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Why do I get a different answer with the parallel axis theorem?

  • Thread starter catmunch
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Why do I get a different answer with the parallel axis theorem? [Solved]

Homework Statement



Imagine four points masses m1 = m3 = 3kg and m2 = m4 = 4 kg. They lie in the xy plane with m1 at the origin, m3 at (0, 2), m2 at (2, 2), and m4 and (2, 0). Each unit on the coordinate plane corresponds to a distance of one meter. What is rotational inertia of the system about the z-axis?


Homework Equations



Rotational inertia = sum of component inertias

inertia for a point mass = m*r^2

parallel axis theorem: rotational inertia around new axis = old rotational inertia + mass of system * (distance of new axis from old axis)^2

The Attempt at a Solution



Because m1 is located on the axis of rotation, I ignore it.
inertia of m3 = 3kg * (2m)^2 = 12kg * m^2
distance of m2 from z-axis = sqrt((2m)^2 + (2m)^2) = 2sqrt(2)m
inertia of m2 = 4kg * (2sqrt(2)m)^2 = 32kg * m^2
inertia of m4 = 4kg * (2m)^2 = 16kg * m^2
total = 60kg*m^2

This seemed to be fine, but my book gave me an answer of 56 (with no explanation). Just for kicks, I decided to calculate the rotational inertia of the system around an axis in the center of the four masses (the line defined by x = y = 1).

Then, the masses are all equidistant from the axis with a radius of sqrt(2)m.
So the rotational inertia (through the new axis) would be (3kg + 3kg + 4kg + 4kg) * (sqrt(2)m)^2, or 28kg*m^2.

Then, I used the parallel axis theorem to find the rotational inertia of the system through the z-axis.

new rotational inertia = 28kg*m^2 + (sqrt(2)m)^2*28kg = 56kg

Can anyone see what I'm doing wrong?
 
Last edited:

Answers and Replies

  • #2
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Could my problem be that I'm applying the parallel axis theorem to a simultaneous shift in both the x and y dimensions?
 
  • #3
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Could my problem be that I'm applying the parallel axis theorem to a simultaneous shift in both the x and y dimensions?
I just realized my problem. When using the parallel axis theorem, you need to plug the rotational inertia around the center of mass into the formula. I just took the line that was equidistant from the points.

And then, with the correct center of mass, I again calculate 60kg*m^2. I guess my book must have had a typo. If the authors of my book had mistaken a 3kg mass for a 4kg mass, with a radius of 2m, that would account for the discrepancy.
 
  • #4
ehild
Homework Helper
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It must be a typo. It happens quite often with books. Your solution is correct.

ehild
 

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