Calculate the moment of inertia of a uniform triangular lamina of mass

[mattgad]

Homework Statement

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.

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.

 

[LowlyPion]

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.

 

[Dr.D]

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.

 

[LowlyPion]

Happily algebraic methods arrive at the same result.

 

[mattgad]

I did actually mean to put mb^2 / 6 in my first post. Thanks for replies. Last night I managed to get it myself as well after spotting errors in my work. Thanks.

 

[kompheak vic]

how about the inertia product of this problem?