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, ... , 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:
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 ... + mn)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, Iy 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 Ry 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. (Iy)
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 Iy :

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.
Note: For rotation about the x-axis, the formulae become:
and
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