- #1
fluidistic
Gold Member
- 3,924
- 261
Homework Statement
Determine the principal moments of inertia of a circular cylinder with radius R and height h.
Homework Equations
Not sure.
The Attempt at a Solution
This is the first problem of this kind I attempt to solve.
From what I've read on wikipedia, the tensor of inertia can be written as [itex]\mathbf{I} = \begin{bmatrix}I_{11} & I_{12} & I_{13} \\I_{21} & I_{22} & I_{23} \\I_{31} & I_{32} & I_{33}\end{bmatrix}[/itex].
And if I'm lucky, I can find a Cartesian system of coordinates such that [itex]\mathbf{I} = \begin{bmatrix}I_{1} & 0 & 0 \\0 & I_{2} & 0 \\0 & 0 & I_{3}\end{bmatrix}[/itex] where [itex]I_i[/itex] are what I'm looking for. Hence I guess I must approach the problem in finding such a Cartesian system of coordinate. By intuition I'd set it centered over the center of mass of the cylinder. Say the y-axis goes along the height of the cylinder and the x-axis goes along a radius of the cylinder (so that the z-axis also goes along a radius of the cylinder).
But I've no idea how to calculate the [itex]I_i[/itex]'s. It likely involve some integral (simple? double? triple?) but I don't know how to set them.
Any explanation/tip would be great.
P.S.:The solution, I think, can be found in wikipedia. I see there that [itex]I_1=\frac{m(3R^2+h^2)}{12}=I_2[/itex]. While [itex]I_3=\frac{mR^2}{2}[/itex].
Edit: I found http://scienceworld.wolfram.com/physics/MomentofInertiaCylinder.html.
At first glance I don't see how to get rid of the "pi" constant in the denominator of rho, the density. Maybe when evaluating the integral with respect to x.
I've never seen a matrix being integrated before, not sure how to tackle this.
Last edited: