system or {{{(T)/(I)}}}. Angular acceleration={{{(463.54)/(84.37)}}}=5.49 rad/sec^2.