1. The problem statement, all variables and given/known data Calculate the moment of inertia of a uniform triangular lamina of mass m in the shape of an isosceles triangle with base 2b and height h, about its axis of symmetry. 3. The attempt at a solution I've tried various things for this and never get the correct answer, 1/2*m*b^2. I'm beginning to think this may involve a double integral. Thanks.
I don't think ½mb² is the right result. Here is a similar example I did earlier: https://www.physicsforums.com/showthread.php?t=278184 In this case I think you would attack the sum of the x²*dm by observing that you can construct m in terms of x as something like h*(1-x/b) so that you arrive at an integral over an expression something like (hx² -x³/b)*dx. At the end you will be able note that the area of the lamina triangle times the implied density ρ yields you an M total mass in the product that defines your moment.
I have coded this problem as a double integral in Maple. > x(y):=b*(1-y/h); > rho:=M/(b*h); > dJ:=int(rho*z^2,z=0..x(y)); > J:=2*int(dJ,y=0..h); In the first line, the right boundary is defined. In the second line, the mass density is expressed. In the third line, the integration in the x-direction is performed from the axis of symmetry to the right edge In the fourth line, the integration is performed in the y-direction from bottom to top. The result is M*b^2/6. It is reasonable that h should not be in the result. The altitude should not affect this function, only the base width which describes how far the mass is distributed off the axis of rotation.
I did actually mean to put mb^2 / 6 in my first post. Thanks for replies. Last night I managed to get it myself aswell after spotting errors in my work. Thanks.