How to find mass moment of inertia in a dynamics problem

  • Thread starter hachi_roku
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  • #1
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Homework Statement


a uniform slender l shaped bar abc is at rest in a vertical plane when two upward forces are applied. knowing that the mass of bar is 20kg, determine the initial angular acceleration of the bar and the initial acceleration of point b


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The Attempt at a Solution


i've attached the solution for this problem. i just don't get how I = 6....
ive tried using the forumula 1/12ml^2 and using the parallel axis theorem but i don't get 6.
 

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  • #3
kuruman
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If I am not mistaken, the solution shows an "L" shaped bar upside down. What is the rectangle extending from "A" to "B"? Are you sure you posted the solution of the problem that you have stated?
 
  • #4
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sorry bout that....the dimension from a to b is also 1.2, and the solution attached is the correct one.
 
  • #5
kuruman
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OK. The center of mass is at point "G". It is the midpoint of the line connecting the two midpoints of the arms of the "L". Use the parallel axes theorem to find the moment of inertia about point "G". Add the moment of inertia about G of one arm to the moment of inertia about "G" of the other arm. That will be the moment of inertia of the "L" about G.
 
  • #6
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so would it be 1/12(20)(1.2)^2 + (20)(.3)^2 for one arm??? i get 4.2

since the distance is .3 on both sides, would it be 4.2(2) = 8.4?

still not sure how they got 6
 
  • #7
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1) The mass of one arm is only 10kg
2) The distance you used when applying parallel axis theorem is wrong. What is the distance between the CoM of the individual arm and point G?
 
  • #8
kuruman
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The parallel axes theorem says
I = ICM+Md2
where d is the distance between the CM axis and the parallel axis. Here 0.3 m is the distance from the rod to point G, not the distance from the midpoint of the rod (where the CM is) to point G. Also, the mass of the whole thing is 20 kg, so one rod is 10 kg. Put it all together and you will get the 6 kg.m2.
 
  • #9
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finally got 6...thanks!
 

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