Calculate the moment of inertia of a right triangle

[homeslice64]

Homework Statement

A right triangle has height 'h' and width 'b.' The right triangle has a constant area density. Calculate the moment of inertia of the triangle rotated around an axis that runs along side 'h.'

Homework Equations

I = integral(r^2*dm) where 'r' is distance from the axis

The Attempt at a Solution

equation of hypotenuse is (h/b)

r^2 dm = r^2 * p * dA where p is area density and dA is the area of the rectangles.

= r^2 * p * (h/b)r * dr = r^3 * p * (h/b) * dr --> integrate =
( (ph)/(3b) )r^3 with the limits of integration being from r=0 to r=b so:
( (ph)/3 )b^2 - 0 so I = ( (ph)/3 )b^2

But all my friends said I was wrong so can someone please tell me why?

 

[tiny-tim]

Hi homeslice64! Welcome to PF!

(have an integral: ∫ and try using the X² tag just above the Reply box )

= r^2 * p * (h/b)r * dr = r^3 * p * (h/b) * dr --> integrate =
( (ph)/(3b) )r^3 with the limits of integration being from r=0 to r=b so:
( (ph)/3 )b^2 - 0 so I = ( (ph)/3 )b^2

erm … ∫r³dr isn't r³/3