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Moment of Inertia of a hollow cone about its base

  1. Nov 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the moment of inertia of a hollow cone of mass M, radius R, and height h about its base is ##\frac{1}{4}M(R^2+2h^2)##

    2. Relevant equations
    ##I=\int r^2dm##
    where r is the perpendicular distance from the axis

    Surface Area of a cone ##= \pi R (R^2+h^2)^{1/2}##

    3. The attempt at a solution

    Mass density ##\sigma = \frac{M}{SA} = \frac{M}{\pi R(R^2+h^2)^{1/2}}##

    Now ##I = \int r^2dm = \int r^2\sigma dA = \sigma \int r^2dA##
    MOI.png

    Now the distance from its base is given by ##r = r(x) = h(1-x/R)## and from the image I made it appears the area element is simply ##dA = 2\pi xds##, and ##ds = \sqrt{dx^2+dy^2} = \sqrt{1+y'(x)^2}dx##. Trivially ##y'(x)^2 = (h/R)^2## giving ##dA = 2\pi x \sqrt{1+(h/R)^2}dx## and $$I = \sigma \int r^2dA =\sigma \int_0^R h^2\left(1-\frac{x}{R}\right)^22\pi x \sqrt{1+\left(\frac{h}{R}\right)^2}dx \\ I = 2\pi\sigma h^2\sqrt{1+\left(\frac{h}{R}\right)^2}\int_0^R \left(1-\frac{x}{R}\right)^2 xdx$$

    which clearly isn't going to give the answer since there's no way ##\sqrt{1+(h/R)^2}## will be cancelled
     
  2. jcsd
  3. Nov 21, 2016 #2

    BvU

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    You sure ? What about the integral to calculate ##\sigma## ?
     
  4. Nov 21, 2016 #3

    Ray Vickson

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    Formulas for surface area are poorly presented on the web, and some of them seem wrong. The more credible sources give
    $$\text{SA} = \pi R \left( R + \sqrt{R^2+h^2} \right), $$
    but one should make absolutely sure by calculating the relevant integral.
     
  5. Nov 21, 2016 #4
    Whoops, I forgot to mention it's an open ended cone. I.e. There is no base portion ##pi R^2##
     
  6. Nov 21, 2016 #5

    BvU

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    Basically you have to calculate $$\int r^2 dm\over \int dm $$
     
  7. Nov 21, 2016 #6
    Could you elaborate on why I need to calculate this quantity when moment of inertia is given by ##I = \int r^2 dm##?
     
  8. Nov 21, 2016 #7

    Ray Vickson

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    Yes, it will cancel because it appears as well in the denominator of your equation for ##\sigma##.
     
  9. Nov 21, 2016 #8
    Hmm, evaluating that integral gives ##I= Mh^2/6## which unfortunately isn't the correct answer
     
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