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Homework Help: Moment of Inertia Ratio

  1. Apr 2, 2017 #1
    1. The problem statement, all variables and given/known data
    What is the ratio of inertia of a 5 meter radius disk to that of a 5 meter radius ring? How would this ratio change if the ring were replaced with a since point mass a distance 5 meters from a pivot point?

    2. Relevant equations
    I = mr2

    3. The attempt at a solution
    The wording in this problem seemed a bit funky to me, but going through it, what I think I figured out is that the inertia would be the mass of the disk times 25. However, since it doesn't give me a mass and just asks me to compare the inertia ratio, I'm also guessing that there is something else that separates the inertia of a disk and a "ring." It could just be mass, since a ring doesn't have mass (assuming he's talking about a theoretical ring, just a circle around an axis essentially), and after that, I don't have any more guesses. Any help would be very appreciated! Thanks in advance!
  2. jcsd
  3. Apr 2, 2017 #2
    Photo shows a ring and a disk. Both have mass.
  4. Apr 2, 2017 #3
    Is it the amount of mass then? For instance, a ring would begin moving faster than a disk if pushed, assuming no outside forces acted on it...
  5. Apr 2, 2017 #4
    Because you are calculating the ratio of 2 inertias, the mass will cancel out.
    Why would a ring begin moving faster than a disk if pushed?
  6. Apr 2, 2017 #5
    Sorry, I meant, the ring would roll farther than the disk in the long run because it has less mass slowing it down...
  7. Apr 2, 2017 #6
    If you are asked to calculate a ratio of two inertias, you have to use equal masses in both cases. But your statement makes me wonder what would cause an object with more mass to slow down faster.
  8. Apr 2, 2017 #7
    I thought it would be because of friction and gravity combining to slow it down. And use equal masses? Then how would there be any difference? :/
  9. Apr 2, 2017 #8
    You wrote an equation in your original post. I = mr2
    You need to understand what that equation applies to - and also to understand that objects that have the same mass do not necessarily have the same moment of inertia. Specifically, a ring that has the same radius and mass as a disk does not have the same moment of inertia as that disk. There are normally tables that people use to look up moments of inertia for various shapes. But you could also derive them yourself - if you are feeling ambitious. :)
  10. Apr 2, 2017 #9
    Ah, so the ratio would be 2:1 in favor of the ring. That makes sense to me, though I'm still slightly unsure about the "moment of inertia" concept, though I'm sure I can learn that eventually. What about the second half of the problem? "How would this ratio change of the ring were replaced with a since point mass a distance 5 meters from a pivot point?" It didn't make sense to me at all, wording wise. A "since point mass" isn't something I have ever heard of..
  11. Apr 2, 2017 #10
    Yes, the ring would have a moment of inertia 2 times greater than the disk. A point mass means simply means that the total mass is contained at a single infinitesimally small point - not spread out over the shape of a ring or a disk or a hollow sphere. The whole mass is contained at one single point in space. But the equation in your original post is valid for any point mass. So what is the moment of inertia for a mass at a single point that is 5 meters away from the pivot point?
  12. Apr 2, 2017 #11
    Well, since I still don't have a specific mass, I may have to put it in terms of m, leaving m as a variable...but at the same time, would the radius become zero, since all mass is concentrated at one point instead of people spread across it? I'm attempting to visualize it, and I'm seeing a circle on a graph plane with one specific dot of mass, kind of like a satellite orbiting the earth.
  13. Apr 2, 2017 #12
    No, the radius would become 5 meters, since that is how far away from the pivot point the problem specified.
    By the way, I assume the original problem statement should have said, "single point mass" instead of "since point mass", true?
  14. Apr 2, 2017 #13
    I guess, yeah. I just always type exactly what my professor sends us. :/ and would the ratio change at all then? Since it remained at 5m? The equation for the disk would remain the same at 1/2mr^2, but the equation for the single point mass would also remain at mr^2...right?
  15. Apr 2, 2017 #14
    Yes, a point mass is same inertia as a ring.
  16. Apr 2, 2017 #15
    Awesome! Thank you so much for the help, as always, Tom! :D
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