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August 29, 2014

August 29, 2014

Posted by **James** on Wednesday, June 20, 2012 at 1:22pm.

The answer is 0.934 rad/s^2. I'm not sure how to set this problem up. Any help is appreciated.

- Physics -
**Elena**, Thursday, June 21, 2012 at 2:32amMA=1 kg, MB =12 kg, MC= 27 kg.

Distances:

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

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