1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Moment of inertia flywheel mass problem

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the moment of inertia of a flywheel of mass M made by cutting four large holes of radius r out of a uniform disk of radius R. The holes are centered at a distance R/2 from the center of the flywheel.

    Here is a very poorly drawn diagram of the disk. Sorry for how crappy it is.
    [​IMG]

    2. Relevant equations
    [tex]I=\int R^{2}dm[/tex]
    (Is this necessary? I don't think it is given the next two equations.)

    [tex]I_{disk}=\frac{1}{2}MR^{2}[/tex]

    [tex]I=I_{CM}+md^{2}[/tex]


    3. The attempt at a solution
    You can just add/subtract moment of inertias of things, right? I'm not sure about that but that's what I tried.
    If the disk didn't have holes, then [itex]I=\frac{1}{2}MR^{2}[/itex]. But it does... the holes are of mass
    [tex]\frac{\pi r^{2}}{\pi R^{2}}M=\frac{r^{2}}{R^{2}}M[/tex]
    and are a distance R/2 from the center. So the total moment of inertia should be

    [tex]I=\frac{1}{2}MR^{2}-4\left(\frac{r^{2}}{R^{2}}\right)M\left(\frac{R}{2}\right)^{2}=\frac{1}{2}MR^{2}-Mr^{2}[/tex]

    right? It makes sense because as r -> 0, I -> I_disk, but I really have no idea if it is right or not. Thanks!

    ::edit:: Just kidding, I know I did that wrong... I'm working on redoing it now.

    ::edit again::
    Okay. This time I think I may have done it more correctly.
    The moment of inertia of each small disk about the center of the big disk should be
    [tex]I_{hole}=\frac{1}{2}\frac{r^{2}}{R^{2}}Mr^{2}+\left(\frac{r^{2}}{R^{2}}M\right)\left(\frac{R}{2}\right)^{2}=\frac{Mr^{4}}{2R^{2}}+\frac{Mr^{2}}{4}[/tex]
    so the total moment of inertia is the big disk's I minus four times the holes' I, right?
    [tex]I=\frac{1}{2}MR^{2}-4\left(\frac{Mr^{4}}{2R^{2}}+\frac{Mr^{2}}{4}\right)=\frac{1}{2}MR^{2}-\frac{2Mr^{4}}{R^{2}}-Mr^{2}[/tex]

    So...

    [tex]I=\frac{1}{2}M\left(R^{2}-\frac{4r^{4}}{R^{2}}-2r^{2}\right)[/tex]

    That seems kind of messy.
     
    Last edited: Nov 20, 2008
  2. jcsd
  3. Nov 20, 2008 #2
    Sometimes things are that way :smile: Blame it on your messy disk.

    You can add/subtract moments as long as they are about the same axis; so your second solution is fine.
     
  4. Nov 20, 2008 #3
    Oh, I didn't read the problem carefully enough. The flywheel has mass M, not the original disk...
     
  5. Nov 20, 2008 #4
    Oh... Does that not work that way? Hm. Well I tried again, finding the mass of the thing if the holes weren't removed and then replacing M in my calculations with that.
    The whole mass would have been
    [tex]M_{whole}=\frac{Mr^{2}}{R^{2}-4r^{2}}[/tex]
    right?

    I got something extremely messy for my final answer then... So messy that I'm pretty sure it can't be right. I think?

    [tex]I_{total}=Mr^{2}\left(\frac{R^{2}}{2R^{2}-8r^{2}}-\frac{2r^{4}}{R^{4}-4R^{2}r^{2}}-\frac{2r^{2}}{R^{2}-4r^{2}}\right)[/tex]

    ::edit:: Yeah, that can't be right... Oh damn! I accidentally switched r and R in my calculation of the mass. Time to do that all over again.

    ::edit 2::
    Okay. With the correct (I hope) mass with MR2 on top instead of Mr2, I got...
    [tex]I_{total}=\frac{MR^{4}-2Mr^{4}\left(2+R^{2}\right)}{2R^{2}-8r^{2}}[/tex]

    [tex]\lim_{r\rightarrow0}I_{total}=\frac{1}{2}MR^{2}[/tex]
    for this answer, so it has the possibility of being correct. Maybe.
     
    Last edited: Nov 20, 2008
  6. Nov 20, 2008 #5
    Yeah, it is a messy problem - you just have to be confident that you haven't made any errors in the process. For example you cannot add 2 to R^2 since the dimensions are all wrong :smile:
     
  7. Nov 20, 2008 #6
    But I got the (2+R2) from factoring 4Mr4 + 2MR2r4... Doesn't that work? Sigh. I give up on this problem. I'll just hope there isn't one like it on the test, hah.
     
  8. Nov 20, 2008 #7
    Your solution is correct, in that you've understood the physics. It is just a matter of going back and checking your math again. Good luck on the test!
     
  9. Nov 20, 2008 #8
    Ah, so finding the mass if there was nothing removed,
    [tex]M_{disk}=\frac{MR^{2}}{R^{2}-4r^{2}}[/tex],
    and calculating everything as I did with that is the right way to do things? Good. I'm not going to bother checking over the math though. Thanks for the help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Moment of inertia flywheel mass problem
Loading...