posted by James on .
The system shown in the figure is held initially at rest. Calculate the angular acceleration of the system as soon as it is released. You can treat MA (1 kg) and MB (12 kg) as point masses located on either end of the rod of mass MC (27 kg) and length L (9 m). (Indicate the direction with the sign of your answer. Let the counterclockwise direction be positive.)
The answer is 0.934 rad/s^2. I'm not sure how to set this problem up. Any help is appreciated.
MA=1 kg, MB =12 kg, MC= 27 kg.
from MA to the pivot point(PP) =3•L/4 m,
from MB to PP = L/4,
from MC to PP = L/4.
Moment of inertia for the rod according to the parallel axis theorem or Huygens–Steiner theorem (moment of inertia of a rigid body about any axis, given the body's moment of inertia about a parallel axis through the object'scentre of mass and the perpendicular distance (r) between the axes.):
I(C) = Iₒ+MC•x² = MC•L²/12 +MC(L/4)² = 7MCL²/48 =7•27•81/48 =318.9 kg•m²,
I(A) = MA•(3L/4) ²= 9•1•81/16 = 45.6kg•m²,
I(B) = MB•( L/4) ²= 12 •81/16 =60.8 kg•m²,
I = I(A) + I(B) = I(C) = 318.9+45.6+ 60.8=425.3 kg•m²,
The torque about PP is
τ = MC•g•(L/4)+MA•g•(3L/4)- MB•g•(L/4) =
=(g•L/4) •(MC+3MA-MB)= (9.8•9/4) •(27+3-12) =397 N•m,
Newton’s 2 law for rotation τ = I•ε
ε= τ/I=397/425.3 = 0.93 rad/s²