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Moment of inertia of a semicircle

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    A uniform disk of radius R is cut in half so that the remaining half has mass M. (Figure (a))

    YF-09-31.jpg

    1. What is the moment of inertia of this half about an axis perpendicular to its plane through point A? Express your answer in terms of the given quantities.

    2. Why did your answer in part A come out the same as if this were a complete disk of mass M? Express your answer in terms of the given quantities.

    3. What would be the moment of inertia of a quarter disk of mass M and radius R about an axis perpendicular to its plane passing through point B? (Figure (b))

    2. Relevant equations

    Moment of inertia of a semicircle.

    [itex]I_{P}=I_{cm}+Md^{2}[/itex]

    [itex]I_{cm}=I=\sum_{i}m_{i}r_{i}^{2}[/itex]

    3. The attempt at a solution

    Is d=R?

    1. [itex]I_{P}=I_{cm}+Md^{2}=MR^{2}+MR^{2}=2MR^{2}[/itex]

    Does this make sense?
     
  2. jcsd
  3. Oct 12, 2011 #2

    vela

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    No, your work doesn't make sense. You're applying the parallel-axis theorem. If you don't know what a variable stands for, look it up in your book or notes. Don't just guess. What do d and Icm correspond to? Does it make sense to apply the theorem to this problem?
     
  4. Oct 12, 2011 #3
    Can I treat it as a solid cylinder with [itex]I=\frac{1}{2}MR^{2}[/itex] since I know that it has mass M?

    I'm not sure how to go about it being a semicircle?
     
  5. Oct 12, 2011 #4
    Okay, I figured out that the answers to 1. and 3. is [itex]I=\frac{1}{2}MR^{2}[/itex] but I really want to understand why.
     
  6. Oct 12, 2011 #5

    vela

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    No, you can't, because it's not a solid cylinder.

    You're expected to derive the answer using
    [tex]I = \int r^2\,dm[/tex]
     
  7. Oct 12, 2011 #6
    Okay, I think I understand. Please correct me if I'm wrong.

    Is it because the mass is always M? We always have to use the moment of inertia of a whole circle with mass M which is [itex]I=\frac{1}{2}MR^{2}[/itex]. But in this case the mass of half of the circle is M so 2M for the whole circle. So the moment of inertia of the whole of the circle is [itex]I=2*\frac{1}{2}MR^{2}=MR^{2}[/itex] and then the moment of inertia of half of the circle is [itex]I=\frac{1}{2}MR^{2}[/itex].

    Same goes for the quarter of a circle.

    Am I right?
     
  8. Oct 12, 2011 #7

    vela

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    Sort of, but you're missing the key point.

    Suppose you had a ring and solid circle, both of mass M and radius R. The ring has a moment of inertia equal to MR2 whereas the circle's is half that. Why are they different?
     
  9. Oct 12, 2011 #8
    I'm not sure. Is it because the ring is hollow?

    From the definition of moment of inertia I don't see why they should be any different. I don't understand this well enough.

    Thank you for all the help. :)
     
  10. Oct 13, 2011 #9

    vela

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    The moment of inertia is the mass of the object times the mass-weighted average of the squared distance from the axis. Because the ring is hollow, all of its mass has to sit at a distance R from the center; hence, you have <r2>=R2 and I = MR2. With a solid disk, the mass is spread out. Only a small amount of the mass is at a distance R from the center, and the rest is closer. When you work out the math, you find <r2>=R2/2, so I=1/2 MR2.

    Can you see why <r2> would be the same for the full circle, the half circle, and the quarter circle?
     
  11. Oct 13, 2011 #10
    Okay I understand the explanation. I want to look at [itex]I=\int r^{2}dm[/itex] that you told me to use in the first place.

    How do I use that to give me the correct answer [itex]I=\frac{1}{2}M R^{2}[/itex]? If I'm supposed to integrate [itex]r^{2}[/itex] then it's [itex]\frac{1}{3}r^{3}[/itex].
     
  12. Oct 13, 2011 #11

    vela

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    What we want to do is look at little strips of constant r. In the full, solid circle, this would be a circle of radius r (not R). If we increase r by the infinitesimal amount dr, we can approximate that the density ρ is constant for the entire strip, and the infinitesimal area dA is simply the length of the strip multiplied by the width dr. The mass dm of that strip is therefore given by [itex]dm=\sigma\,dA = \sigma (2\pi r)dr[/itex].

    That's an example of the basic procedure you need to follow to figure out dm for various shapes. You find the shape corresponding to constant distance r from the axis of rotation. Multiply its length by dr to get the area, and then multiply by the density to get dm. If you're working with a three-dimensional shape, you'll get a surface of constant r, e.g. a cylinder. You find its area and then multiply by dr to get a volume. Again, you multiply by the density to get dm. If you go through the examples in your book, you'll see that's all they're doing each time.

    To get the moment of inertia, you need to integrate
    [tex]I = \int r^2\,dm = \int_0^R r^2\sigma(2\pi r)dr = \frac{1}{2}\sigma\pi R^4 = \frac{1}{2} (\sigma \pi R^2) R^2 = \frac{1}{2} MR^2[/tex]where [itex]M=\sigma A = \sigma \pi R^2[/itex].

    To connect this back to what I said in my previous post, suppose we were to pull M out right at the start, so we have
    [tex]I = M\int r^2\,\frac{dm}{M}[/tex]The quantity dm/M is the fraction of the total mass M that is at a constant distance r from the axis. Can you see why the integral gives you the mass-weighted average <r2>?

    In the case of the circle, dm/M is equal to
    [tex]\frac{dm}{M} = \frac{\sigma(2\pi r)dr}{\sigma\pi R^2} = \frac{2r\,dr}{R^2}[/tex]Show that for the half circle and the quarter circle, you get the same expression for dm/M. Since dm/M is the same for all three shapes, you get the same average for r2 and hence the same moment of inertia.
     
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