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I'm not understanding the very basics of Moment of Inertia

  1. Apr 21, 2008 #1
    1. The problem statement, all variables and given/known data
    A 13cm diameter CD has a mass of 25 g.
    What is the CD's moment of inertia for rotation about a perpendicular axis through its center?

    2. Relevant equations
    [tex]I = \int r^2 dm[/tex]

    and possibly...

    Moment of inertia of cylinder or disk, about center = [tex]\frac{1}{2}MR^2[/tex]


    3. The attempt at a solution
    I'm given certain values. The mass of the CD is .025 kg and it's radius is .065 m. Other than this, I am very confused. I tried just plugging in (.025)(.065)^2, but that didn't work.

    Now I've checked through other threads regarding moment of inertia, but it's not exactly helpful unless I know the very basics. In all honesty, I've never even done integrals before; my calculus class is supposed to parallel what we are doing in physics, but they are doing a very poor job at it. I think I might have the hang of them from looking it up quickly online, though (I'm just a pain, huh?).

    I know with such a very limited amount to work with here it can be difficult, but I catch on really quickly (I swear!), but for some reason this is a little tricky. Any pointers in the right direction would be greatly appreciated!
     
  2. jcsd
  3. Apr 21, 2008 #2
    Wikipedia is your friend.

    The moment of inertia represents an object's resistance to angular acceleration about an axis. If you have two wheels of the same dimensions, but one is made out of lead and the other out of styrofoam, and you apply the same moment to each (by, for example, pushing on both with the same force at the same point), the lead wheel will spin much slower. This is because it has a much larger moment of inertia.

    It is very important to keep track of what axis you are rotating the object around, as the same object will have a different moment of inertia depending on which axis it is rotating about. If you think about it, this makes sense--it's much easier to spin a dumbbell along the bar's axis than it is to spin it end-over-end.


    edit: Also, since doing that integral you posted is usually a big pain, there are tables of moments of inertia for commonly-shaped objects. Using these tables along with the Parallel Axis Theorem usually lets you calculate moments without having to resort to ugly integrals.

    Also, for the problem you posted, you forgot to multiply by 1/2
     
    Last edited: Apr 21, 2008
  4. Apr 21, 2008 #3
    Also, for the problem you posted, you forgot to multiply by 1/2

    That I did. So I have the correct answer now, and I'm pretty sold on the concept of Moment of Inertia. I had already read up a bit on it, but it was a simple calculation error that had me thrown off the whole time.

    What if the axis is at the edge of the disk rather than the center of the disk?
     
  5. Apr 22, 2008 #4
    That's where the Parallel Axis Theorem comes into play.
     
  6. Apr 22, 2008 #5
    Got it. Thanks.
     
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