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Find its moment of inertia about its axis of symmetry

  1. Nov 6, 2008 #1
    1. The problem statement, all variables and given/known data
    A cone of mass M has a height h and a base diameter R. Find its moment of inertia about its axis of symmetry

    2. Relevant equations
    [tex]I=\int R^{2}dm[/tex]

    3. The attempt at a solution

    I tried starting by finding dm. It's equal to the mass per unit volume, [tex]\frac{M}{\frac{1}{3}\pi R^{2}h}[/tex], times a small dV. I broke up the cone into small disks of area [itex]\frac{1}{2}\pi x^{2}[/itex] (x being the radius of the disk) and height dy. I then found x in terms of y through similar triangles: x/y = R/h, so x = (Ry/h). The equation for dV then becomes [tex]dV=\pi\left(\frac{yR}{h}\right)^{2}dy[/tex]. dm then becomes (after simplifying) [tex]dm=\frac{3My^{2}}{h^{3}}dy[/tex], so the whole integral for I is [tex]\int_{0}^{h}R^{2}\frac{3My^{2}}{h^{3}}dy=\frac{3R^{2}M}{h^{3}}\int_{0}^{h}y^{2}dy[/tex], which evaluates to just MR2. And I definitely don't think that's right, because it seems like a cone would have a smaller moment of inertia than a hoop.

    So because that method didn't work for some reason, I tried polar coordinates, breaking up the cone into a series of right triangles rotated around the center. The area of the triangle is just .5*R*h, so [tex]dm=\left(\frac{3M}{\pi R^{2}h}\right)\left(\frac{1}{2}Rhd\theta\right)[/tex] and the integral becomes [tex]\int_{0}^{2\pi}R^{2}\frac{3M}{2\pi R}d\theta=\frac{3RM}{2\pi}\int_{0}^{2\pi}d\theta=3RM[/tex] which doesn't seem right at all to me.

    I'm sure I screwed up the polar coordinates somehow (I've never actually learned them... I've just kind of seen them here and there and heard a little about how they work) but it seems like my first attempt should have worked... Can someone tell me what I did wrong? Thanks!
    Last edited: Nov 6, 2008
  2. jcsd
  3. Nov 7, 2008 #2


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    Homework Helper

    Hi bowma166,

    This last equation is not correct. In the formula you are following ([itex]I = \int r^2\ dm[/itex]), r2 is the distance from dm to the axis of rotation. But each part of your small disks is not a distance R from the axis, so this is not true.

    I would say you have two main choices. The first (and what I would do) is to use the formula:

    I = \int dI

    where dI is the moment of inertia of each of your very thin disks. (So you need to know the moment of inertia of a disk, which you can find in a table.)

    To use

    I = \int r^2 dm
    you can integrate over every piece of the object (not break it up into small disks), which will give you a three dimensional integral, and you will need to figure out the limits for the integral. For this approach, I would say that cylindrical coordinates are best.

    Both ways are straightforward, but the first is simpler. Does this help? What do you get?
  4. Nov 7, 2008 #3
    Oh... very true. I was being stupid there, sorry.

    I'm having trouble setting up the integral [itex]\int dI[/itex]. It seems like it should be easy to do but I just can't figure out how to do it... I'll blame my lack of sleep to make me feel better. But anyway. How would I go about doing that? I of a disk is [itex]\frac{1}{2}mr^{2}[/itex], so I tried integrating [tex]\int_{0}^{R}\frac{1}{2}mr^{2}dr[/tex] but I got 1/6 MR3. Where did that dr come from anyway? I just kind of threw it in there to try and make an integral work.

    Trying to integrate with respect to h then... Because [tex]h=\frac{Hr}{R}[/tex] the integral becomes [tex]I=\int_{0}^{H}\frac{Mh^{2}R^{2}}{H^{2}}dh=\frac{MR^{2}H}{6}[/tex] which also isn't right. I don't know why I'm having such problems with this...

    I have no idea how to do a three dimensional integral or use cylindrical coordinates so I don't think I'll attempt that.

    Thanks for your help!
  5. Nov 7, 2008 #4


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    Right, you cannot just throw a dr in there (for one thing it would have the wrong units).

    Instead, remember that you want to find dI. That is, your disk has a radius of x, and a height dy. The formula for the moment of inertia for a disk is:

    I = \frac{1}{2} M x^2

    Now speaking informally: for this very thin disk, we have a very small moment of inertia dI. In the formula just above, what is very small on the right hand side? It's not x, because the radius can be large. What is small is the mass, or in other words

    dI = \frac{1}{2}\ dm\ x^2
    and you already found an expression for dm and for x in your first post.
  6. Nov 7, 2008 #5



    Awesome. Thanks for the help.
  7. Nov 7, 2008 #6


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    You're welcome! And I'm fairly sure it was just a typo, but the above should be:

    I =\frac{3}{10} M R^2
  8. Nov 8, 2008 #7
    Oops. Yup, that was just a typo.
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