Moment of Inertia of a Solid ConeDate: 02/03/99 at 04:25:35 From: Anonymous Subject: MI of Solid Cone How do you find the Moment of Inertia of a solid cone, in terms of height h and base radius r? I know how to find the MI of a solid sphere; do I use the same method? If so, what do I take as the arbitrary disc radius? Should I model the cone as sitting on the axis, with its apex at (0,0) and its height as the maximum x along the x-axis? Date: 02/03/99 at 14:37:05 From: Doctor Anthony Subject: Re: MI of Solid Cone You must, of course, specify about which axis you want the moment of inertia. For the MI about the major axis you would use the MI of a disk about an axis perpendicular to the disc and integrate from 0 to h. Have the cone with its axis along the x-axis and its vertex at the origin. If base radius is r and height h, then the radius at distance x from the origin is given by (r/h)x. The MI of a disc of mass m and radius a perpendicular to its plane is ma^2/2 For our elementary disk at point x the volume is pi.radius^2.dx = pi.(r/h)^2.x^2.dx and the MI of this elementary disc is (1/2)pi.(r/h)^2.x^2.(r/h)^2.x^2.dx = (pi/2)(r/h)^4.x^4.dx and integrating between 0 and h we get = (pi/2)(r/h)^4.h^5/5 = (pi/2)r^4.(h/5) The volume of the whole cone is (1/3)pi.r^2.h So we have MI = (1/3)pi.r^2.h[(3/10).r^2] = (3/10)M.r^2, where M = mass of cone --------------------------------------------------------------------- Moment of Inertia about axis through Vertex perpendicular to axis of cone --------------------------------------------------------------------- Now we use the MI of a flat disk about a diameter and then use the parallel axis theorem to get MI about the y-axis. The MI of a disk about a diameter is ma^2/4 So the MI of the elementary disk about the diameter is (1/4)pi.(r/h)^2.x^2.(r/h)^2.x^2.dx and the MI about the y-axis is (pi/4)(r/h)^4.x^4.dx + pi.(r/h)^2.x^2. x^2.dx = (pi/4)(r/h)^4.x^4.dx + i.(r/h)^2.x^4.dx and integrating, = (pi/4)(r/h)^4.(h^5/5) + pi.(r/h)^2.h^5/5 = (pi/20)r^2.h[r^2 + 4h^2] = (3/20)M[r^2 + 4h^2] - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/