Moment of Inertia of Shapes

[salman213]

Ok I am try to understand how to find the moment of inertia of different shapes by using direct integration. In a lot of the solutions I see they always have this one thing that I don't understand. Example:

Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.

http://img406.imageshack.us/img406/4816/64329189gz9.jpg


WHERE DOES THE 1/3 COME FROM??

I know the formula is

Iy = INTEGRALOF(x^2 dA)

doesnt da = xdy

so that makes it

Iy =INTEGRALOF(x^2 * x *dy)

 

[alphysicist]

Hi salman213,

It appears to me that the way they are setting up the integral is by dividing the area into infinitesimal strips of length x and height dy. Each of these strips has the form of a thin rod which is rotating about one end, and that is where the (1/3) comes from. (The moment of inertia of a thin rod rotating about one end is $\frac{1}{3}ML^2$.)

The integral then sums up the moments of inertia of these strips.

 

[Moetasim]

The 1/3 comes from the parallel axis theorem, which states that the moment of inertia of a shape about an axis parallel to its centroid is equal to the moment of inertia about its centroid plus the product of the area and the square of the distance between the two axes. In this case, the shaded area is not centered about the y-axis, so we need to use the parallel axis theorem to find the moment of inertia about the y-axis.

To understand this better, let's break down the formula for moment of inertia about the y-axis:

Iy = INTEGRALOF(x^2 dA)

First, we need to find the differential element of area, dA. In this case, it is a small strip of width dx and height x, so dA = xdx. This is because we are integrating along the y-axis, so we need to express the area in terms of y.

Next, we need to find the distance between the y-axis and the centroid of the shaded area. This distance is x, since the centroid is located at (x, x^2/2) in the coordinate system shown.

Now, using the parallel axis theorem, we can rewrite the formula as:

Iy = INTEGRALOF(x^2 dA) + Ax^2

where A is the area of the shaded region. Plugging in dA = xdx and A = 1/2, we get:

Iy = INTEGRALOF(x^2 * x *dx) + (1/2)x^2

which simplifies to:

Iy = (1/3)x^3 + (1/2)x^2

This is the same as the formula you mentioned, but with the added term (1/2)x^2 to account for the distance between the centroid and the y-axis.

In summary, the 1/3 comes from the parallel axis theorem, which is necessary to find the moment of inertia about an axis that is not passing through the centroid of the shape.