Question

Give the statement of theorems of moment of inertia and prove them.

Answer 1

Theorems of Moment of inertia

In the case of the rigid bodies having regular shapes, the axis through the center of mass is usually a symmetric axis. It will be seen that it is always convenient to find out moment of inertia of the body about the axis through its center of mass. However, moment of inertia about some other axis can be found with the help of theorems of parallel and perpendicular axis.

(a) Theorem of Perpendicular Axis : The perpendicular axis theorem (or plane figure theorem) can be used to determine the moment of inertia of a rigid body that lies entirely with in a plane.

This theorem states that the moment of inertia about an axis perpendicular to the plane is equal to the sum of the moments of inertia of two perpendicular axis through the same point in the plane of the object.

Explanation : Let OZ be the axis perpendicular to the plane lamina and passing through point ‘O’. Let OX and OY be two mutually perpendicular axis in the plane of lamina and intersecting at O’. It I_X, I_Y and I_Z are the moments of inertia of the plane lamina about the axis OX,OY and OZ respectively, then according to the theorem of perpendicular axis :
I_Z = I_X + I_Y
Proof : Let the lamina consists of n number particles,of masses m₁, m₂, m₃ …….. m_n. Their positions are at the distances r₁, r₂, r₃ ………. r_n respectively from the origin O. For simplicity, let a particle of mass m₁ is at point P(x₁, y₁) and its position is denoted by r₁ 

The moment of inertia of the particle of mass m₁, about OZ axis is :
I₁ = m₁r₁² = m₁ (x₁² + y₁²)
Similarly the inertia of the particle of mass m₂ about OZ axis is
I₂ = m₂ (x₂² + y₂²)
Therefore, the moment of inertia of the whole lamina about OZ axis is :
I_Z = I₁ + I₂ + I₃ + …………… I_n
= m₁ (x₁² + y₁²) + m₂ (x₂² + y₂²) + ………… + m_n (x_n² + y_n²)
or I_Z = (m ₁x₁² + m₂x₂² + m₃x₃² + …………. + m_nx_n²) + (m₁y₁² + m₂y₂² + m₃y₃² + …………. + m_ny_n²)
or I_Z = I_P + I_Y Proved

(b) Theorem of Parallel Axis : It states that the moment of inertia of rigid body about any axis is equal to its moment of inertia about a parallel axis through its center of mass plus the product of mass of the body and the square of the perpendicular distance between the two axis.

Let I_C be the moment of inertia of a body of mass M about an axis A’ B’ passing through its center of mass C. Let I be the moment of inertia of the body about an axis AB parallel to the axis A’ B’ and at a distance I

Then according to the theorem of parallel axis,
I = I_C + MR²
Proof : Consider that t^th particle located at the point P in the body, is of the mass m_i and lies at a distance r_i from the axis A’B’. Then, the distance of the i^th particle from the axis AB is (r_i + R).
The moment of inertia of the ith particle about the axis A’B’ is (m_ir_i²).
Therefore, moment of inertia of the body about the axis A’B’ is given by
I_C = \(\sum_{i=1}^{n} m_{i} r_{i}^{2}\) ………… (1)
Also, moment of inertia of the particle about the axis AB is m_i (r_i + R)², so that moment of inertia of the body about the axis AB is given by

\(\sum_{i=1}^{n} m_{i} r_{i}\) = Sum of the moments of the masses of the particles constituting the body about the axis’ through its center of mass. Since, sum of the moment of the masses of the particles constituting the body
about an axis through its center of mass must be zero, n
∴ \(\sum_{i=1}^{n} m_{i} r_{i}=0\)
In equation (2) substituting the value of the two factors, we have 
I = I_C + MR² …………. (3)
It proves the theorem of parallel axis for moment of inertia.