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posted by  nightstar918 on 11/7/2009 1:02:00 AM  |  status: Closed  |  Earned Karma: 35

Rotation, parallel-axis theorem

Course Textbook Chapter Problem Needs by
Calculus Based Physics Fundamentals of Physics Extended (8th) by Halliday, Resnick, Walker 10 116 11/7/2009 at 11:00:00 PM
Question Details:
Three 0.50 kg particles form an equilateral triangle with 0.60 m sides. The particles are connected by rods of negligible mass. What is the rotational inertia of this rigid body about (a) an axis that passes through one of the particles and is parallel to the rod connecting the other two, (b) an axis that passes through the midpoint of one of the sides and is perpendicular to the plane of the triangle, and (c) an axis that is parallel to one side of the triangle and passes through the midpoints of the other two sides?
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posted by Kevin_08 on 11/7/2009 1:17:35 AM  |  status: Live
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Response Details:
(a) One particle is on the axis, so r = 0 for it. For each of the others,
   the distance from the axis is r = (0.60m)sin60o = 0.52m 
    Therefore, the rotational inertia is I =Σmiri2
                                                                 
                                                         = 0.27 kg.m2
 

  (b) The two particles that are nearest the axis are each a distance of r = 0.30 m from it.
  The particle “opposite” from that side is a distance r = (0.60 m) sin 60° = 0.52 m from the
   axis. Thus, the rotational inertia is I =Σmiri2
                                                       = 0.22 kg .m2

  (c) The distance from the axis for each of the particles is r=1/2(0.60m) sin600 . Now,
                                 I = 2(5.0 kg)(0.26m)2
  Simplifying we get the answer.
I hope this helps! Best of luck with the rest of your coursework.
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