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

1. Dec 10, 2008

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.

2. Dec 10, 2008

### LowlyPion

I don't think ½mb² is the right result.

Here is a similar example I did earlier:

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.

3. Dec 10, 2008

### 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.

4. Dec 11, 2008

### LowlyPion

Happily algebraic methods arrive at the same result.

5. Dec 11, 2008