Revision: Motion of System of Particles and Rigid Bodies Physics HSC Science Class 11 Tamil Nadu Board of Secondary Education

"Centre of mass is a point about which the summation of moments of masses in the system is zero."

Torque is defined as the moment of the externally applied force about a point or axis of rotation. The expression for torque is, `vectau = vecr xx vecF`.

Parallel axis theorem:

Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is – given by the relation,
I = I_C + M d²
Let us consider a rigid body as shown in the figure. Its moment of inertia about an axis AB passing through the center of mass is I_C. DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of I_C. For this, let us consider a point mass m on the body at position x from its center of mass.

Parallel axis theorem

The moment of inertia of the point mass about the axis DE is, m (x + d)². The moment of inertia I of the whole body about DE is the summation of the above expression.

I = ∑ m (x + d)²

This equation could further be written as,

I = ∑ m(x² + d² + 2xd)

I = ∑ (mx² + md² + 2 dmx)

l = ∑ mx² + md² + 2d ∑ mx

Here, ∑ mx² is the moment of inertia of the body about the center of mass. Hence,I_C = ∑ mx²

The term, ∑ mx = 0 because x can take positive and negative values with respect to the axis AB. The summation (∑ mx) will be zero.

Thus, I = I_C + ∑ m d² = I_C + (∑m) d²

Here, ∑ m is the entire mass M of the object (∑ m = M).

I = I_C + Md²

Perpendicular axis theorem:

This perpendicular axis theorem holds good only for plane laminar objects. The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.

Let the X and Y – axes lie in the plane and Z-axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are I_X and I_Y respectively – and I_Z is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
I_Z = I_X + I_Y

To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (O). The X and Y – axes lie on the plane and Z-axis is perpendicular to it as shown in the figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O. 

Perpendicular axis theorem

The moment of inertia of the particle about the Z-axis is, mr².

The summation of the above expression gives the moment of inertia of the entire lamina about the Z-axis as, I_Z = ∑ mr²

Here, r² = x² + y²

Then, I_Z = ∑ m (x² + y²)

I_Z = ∑ m x² + ∑ m y²

In the above expression, the term ∑ m x² is the moment of inertia of the body about the Y-axis, and similarly the term ∑ m y²is the moment of inertia about the X-axis. Thus,

I_X = ∑ m y² and I_Y = ∑ m x²

Substituting in the equation for Iz gives,

I_Z = I_X + I_Y

Thus, the perpendicular axis theorem is proved.