Moment of Inertia question

  • Thread starter pchalla90
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  • #1
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Homework Statement


Show that the rotational inertia of the rod about an axis through its center and perpendicular to its length is ML^2/12, where M is Mass and L is length from axis of rotation.


Homework Equations


∫r^2dm=Mr^2


The Attempt at a Solution



Well, you see the integral that I did in the relevant equations? I got most of the work done there, assuming that R=L. My problem is understanding where the /12 comes from in the question. There were no numbers given in the initial question, just M and L.
 

Answers and Replies

  • #2
Dick
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Great. dm should have a dr in it. What's R? What are the limits on your integral? Take rho to be the density of the rod per unit length. What's M in terms of rho and L? What's the moment of inertia in terms of rho and L? Now eliminate rho.
 
  • #3
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well, dm=pdL is what im assuming...

so ∫dm=∫pdL
M=pL

but shouldnt the right side of the above equation be a definite integral while the left is an indefinite integral?

i think the right side of the above equation should go from -L/2 to L/2, but that raises the question of the limits of the left side of the equation...

either way, assuming both sides are indefinite integrals, substituting pL for M in ML^2 gives me

pL^3=I

but thats as much as i think i understood of your post.

thanks for the quick reply.
 
  • #4
Dick
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Now you want to integrate r^2*rho*dr for r from -L/2 to +L/2 to get the moment of inertia. Then use M=rho*L to get rid of the rho.
 
  • #5
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Okay, so....

∫(-L/2,L/2) pR^2dr= (1/3)R^3p|(-L/2,L/2)
=(1/3)(L/2)^3p-(1/3)(-L/2)^3p
=(1/3)(L/2)^3p+(1/3)(L/2)^3p
=(2/3)(L^3/8)p
=L^3/12p

p=M/L

=(L^3/12)(M/L)
=ML^2/12


Thank you so much for your help.
 

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