Parallel and Perpendicular Axis Theorems

Parallel Axis Theorem Statement

We call the parallel axis theorem the Huygens-Steiner theorem. It was named after two persons named Christiaan Huygens and Jakob Steiner.

According to the parallel axis theorem, the moment of inertia of a rigid body about an axis YY is equal to the moment of inertia about another axis  Y’Y’ passing through the centre-of-mass G of the body in a direction parallel to YY, plus the product of total mass M of the body and square of the perpendicular distance between the two parallel axes.

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From Fig.1, if d is the perpendicular distance between the axes YY and Y’Y’, then according to the following parallel axis theorem formula:

State and Prove the Theorem of Parallel Axis

The theorem of parallel axis states that the moment of inertia of a body about an axis passing via the centre of mass is equivalent to the sum of the moment of inertia of the body passing via the centre of mass and the product of a square of the distance between them.

Proof

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Let 1cm be the moment of inertia of an axis passing via the centre of the mass i.e., about axis AB I is the moment of inertia about an axis A’B’ at a distance h.

Then,

1 cm = I + Mh² ……..(1)

Consider a particle of mass m at a distance ‘r’, from the centre of gravity of the body (as shown in Fig.4).

So the distance from the axis A’B’ = r + h ……(2)

I = Σ m(r + h)² …..(3)

Since (r + h)² = r² + 2rh + h² ...putting it in (3)

I = Σ mr² + Σ mh² + Σ 2mrh = Icm + h² Σ m + 2h Σ mr   

I = Icm + Mh² + 0

Here Σ mr = 0 because the algebraic sum of the moments of weights of all the particles about an axis passing via C must be zero.

Parallel Axis Theorem Derivation

Now, we will do the parallel axis theorem derivation. For this, let’s take an example to explain how to use a parallel axis theorem.

Consider two blocks having their respective masses m1 and m2 of 10 kg each and these two blocks are separated by a distance ‘d’ which is equal to 10 m  as shown in the figure below:

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Draw the axis of rotation at the centre of the mass of the system so each block is separated by a distance of 5m from the axis of rotation.

What Would be the Moment of Inertia of this System?

The total inertia of the system will be the inertia of every point, so we get the inertia of the first object as:

Here, d = 5 m and m1 = 10 kg

I = 10 x 5² = 250 Kgm²

Similarly,  the moment of inertia for the second body of mass m1 will be:

So the total moment of inertia of the system when an axis of rotation is at the centre of mass will be:

Now, if I move 5 m to the left and the axis of rotation is at the centre of mass of the body 

of mass m1, then the moment of inertia of the system will be:

I1 = 10 x 0 = 0

Since the axis of rotation lies on the centre of the mass of the body with mass m1, the distance d will be zero, so I will also be zero.

Now, if the distance of the centre of mass of body m1 is at the distance of 10 m from the axis of rotation, the moment of inertia will be:

I2 = 10 x 10² = 1000

Now, the new moment of inertia will be:

Since equation (1)  works if the new axis is parallel to the original axis and the system has to be symmetrical. 

Note that this theorem is applicable only when the masses: m1 and m2 are of the same magnitude i.e., m1 and m2 must be equal.

We concluded that is the moment of inertia of the whole system and d, the displacement of the axis of rotation.

From Fig.2,

Moment of inertia at the centre of mass will be given as:

I cm = 10 x 5² = 500 Kgm² ….(2)

M = 10 kg

As we displaced the axis of rotation by d = 5 m.

Md² = 10 x 5² = 1000 Kgm² …(3)

Adding eq (2) and (3) we get,

= 500 + 500 = 1000 Kgm² which is equal to in I new in eq(1)

Hence the equation for the parallel axis theorem is:

Parallel Axis Theorem and Perpendicular Axis Theorem

According to this theorem, the moment of inertia of a plane lamina (two-dimensional body of any shape or size) about any axis is perpendicular to the plane of the lamina is the total of the moments of inertia of the lamina (regular-shaped) about any two mutually perpendicular axes OX and OY in the plane of the lamina, meeting at a point where the given axis OZ passes via the lamina.

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In Fig.3, suppose the lamina is in the XY plane.

Ix = Moment of inertia of the lamina about an axis OX.

Iy = Moment of inertia of the lamina about an axis OY.

Iz = Moment of inertia of the lamina about an axis OZ.

According to the perpendicular axis theorem: