Stanton D. answered • 02/04/20
Tutor to Pique Your Sciences Interest
This is a quite complex problem, first from the standpoint of balancing the cone, and then from setting up the integrations for moment of inertia. So I'll present it in several pieces of "answers", corresponding to steps of the task. (That's also operationally wise, since the text editor here has a tendency to perceive inadvertant mouse pad touches, and shift away from text entry, losing all entered text!)
So, the first task: determine the center of mass, and balance characteristics of the cone as stated in the question.
Designate the height of the cone as h, the area of the base as b, then volume V=(1/3)bh. More specifically for the present cone, with a central angle of 60°, hence angle-from-axis-to-side of 30°, this is V = (1/3)h*π(h/3^0.5)^2 or V=πh^3/9 .
Since the volume (and thus the mass) of the cone depends on h^3, then a cone of (1/2) the initial mass will have height = h/2^(1/3), and the CM will lie at h/2^(1/3) from the apex. (Since 1/2 the initial volume lies above, 1/2 the initial volume must lie below.)
Now, in order to balance this cone on a frictionless support, the cone surface must be parallel with the ground. This requires rotating the cone to this position. The apex, CM, and point of support now describe a 30-60-90° triangle with the 60° angle at the CM; hence the distance from the apex to the point of support (down the cone surface) is (h/2^(1/3))*3^0.5/2, or h2^(-2/3)3^0.5 . Note that the cone requires an initial "boost" to get it balanced, but that isn't part of the "frictioness" requirement; if you initially visualized the cone axis as parallel to the ground, that was arbitrary. The balance point is only a metastable one, that's inherent to a cone!
Now, we're still not out of the woods, so to speak, with respect to setting up our integration: it's only convenient to integrate the cone as slices parallel to its base, so we must set up yet another 30-60-90° triangle, this one from the point of support, in a line perpendicular to the central axis; that intersects the central axis at a distance of h2^(-2/3)3^0.5*(3^0.5/2) or 3h*2^(-5/3) from the apex.
That gives us the essential basis for setting up the moment of inertia calculation, which is the next answer in the set!
Stanton D.
OK. Now to set up the integration for moment of inertia. It's convenient to do this in 2 pieces, one towards the apex, and one towards the base, of the cone, from the plane perpendicular to the axis from the supporting tangent (designate this as plane P for reference). The most natural coordinate grid would appear to be with x coordinate axis along the cone axis, from 0 at the apex; the y-coordinate axis, from the point of tangency perpendicularly to the axis, with 0 at the point of tangency; and the z coordinate axis, distance laterally from the cone axis perpendicular to the other 2 axes (i.e., parallel to the axis of rotation of the problem). This enables the moment of inertia to be integrated in slices from the apex to the plane P (and eventually likewise for the other portion of the cone). Now, the moment of inertia calculation uses r^2 (distance from the axis of rotation , squared) as its weighting factor for the mass increments. Therefore, only x and y distances will figure into the calculation of r, though the range of z for the needle-like slice will be a multiplier for each particular value of x and y. Therefore, the x-coordinate integration limits are 0 and 3h*2^(-5/3); the y-coordinate limits are 0 and x*3^(-1/2)*2, and the z-coordinate slice length subtended = 2*((x^2/3)^2 - (x*3^(-1/2)-y)^2 )^0.5 [derivation: 2 accounts for +- limits on z; z^2+yslice^2 = r^2 for the total circular slice outline, and r = (x*3^(-1/2)) which also serves as the center of the slice circle for measuring yslice == the re-positioning of y vs. the center of the slice circle]. So putting this together, we have I (moment of inertia) = density * int[x=0 to 3h*2^(-5/3)] int [y=0 to x*3^(-1/2)*2] * 2*((x^2/3)^2 - (x*3^(-1/2)-y)^2)^0.5 * ((3h*2^(-5/3)-x)^2 + y^2) dy dx At this point, given the non-separation of variables, the prudent course would be numerical integration! (This integral does exceed Wolfram Alpha basic time limits, for a definite integral solution, even for a unit value for h). The other portion of the cone would be similarly treated, with appropriate redesignation of limits and distance function (measuring x displacement from plane P as a positive number). I'm going to leave the setup of that to you readers! Those of you in engineering might wonder, aren't moments of inertia of cones already solved out there? Of course they are, but only for rotations around the axis of the cone; this was quite a more ambitious task, with the cone axis skew with respect to the rotation axis, giving much more involved functions for distance from the axis of rotation, and for integration limits. -- Cheers, -- Mr. d.02/04/20