6. Moments of Inertia

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 = md²

where d is the radius of rotation.

Inertia for a Collection of Particles

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

I = m₁d₁²+ m₂d₂²+ m₃d₃² ... + m_nd_n²

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

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

I = (m₁+ m₂+ m₃ ... + m_n)R²

R is called the radius of gyration.

Example:

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)

Answer: 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 ... + m_n)R²

211 = (6 + 5 + 9 + 2)R²

So R ≈ 3.097

This means a mass of 22 units placed at (3.1,) 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, I_y of the given area, which is rotating around the y-axis.

Each "typical" rectangle indicated has width dx and height y₂ − y₁, so its area is (y₂ − y₁)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:

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

where m is the mass of the area.

Example:

For the first quadrant area bounded by the curve

y = 1 − x²,

find:

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

b) The mass of the area

c) Hence, find the radius of gyration

Answer: Of course, we sketch the part of the circle in the first quadrant first.

Here, y₂ = 1 − x², y₁ = 0, a = 0 and b = 1.

a) Finding I_y :

b) The mass of the area, m.

Now m = kA, where A is the area.

Now

So m = kA = 2k/3

c) The radius of gyration:

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.