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

 

[nlightner]

Your approach is correct, however, your final answer is incorrect. The moment of inertia should have units of mass times length squared, whereas your answer has units of length squared. This is because you forgot to include the density in your final answer. The correct moment of inertia would be (phb/3) * b^2, where p is the area density. This can also be written as (p/3) * h * b^3, which has the correct units.