Question 1831
A uniform rod with a 3.0 m length and a mass 
of 14.0 kg is supported by a hinge at its left end and held in a 
horizontal position. A mass of 6.0 kg is attached to the rod 0.36 m 
from the hinge. A second mass of 7.5 kg is attached to the rod 2.80 m 
from the hinge. (A) Find the moments of inertia for each of the three 
masses about an axis at the hinge. (B) At the instant the right end of 
the rod is released, what is the net torque acting on this system? 
(C) What is the angular acceleration of this system the instant it is released?
 Sol: (a) Center of mass of the rod is 3/2 = 1.5 m from the hinge,
          The moments of inertia of the rod about the hinge = 14* (1.5)^2 = 31.5  kg*m^2
          The moments of inertia of the mass- 6 kg = 6* (0.36)^2 = 0.7776  kg*m^2
          The moments of inertia of the mass- 7.5 kg = 7.5* (0.28)^2 = 0.588  kg*m^2
          (Total  of 3 moments of inertia = 32.8656 kg*m^2)
      (b) Net Torque = r x F = 1.5 * 14 g + 0.36* 6 g + 0.28* 7.5g 
                     = (1.5 * 14 + 0.36* 6 + 0.28* 7.5)*9.8 = 247.548 meter X kg * meter/sec^2 
      (c) The angular acceleration =  Net Torque/ (Sum of The moments of inertia in (a))
          = 247.548/32.8656 = 7.53 (radians/ sec)


 This is a physics problem not in the category of algebra.

 Kenny