Calculation of moment of inertia
Each and every single physical system has associated itself with a certain point whose motion characterises the motion of the the system. When the system moves by influence of an external force, then this point moves as if the entire mass of the system were concentrated at it, and also the external force was applied at it. This point is therefore called the ‘center of mass’ of the system. The motion of a system or any object can be discussed in terms of the motion of its center of mass.
The center of mass can be calculated easily by taking the masses we are trying to find the center of mass between and multiplying them by their positions. Then, we have to add these together and divide that by the sum of all the individual masses.
1. Thin uniform rod (about an axis passing through centre of mass perpendicular to its length) :
Let length of the rod = l
mass per unit length= λ
so M = λl
Let us take an elementary length dx at a distance x from c.
Moment of Inertia about an axis due to element of length
Moment of Inertia of the whole rod about centre of mass axis
About an axis through the end and perpendicular to the rod
According to parallel axes theorem,
here h= l/2 and
About an axis passing through a point dividing the length in 1:3 ratio and perpendicular to the rod :
and h = l/4
According to parallel axis theorem,
2. Rectangular lamina – about an axis perpendicular to its plane and passing through the centre of mass
Let length = a , breadth= b , mass per unit area = 𝛿
so, M = 𝛿ab
Let us take a small strip which have a width dx at a distance x from YY’ axis.
So, Moment of Inertia of the strip about YY’ axis
Moment of Inertia of the whole lamina about the YY’ axis,
Calculation of Moment of inertia of a a solid cylinder :-
1. About an axis passing through its centre of mass and perpendicular to its own length
Let, radius of the solid cylinder = a
length of the solid cylinder = l
mass per unit volume = ρ .
A solid cylinder can be divided into a number of coaxial discs . Let us consider such an elementary disc of thickness dx at a distance x from o.
Mass of the elementary disc =
Moment of Inertia of the elementary disc about its own diameter (CD)
According to parallel axis theorem
Thus Moment of Inertia of the elementary disc about axis AB,
Moment of Inertia of the whole cylinder about the given axis,
2. About its own axis :
A solid cylinder is nothing but a number of thin circular discs piled one above the other perpendicular to its axis. Let us consider such a disc of mass m and radius a .
Moment of Inertia of the elementary disc about its own axis passing through centre of mass =
Summing up all such coaxial discs, we get Moment of Inertia of the whole cylinder about its own axis ,
Where Σm = M, mass of the cylinder
Moment of inertia of a hollow cylinder
Moment of inertia of a hollow cylinder that is rotating on an axis passing through the centre of the cylinder where it has an internal radius a1 and external radius a2 with mass M can be expressed in the following way.
Let length of the cylinder = l
Inner radius = a1
Outer radius = a2
Mass per unit volume = ρ
Total mass = M
a) About an axis passing through Moment of inertia and perpendicular to its length
Instead of dealing with the whole cylinder , an easy approach would be to divide the cylinder into infinitesimally thin disks. Then, using the moment of inertia of the disk along the central axis, extend it to the transverse axis by using the perpendicular axis theorem and parallel axis theorem. Once you get the resultant moment of inertia for the disk along the transverse axis,do integration over the entire length of the cylinder to finally arrive at the moment of inertia of the cylinder about its transverse axis.
Let us consider an co- axial annular disc of thickness dx at a distance x.
Mass of the elementary annular disc =
Moment of inertia of the elementary disc about its diameter, (CD) :
Moment of inertia of the elementary annular disc about the given axis (AB) according to parallel axis theorem,
So whole of the hollow cylinder about the given axis have the Moment of inertia,
b) About its own axis
If we take a hollow cylinder it will consist of inner radius a1 and outer radius a2 with mass M, and length L. We will calculate its moment of inertia about central axis.
However, before we get into the derivation we have to be aware of certain things. These are;
- The cylinder is split into infinitesimally thin rings.
- Each ring will have a thickness of dr with length L.
It is similar to the derivation of the moment of inertia of a solid cylinder.
Read more –
MOMENT OF INERTIA – Radius of gyration or Swing radius,Kinetic energy of a rotating body, Angular momentum for a rotating body, Torque of a rotating body, State and proof perpendicular axis theorem, State and proof parallel axis theorem