Physics - Moment of Inertia

Question

Consider an equilateral triangle cut from a thin board. The triangle has a mass M and an edge length L. Determine the moment of inertia of the triangle for rotation about an axis that bisects one of its angles.

Answer 1

Hi Megan!

 

The rotation axis described divides the equilateral triangle into two symmetric halves.  Around that axis, we need to evaluate:

 

I = ∫r²dm

 

which requires integrating over all the differential mass units.  If we work in Cartesian coordinates, and the equilateral triangle is of uniform mass distribution (which we have to assume, in the absence of any position dependence on the density), then we can define dm as:

 

dm = μdxdy

 

where μ is the mass per unit area, assumed constant.  So we get:

 

I = ∫∫r(x,y)(μdxdy) = μ∫∫r(x,y)dxdy

 

The r(x,y) represents the perpendicular distance of each mass element from the rotation axis.  If we choose to call the rotation axis the "y" direction, then r actually only has an x dependence (r(x)).  In fact, if we start from the axis as x = 0, r(x) = x out to the edge of the triangle.  There are a couple ways to proceed with the integral, then, and here is one.

 

The height of an equilateral is equal to the side length times the square root of 3, all divided by 2 [(L√3)/2].  At every y value, then, the triangle extends away from the axis by an amount equal to [(L√3)/2 - y]tan(30^o).  So, we have to integrate the x direction from 0 to that value, then integrate y from 0 to (L√3)/2.

 

(Someone more mathematically inclined may have a more elegant way to do this, but this is what I see right away)

 

So we end up with:

 

I = ∫₀^(L√3)/2∫₀^{{[(L√3)/2]-y}tan(30)}xdxdy

 

Evaluate this integral and then multiply it by two to account for both sides of the triangle.

 

The area of the triangle is (1/2)bh,  with b = L and h = (L/2)√3.  So M = μA = μ(L²√3)/4.  So, if you have that expression tucked inside the result of the integral, you can convert it to M.

 

I hope this helps get you started!  If you have any question, just let me know, or if you would like to check an answer.