Expression for the moment of inertia

  • Thread starter Firben
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  • #1
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Homework Statement


The thin, homogeneous bent rod has the mass m and the total length of 4b. It rotates with the angular speed of ω = ω0(24i + 12j - 6.0k) (only rotation)

Determine the expression for the moment of inertia with consideration of the center of mass of the rod

The figure:
http://s716.photobucket.com/user/Pitoraq/media/Mek4_zps8205c021.png.html

Homework Equations



H = (Ixxωx-Ixyωy-Ixzωz)i+(-Iyzωx+Iyyωy-Iyzωz)j+(-Izxωx-Izyωy+Izzωz)k

m = ρb


The Attempt at a Solution

¨


I plotted the figure in yx-axis and could see that

Iyz=Iyx=Iyz=Izy=Ixz=Izx=0 is symmetric

But i'm not sure about Ixx, Iyy and Izz
 

Answers and Replies

  • #2
adjacent
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Can you post your graph?
 
  • #4
SteamKing
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The Attempt at a Solution

¨


I plotted the figure in yx-axis and could see that

Iyz=Iyx=Iyz=Izy=Ixz=Izx=0 is symmetric

But i'm not sure about Ixx, Iyy and Izz

You might have to, you know, calculate Ixx, Iyy, and Izz. It's a shocking suggestion, I'm sure.
 
  • #5
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The centre of mass is easy: it is at the origin of y-z axis.

There are three parts to this problem: the rotating center bit, and the two sides. By symmetry, the MOI of the left and right side are equal.

Calculate the MOI for rotation in the ##i## and ##j## and ##k## axes. Use parallel axes theorem to calculate the MOI of the sides.
 

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