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Moment of intertia of a W figure

  • Thread starter arnab321
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  • #1
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Homework Statement



A W figure.
each of the 4 arms of the W has mass M, length L, and are at right angles.
axis is in the plane of W, passing through the middle, vertically.


Homework Equations



i used the result I=M(L sinθ)²/3 for a rod inclined at angle θ with the axis, and axis passing through edge, from a previous sum

The Attempt at a Solution


attachment.php?attachmentid=46275&stc=1&d=1334553599.png

in the figure i found out the MI only for the left part, so for the whole figure will be twice of that result, i.e. 5ML²/3

sinθ is 1/√2 here,
so i found out the MI for the leftmost rod through the edge, shifted its axis to the center of the W and added it to the MI of the second rod.
 

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Answers and Replies

  • #2
tiny-tim
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hi arnab321! :smile:
so i found out the MI for the leftmost rod through the edge, shifted its axis to the center of the W and added it to the MI of the second rod.
i don't understand what you've done here :confused:

if you're intending to use the parallel axis theorem, you must start with the moment of inertia about an axis through the centre of mass

(btw, there is also a much easier way of doing it :wink:)
 
  • #3
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hi arnab321! :smile:


i don't understand what you've done here :confused:

if you're intending to use the parallel axis theorem, you must start with the moment of inertia about an axis through the centre of mass

(btw, there is also a much easier way of doing it :wink:)
wow nvr knew that this forum also has study material :eek:

well, what i was saying was that, the MI of leftmost rod about the grey line would be M(L sin 45)²/3 = M(L)²/6. this result was already derived from the CM of the rod, so i applied parallel axis theorem to shift that axis to a distance L/√2

(btw i really dunno if my attempt at this problem is correct or not because i posted the wrong question, whose ans. i dont know. anyways let me know)
 
  • #4
tiny-tim
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still not completely following what you've done :confused:

you found the moment of inertia about the centre of mass was ML²/24, then you used the parallel axis theorem and added ML²/8, making ML²/6

did you then use the parallel axis theorem again, starting from the new axis, adding 4ML²/8, making 2ML²/3 ?

you can't do that, when you use the parallel axis theorem, you can only start from the centre of mass …

you must go back there, and add 9ML²/8, not ML²/8 + 4ML²/8 :wink:
 
  • #5
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you can't do that, when you use the parallel axis theorem, you can only start from the centre of mass …

you must go back there, and add 9ML²/8, not ML²/8 + 4ML²/8 :wink:
oh, thanks!
 
  • #6
tiny-tim
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now that you're clear about the parallel axis theorem, can you see that you could have got the same result simply by treating it as a straight line of length 4L (at 45°) ? :wink:
 
  • #7
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I kinda can't get how to do that transformation. I can see that W and that with one on its sides inverted, something like v^ will have the same MI as they are still at the same dist. from the axis.
but how to know that straightening that v^ into a rod of 4L will still have the same MI?
 
  • #8
tiny-tim
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because all that matters (for the moment of inertia about the vertical axis) is the (horizontal) distance of each point from the axis, not how far along the axis it is …

so you can rearrange it in any way that preserves the distances :smile:
 

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