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Polar Moment on Inertia "J"

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WardHollowayPE

Mechanical
Oct 19, 2000
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Im not sure if this is a trick question, but i have gotten more than one answer.
What is the equation of Polar Moment of Inertia of a uniform beam supported at one end with a length of 2L?
So far I have gotten...
mL^2 and 4/3ml^2 The first answer treats the problem as a pendulum with a lumped mass. The second uses the equation 1/3ml^2 and sets l=2L. Which is right?
 
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Depends on how you are rotating the object. I would use

J = m * 2[sup]2[/sup]* L[sup]2[/sup] /3

But this is valid if your uniform beam is a thin rod


J = m * 2[sup]2[/sup]* L[sup]2[/sup] /12
or J = m * L[sup]2[/sup] /3

is valid when the center of rotation is in the middle of the beam

regards Steven van Els
SAvanEls@cq-link.sr
 
Yep, first identify how the moments are going to happen or happening, if any, then determine the right equation from there!

lets face it, there's a Polar Moment for just about anything right? it's in the books and the items are not moving in them, so first determine or even "hey, is this thing subjected to a moment"? because if not, then your answer will be wrong.

you said:

"What is the equation of Polar Moment of Inertia of a uniform beam supported at one end with a length of 2L?"


is this a uniform beam or uniformly loaded? is it light loaded, intermediate loaded, is it center loaded, etc.?

you also said: (supported at one end), shouldn't you be taking it to be a cantilevered beam? M1= -wl^2/2 M = w/2(l-x)^2 ?


Joseph Moylan
moygr1@home.com
 
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