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What is the moment of inertia I for rotation around r_cm?

  • #1
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Homework Statement


A massless rod of length L connects three iron masses. If the first mass (at x=0) has a mass of 1.90 kg and the second mass (at x=L) has a mass of 8.77 kg and the third mass (located at the center of mass of the system) has a mass of 148 kg, what is the moment of inertia I for the free rotation about the center of mass?

Homework Equations


I= Sum(mi(ri^2))

The Attempt at a Solution


I= Sum(mi(ri^2)) = 1.9(0^2) + 8.77(L^2) + 148(r_cm^2)

Now i'm assuming i'm going to be wanting to find what r_cm is (it'll probably be a function of L)
To find the center of mass for i have to take an integral? or is there another way to solve this??
This isn't a uniform rod because of the different weights of the masses correct?
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


A massless rod of length L connects three iron masses. If the first mass (at x=0) has a mass of 1.90 kg and the second mass (at x=L) has a mass of 8.77 kg and the third mass (located at the center of mass of the system) has a mass of 148 kg, what is the moment of inertia I for the free rotation about the center of mass?

Homework Equations


I= Sum(mi(ri^2))

The Attempt at a Solution


I= Sum(mi(ri^2)) = 1.9(0^2) + 8.77(L^2) + 148(r_cm^2)

Now i'm assuming i'm going to be wanting to find what r_cm is (it'll probably be a function of L)
To find the center of mass for i have to take an integral? or is there another way to solve this??
This isn't a uniform rod because of the different weights of the masses correct?
The rod is massless according to the problem statement, so you don't have to worry about it.

If you have three iron balls of different masses and different locations, how would you normally find the location of their center of mass? Don't you know what a weighted average is?
 
  • #3
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Okay so using equation
MR=m1r1+m2r2
MR=0 + 8.77kg*L
And then divide by the total mass
So 1.90kg+8.77kg=10.67 kg

So the center of mass is at .822L?
 
  • #4
SteamKing
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Okay so using equation
MR=m1r1+m2r2
MR=0 + 8.77kg*L
And then divide by the total mass
So 1.90kg+8.77kg=10.67 kg

So the center of mass is at .822L?
You've got three masses to consider, not two.
 
  • #5
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You've got three masses to consider, not two.
I know ive got three masses but i thought since the third mass was at the center of mass that i wouldnt have to consider it when finding the center, why would this be a wrong assumption?
 
  • #6
SteamKing
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I know ive got three masses but i thought since the third mass was at the center of mass that i wouldnt have to consider it when finding the center, why would this be a wrong assumption?
I've checked your calculations. You got the correct distance r. Now, proceed and calculate the inertia of the system.
 
  • #7
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I've checked your calculations. You got the correct distance r. Now, proceed and calculate the inertia of the system.
1.9(0^2) + 8.77(L^2) + 148(r_cm^2)
r_cm=.822L
therefore, 8.77L^2+121.656L^2
I=130.426L^2
 
  • #8
SteamKing
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1.9(0^2) + 8.77(L^2) + 148(r_cm^2)
r_cm=.822L
therefore, 8.77L^2+121.656L^2
I=130.426L^2
The distances x = 0 and x = L are not measured relative to the center of mass of the rod and the iron balls. The problem asks you specifically to calculate MOI about the center of mass.

Also, since you know the mass of each ball and the material it is composed of, you may not be able to treat each ball as a point mass for calculating inertia, especially the 148-kg ball.
 

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