# Moments of inertia by integration

1. Jan 23, 2012

### Lucy Yeats

1. The problem statement, all variables and given/known data

A mass distribution in the positive x region of the xy-plane and in the shape of a semicircle of radius a, centred on the origin, has mass per unit area k. Find, using plane polar coordinates, its moments of inertia about the x and y axes.

2. Relevant equations

3. The attempt at a solution

I know that I=∫r^2.dm
dm=k.dA=k.r.dr.dθ in polar coordinates.
So I=k∫dθ∫r^3.dr
But I'm not sure about the limits, or how to differentiate between the moments of inertia about the x and y axes.

Any help would be brilliant.

2. Jan 23, 2012

### I like Serena

For the moment of inertia about the x-axis, the "r" in your formula for I is the y-coordinate, and not the polar distance to the origin.

Same thing for the moment of inertia about the y-axis, where "r" is the absolute value of the x-coordinate.

3. Jan 23, 2012

### A.V

For a planar object which lies in the x − y plane, Ix = ∫ x^2dm and Iy = ∫ y^2dm and according to perpendicular axis theorem , Iz = ∫ (x^2 + y^2)dm = ∫ r^2dm is the rotational inertia around the z axis and r is the distance from the z axis.
In that case r changes from 0 to a and since the disk is semicircular , θ will change from 0 to Pi .

Last edited: Jan 23, 2012
4. Jan 23, 2012

### I like Serena

I think you mean Ix = ∫ y^2dm and Iy = ∫ x^2dm.

5. Jan 27, 2012

### Lucy Yeats

Oh, I see. I wrote rcosθ instead of r for the moment of inertia around the x axis, and rsinθ for the y axis, and got the right answers.

Thanks!