# 6. Moments of Inertia by Integration

by M. Bourne

The moment of inertia is a measure of the resistance of a rotating body to a change in motion.

The moment of inertia of a particle of mass m rotating about a particular point is given by:

"Moment of inertia" = md^2

where d is the radius of rotation.

## Inertia for a Collection of Particles

If a group of particles with masses m1, m2, m3, ... , mn is rotating around a point with distances d1, d2, d3, ... dn, (respectively) from the point, then the moment of inertia I is given by:

I = m1d12 + m2d22 + m3d32 +... + mndn2

If we wish to place all the masses at the one point (R units from the point of rotation) then

d1 = d2 = d3 = ... = dn = R and we can write:

I = (m1 + m2 + m3 ... + mn)R2

R is called the radius of gyration.

### Example 1

Find the moment of inertia and the radius of gyration w.r.t. the origin (0,0) of a system which has masses at the points given:

 Mass 6 5 9 2 Point (-3, 0) (-2, 0) (1, 0) (8, 0)

The moment of inertia is:

I = 6(-3)2 + 5(-2)2 + 9(1)2 + 2(8)2

= 54 + 20 + 9 + 128

= 211

To find R, we use:

I = (m1 + m2 + m3 ... + mn)R2

211 = (6 + 5 + 9 + 2)R2

So R ≈ 3.097

This means a mass of 22 units placed at (3.1, 0) would have the same rotational inertia about O as the 4 objects.

## Moment of Inertia for Areas

We want to find the moment of inertia, Iy of the given area, which is rotating around the y-axis.

Each "typical" rectangle indicated has width dx and height y2y1, so its area is (y2y1)dx.

If k is the mass per unit area, then each typical rectangle has mass k(y2 y1)dx.

The moment of inertia for each typical rectangle is [k(y2 y1)dx] x2, since each rectangle is x units from the y-axis.

We can add the moments of inertia for all the typical rectangles making up the area using integration:

I_y=kint_a^bx^2(y_2-y_1)dx

Using a similar process that we used for the collection of particles above, the radius of gyration Ry is given by:

R_y=sqrt((I_y)/m

where m is the mass of the area.

### Example 2

For the first quadrant area bounded by the curve

y = 1 − x^2,

find:

a) The moment of inertia w.r.t the y axis. (Iy)

b) The mass of the area

c) Hence, find the radius of gyration

As usual, first we sketch the part of the curve in the first quadrant. It's a parabola, passing through (1, 1) and (0, 1). A "typical" rectangle is shown.

In this example,

y_2= 1 − x^2, and y_1= 0, a = 0 and b = 1.

a) Finding Iy:

I_y=k int_a^b x^2(y_2-y_1) dx

=k int_0^1x^2 [(1-x^2)-(0)] dx

=k int_0^1 [x^2-x^4] dx

=k[(x^3)/(3) - (x^5)/(5)]_0^1

=k(1/3-1/5)

= (2k)/15

b) The mass of the area, m.

Now m = kA, where A is the area.

Now

A=int_0^1 (1-x^2) dx

=[x-(x^3)/(3)]_0^1

=1-1/3

 = 2/3

So m = kA = (2k)/3

R_y=sqrt{(I_y)/(m)}

=sqrt{(2k//15)/(2k//3)}

=sqrt[1/5]

 approx 0.447

This means that if a mass of (2k)/3 was placed 0.447 units from the y-axis, this would have the same moment of inertia as the original shape.

Get the Daily Math Tweet!

For rotation about the x-axis, the moment of inertia formulae become:

I_x=kint_c^(d)y^2(x_2-x_1)dy

and

R_x=sqrt((I_x)/m

### Example 3

Find the moment of inertia and the radius of gyration for the area y=x^2+1 from x=1 to x=2, and y>1, when rotated around the x-axis. The mass per unit area is 3 kg m−2.

We sketch the parabola. The shaded area is the part that's rotating around the x-axis, and we have indicated a "typical" rectangle.

In this example, we are rotating the area around the x-axis. We need to express our function in terms of y.

Since we started with y=x^2+1, we solve for x and obtain:

x=sqrt(y-1)

We take the positive case only, as we are dealing with the first quadrant.

So the required values which we can use in the formula are:

x_2= 2 (the "curve" furthest from the y-axis),

x_1= sqrt(y-1) (the curve closest to the y-axis), and

c = 1 and d = 5; and k=3.

a) Finding Ix:

I_x=k int_c^d y^2(x_2-x_1) dy

=3 int_1^5 y^2 [(2) - sqrt(y-1)] dy

=3 int_2^5 (2 y^2 - y^2 sqrt(y-1)) dy

We use a computer algebra system to obtain:

=45.5 kg m2

b) Now to find the mass of the area, m.

m = kA, where A is the area.

Now

A=int_1^2 (x^2+1) dx - 1 (We need to subtract the area of the 1xx1=1 square below the shaded area.)

=[(x^3)/(3)+x]_1^2 - 1

=(8/3+2)-(1/3+1) - 1

 = 7/3

So m = kA = 3xx7/3 = 7 kg

R_x=sqrt{(I_x)/(m)}
=sqrt{(45.5)/(7)}
 ~~ 2.6 m
This means that if a single mass of 7 kg was placed 2.6 m from the x-axis, this would have the same moment of inertia as the original shape when rotating around the x-axis.