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Finding the Moment of Inertia

  1. Moments of Inertia are always determined relative to a specific axis of rotation.

  2. For particles (and only particles),

    .

    Where i is the label assigned to a particular particle, mi is the mass of that particle, and ri is the shortest distance from the particle's position to the axis of rotation.

  3. The Moments of Inertia of a simple 3D shapes (plates, cylinders, spheres, etc.) is given in the text in a table of your textbook. Note that the axis of rotation is specified for each shape.

  4. A complex shape can be reduced to more simply-shaped pieces such as are given in your textbook's Table of Moments of Inertia. Holes may be treated as objects with a negative moment of inertia.

  5. The Moment of Inertia of the complex object about a specified axis of rotation is equal to the sum of the Moments of Inertia of each smaller piece about the specified axis of rotation.

  6. If the specified axis of rotation is not the one given in the your textbook's Table of Moments of Inertia, the Parallel Axis Theorem must be applied.

  7. The Parallel Axis Theorem states that the Moment of Inertia of an object about a specified axis of rotation equals the moment of inertia about a parallel axis through the Centre of Mass plus the mass of the object times the square of the distance between the axes,

    .

Example #1.

Find the Moment of Inertia of an NO2 molecule about an axis through the centre of the Nitrogen atom and perpendicular to a line joining the Oxygen atoms. The mass of a Nitrogen atom is 2.345 × 10-26 kg and of an Oxygen atom is 2.657 × 10-26 kg. The Oxygen atoms are 1.95 × 10-10 m from the Nitrogen atom.

Atoms are definitely particles, so we may use . Constructing a table:

ATOM
MASS (mi)
DISTANCE (ri)
mi(ri)2
O
2.657 × 10-26
-1.95 × 10-10
1.01 × 10-45
N
2.345 × 10-26
0
0
O
2.657 × 10-26
+1.95 x 10-10
1.01 × 10-45
Total:
2.02 × 10-45

The Moment of Inertia about the indicated axis is 2.02 × 10-45 kg-m2.

Example #2

A weirdly constructed winch is shown in the diagram below. The ball bearings are welded to each other and to the outer shell and inner disk. Determine the moment of inertia of the winch about an axis through the centre of the winch and out of the paper.

The moment of inertia of the is equal to the sum of the moments of inertia of the cylindrical shell, the solid disk, and the ball bearings about the axis of rotation,

.

The axis is through the centre of mass of the cylindrical shell and the solid disk. The Table of Moments of Inertia gives values for these:

and

The ball bearings are not rotating about their centre of mass so we must use the Parallel Axis Theorem. The Table gives the moment of inertia of a solid sphere about it's centre of mass as . The distance between the centre of mass of one of the ball bearings and the axis of rotation is . The moment of inertia of the ball bearing about the axis of rotation is therefore:

.

The moment of inertia of the winch is therefore:

.

This has a numerical value of 171.6 kg-m2.

Example #3:

A 2.00 m by 0.50 m uniform board has a hole of radius 0.20 m drilled near one end. The mass of the board before the hole was drilled was 4.00 kg. Find the moment of inertia of the board through an axis through its centre and out of the page.

Solution: Treat the board as the solid board plus a disk of negative inertia.

As mass is proportional to area,

Therefore,

.

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