Moment of Inertia : Definition, Unit, Dimension and Important Theorems

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Relation Between Torque And Angular Acceleration : Moment of Inertia or Rotational Inertia

When a force is applied on a body, a linear acceleration is produced in that body. Similarly, when a torque is applied on a body, an angular acceleration is produced in it. So it can be said that, torque plays the same role in rotational motion as that of force in the case of linear motion. Hence, torque is the rotational analogue of force.

Suppose the body PQR is revolving with a uniform angular acceleration α about the axis AB [Fig.]. The body is assumed to be made up of innumerable point masses m₁, m₂, m₃, …, etc. These point masses are at distances r₁, r₂, r₃,……. etc. respectively from the axis of rota-tion AB. In case of pure rotation, the axis of rotation remains fixed and the angular acceleration of each point mass remains the same. But due to the difference in distances of the point masses from the axis of rotation, their linear accelerations are different. If the linear acceleration of the particle m₁ is a₁, then a₁ = r₁α and the force acting on it is F₁ = m₁r₁α.
The moment of force F₁ about the axis of rotation,
G₁ = force × perpendicular distance of the particle from the axis of rotation
= F₁r₁ = \(m_1 r_1^2 \alpha\)
In this way, the moment of force can be found out for every particle. The couple or torque acting on the entire rigid body is the algebraic sum of the moments of the forces acting on individual particles.

Hence, torque r = G₁ + G₂ + ………. = m₁\(r_1^2 \alpha\) + m₂\(r_2^2 \alpha\) + ………
= (m₁\(r_1^2\) + m₂\(r_2^2\) + ……..) × α = \(\sum_i m_i r_i^2\)
[m_i is the mass of the i-th particle and r_i is its perpendicular distance from the axis of rotation] = Iα ………. (1)
Here, I = \(\sum_i m_i r_i^2\) ……….. (2)
= moment of inertia of the body about the axis of rotation
So, I = \(\frac{\tau}{\alpha}\)
i.e., momentum of inertia \(=\frac{\text { torque }}{\text { angular acceleration }}\)

Definition: Moment of inertia of a body about an axis of rotation is defined as the torque acting on the body divided by the corresponding angular acceleration thus generated about the same axis of rotation.

In calculus, equation (2) can be represented as
I = ∫r² dm ………. (3)

Unit and dimension of moment

Dimension of moment of inertia
= dimension of mass × (dimension of distance)²
= ML²

Some important points about moment of inertia:

i) Moment of inertia not only depends on the mass of a body, but also depends on the perpendicular distance of the particles constituting the body from the axis of rota-tion, i.e., on the distribution of mass of the body.

ii) In case of translational motion,
force = mass × acceleration (F = ma)
Again, in case of rotational motion,
torque = moment of inertia × angular acceleration
(\(\tau\) = Iα)

Hence, the equation \(\tau\) = Iα is the rotational analogue of the equation F = ma. Moreover, we know that rota-tional analogues of force and linear acceleration are torque and angular acceleration, respectively. So, com-paring the above two equations, we can say that the rotational analogue of the mass of a body is its moment of inertia. Hence, the moment of inertia in rotational motion plays the same role as the mass in case of translational motion.

iii) Moment of inertia of a rigid body about a specific axis does not depend on the total mass (M = \(\sum_i m_i\)) of the body but on distribution of mass of the contituent particles i.e., \(\sum_i m_i r_i^2\) of the body. As the distribution of masses front the axis of rotation changes, the moment of inertia due to the change of the axis of rotation in its position. Except those cases, the moment of inertia of rigid body about a specific axis of rotation. It can safely assume to be a scalar quantity.

Concept of moment of inertia; It has been said that the moment of inertia in rotational motion plays the same role as the mass in translational motion. It is evident from the following discussion.

We know that the mass of a body in translational motion can be called its translational inertia. This is because mass is nothing but the hindrance that is generated in a body to resist any change in its translational motion. In case of rotational motion, a body is compelled to change its state of motion when an external torque (rotational analogue of force) acts on it. In absence of external torque, the body either remains at rest or executes uniform circular motion. It means that the moment of inertia of a body can be called its rotational inertia. It resists any change in the rotational motion of the body. To sum up it can be said that the relation between moment of force (torque) and moment of inertia is similar to the relation between force and mass.

It is clear that more the moment of inertia of a body about an axis, more the torque necessary to rotate the body about that axis or to stop the body from rotating.

Two Important Theorems Regarding Moment of Inertia

A regular-shaped body usually has some axis of symmetry. When the body rotates about such an axis, it undergoes just a spinning motion; during this spin, the entire body remains confined in the same region of space. A few examples of such axes of symmetry are:

1. Circular ring or circular disc: The axis passing through the centre of the circle and perpendicular to its plane is the axis of symmetry.
2. Sphere: Any diameter is a axis of symmetry.
3. Right circular cylinder: The axis passing through the centres of the two circular faces is the axis of symmetry.

Now, it should be mentioned that, the symmetry axis is not the only possible axis of rotation of a rigid body; a body may rotate about any other axis as well. For example, the diurnal motion of the earth (a sphere) is about its diameter, which is of course an axis of symmetry. In addition, the earth rotates around the distant sun; the axis of rotation passing through the sun is certainly not an axis of symmetry of the earth. Earth has a different moment of inertia about that axis also.

From the above discussion, it is evident that a rigid body may rotate about many possible axes. Fortunately, it is not necessary to tabulate the formulae for moments of inertia corresponding to all those axes. The following two theo-rems help us to find the moments of inertia of a body about some special axes of rotation, provided that the expression for the moment of inertia about a symmetry axis is known beforehand.

i) Parallel-axes theorem: This theorem is applicable for a body of any shape.

The moment of inertia (I) of a rigid body about any axis is equal to the sum of its moment of inertia (I_cm) about a parallel axis through its centre of mass and the product of the mass (M) of the body with the square of the perpendicular distance (d) between the two axes [Fig.].

The mathematical form of the theorem,
I = I_cm + Md² ………. (1)

Explanation: Let a body is composed of infinite number of straight line segment parallel to z-axis. The masses of the segments are m₁, m₂, m₃, ……… [Fig]. The part at which the body intercepts the xy -plane is shown in Fig.

Let, the total mass of the body M is concentrated at that intersection and mass m₁ is at a distance r from the z-axis.

Now, the moment of inertia of m j about z-axis,

Therefore, the moment of inertia of the whole body about z-axis,

ii) Perpendicular-axes theorem: The moment of inertia (I_z) of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moments of inertia (I_x + I_y) of the lamina about two mutually perpendicular axes lying on the plane of the lamina and intersecting each other at the point through where the perpendicular axis passes [Fig.].
The mathematical form of the theorem,
I_x + I_y = I_z …… (3)

Explanation Let the plane lamina be composed of infinite number of particles and the masses of the particles are m₁, m₂, m₃, …… [Fig.]. Let the distance of the particle of mass m₁ from the axes x, y and z are y₁, x₁ and d₁ respectively.

Now, the moments of inertia of the particle of mass m₁ about x, y and z -axes,

It is to be noted that this theorem of perpendicular axes is applicable only for plane sheets of small thickness.

Determination of moment of inertia of some uni-form symmetrical objects:
i) Moment of Inertia of a uniform rod about the perpendicular axis to its length passing through its centre of mass:

Let PQ is a uniform rod of mass m and length l [Fig.]. The centre of mass is at the midpoint O of the rod. Considering O as origin (0, 0) the x-axis along the length of the rod, the position coordinates of the points P and Q are (\(-\frac{l}{2}\), and \(\frac{l}{2}\) respectively.
The moment of inertia about the axis CD passing through the point O and perpendicular to the rod is to be determined.
Mass per unit length of the rod \(=\frac{m}{l}\)
Let us consider a small segment dx which is at a distance x from point O.

So, the mass of length dx = \(\left(\frac{m}{l} d x\right)\)
Moment of inertia of this small segment dx about CD = \(\left(\frac{m}{l} d x\right) x^2\)
Hence, the moment of inertia of the whole rod about CD.

ii) Moment of Inertia of a uniform rod about the perpendicular axis to its length passing through one end of
the rod (application of parallel-axes theorem):
Suppose,m ass of the rod = m, length of the rod = l. Moment of inertia of the rod about the axis CD passing through m
its centre of mass and perpendicular to its length,
I_CD = \(\frac{1}{12} m l^2\)

Now by the parallel-axes theorem we can write,
I_AB = I_CD + m\(m\left(\frac{l}{2}\right)^2\)
= \(\frac{1}{12} m l^2\) + \(\frac{1}{4} m l^2\) = \(\frac{1}{3} m l^2\) ………. (1)

iii) Moment of Inertia of a uniform rectangular lamina about an axis parallel to its length and breadth passing through its centre of mass:

Suppose, the mass of the lamina =m, length =1, breadth = b [Fig.], The centre of mass of the lamina is O. The moment of inertia of the lamina about CD parallel to its breadth and passing through O is to be determined.

The mass per unit area of the rectangular lamina = \(\frac{m}{l b}\). Let us imagine a small rectangular strip of width dr at a distance r from CD [Fig.]

Area of this strip = bdr
Mass of this strip = \(\left(\frac{m}{l b}\right)\)bdr = \(\frac{m}{l}\)dr
Therefore, moment of inertia of the whole lamina about the axis parallel to its breadth and passing through the centre of mass,

Similarly, moment of inertia of the lamina about an axis parallel to its length and passing through the centre of mass,
I_y = \(\frac{1}{12} m b^2\)

iv) Moment of inertia of a uniform rectangular lamina about an axis perpendicular to its plane passing through its centre of mass (application of perpendicular-axes theorem):

Suppose,mass of the lamina = m; length of the lamina = l; breadth of the lamina = b
In Fig., suppose O be the centre of mass of the lamina. OX and OY are the two axes lying on the plane of the lamina, mutually perpendicular to each other. The axis OZ is perpendicular to the plane of the lamina.

We know, the moment of inertia of the lamina about an axis passing through its centre of mass and parallel to its length,
I_x = \(\frac{1}{12} m b^2\)
The moment of inertia of the lamina about an axis pass-ing through its centre of mass and parallel to its breadth,
I_y = \(\frac{1}{12} m l^2\)
Now, by the perpendicular-axes theorem we can write,
I_z = I_x + II_y = \(\frac{1}{12} m b^2\) + \(\frac{1}{12} m l^2\)
= \(\frac{1}{12} m\left(b^2+l^2\right)\) ……… (2)

v) Moment of Inertia of a ring about an axis passing through its centre and perpendicular to the plane of the ring:
The mass of the circular ring is m and the radius of the ring is r, whose centre is O [Fig. ]

The moment of inertia about AB passing through the point O and perpendicular to the plane of the ring to be calculated.
Let us imagine a small element of length dx on the cir-cumference of the ring.
Mass per unit length of the ring = \(\frac{m}{2 \pi r}\)
∴ Mass of that small element = \(\frac{m}{2 \pi r} d x\)
∴ The moment of inertia of the element of length dx about AB = \(\left(\frac{m}{2 \pi r} d x\right) r^2\) = \(\frac{m r}{2 \pi} d x\)
∴ The moment of inertia of the ring about AB,
I = \(\int_0^{2 \pi r} \frac{m r}{2 \pi} d x\) = \(\frac{m r}{2 \pi}\)[2πr – 0] = mr²

vi) Moment of inertia of a ring about its diameter (application of perpendicular-axes theorem):

Let AB and CD be the axes along two mutually perpendicular diameters of the ring [Fig.].

Now, by the theorem of perpendicular-axes we can write, moment of inertia of the ring about the axis AB + moment of inertia of the ring about the axis CD = moment of inertia of the ring about an axis through the centre of the ring O and perpendicular to its plane,
i.e., I_AB + I_CD = mr² [where m = mass of the ring, r = radius of the ring]
For symmetry of the ring, I_AB = I_CD = I (say)
∴ I + I = mr², I = \(\frac{m r^2}{2}\) …… (3)
So, moment of inertia of a ring about its diameter = \(\frac{m r^2}{2}\).

vii) Moment of inertia of a circular disc about an axis passing through its centre and perpendicular to the plane of the disc:

P is a circular disc of mass m and radius r with centre O [Fig.], The moment of inertia about AB passing through the point O and perpendicular to the plane of the disc is to be calculated.

Mass per unit area of the disc \(=\frac{m}{\pi r^2}\)
Let us imagine an annular ring of width dx at a distance x (x < r) from the centre of the disc.

viii) Moment of inertia of a circular disc about its diameter (application of perpendicular-axes theorem):
Since the disc is symmetrical with respect to all diameters, its moment of inertia about every diameter is the same.
Let AB and CD be the axes along two mutually perpendicular diameters of the circular disc [Fig.],

Now, by the perpendicular-axes theorem we can write, moment of inertia of the disc about the axis AB + moment of inertia of the disc about the axis CD = moment of inertia of the disc about an axis through the centre of the disc O and perpendicular to its plane,
i.e., I_AB + I_CD + I = \(\frac{m r^2}{2}\)
[where, m = mass of the circular disc, r = radius of the circular disc]
For symmetry of the disc I_AB = I_CD = I (say)
or, I + I = \(\frac{m r^2}{2}\)
I = \(\frac{m r^2}{4}\) ….. (4)
So, moment of inertia of a circular disc about its diameter = \(\frac{m r^2}{4}\)

ix) Moment of inertia of a circular disc about a tangent in the plane of the disc (application of parallel-axes theorem):

Let CD be a tangent in the plane of the circular disc and AB be an axis along the diameter parallel to CD [Fig.],

Let the mass of the disc be m and its radius be r. By parallel-axes theorem we can write,
moment of inertia of the disc about CD = moment of inertia of the disc about AB + mr²
i.e., I_CD = I_AB + mr²
= \(\frac{m r^2}{4}\) + mr²
= \(\frac{5}{4} m r^2\) …… (5)
So, the moment of inertia of a circular disc about a tan-gent on the plane of the disc = \(\frac{5}{4} m r^2\).

x) Moment of inertia of a circular disc about a tangent perpendicular to the plane of the disc (application of parallel-axes theorem):
Let CD be a tangent to the circular disc perpendicular to its plane and AB be an axis passing through the centre O of the disc and parallel to CD.

Let the mass of the disc be m and its radius be r. By parallel-axes theorem we can write, moment of inertia of the disc about CD = moment of inertia of the disc about AB + mr²
= \(\frac{m r^2}{2}\) + mr².

Table – 1

Moment of inertia of some uniform bodies of mass m

Radius of gyration: Notice from the above table, that in all cases moment of inertia of an extended body rotating about specific axis depends not on total mass but on the mass distribution of the body from that very axis.

We shall now find a measuring way in which the mass of a rotating rigid body is related to the moment of inertia. For this, a new parameter, the radius of gyration (k) is intro-duced. From Table-1, we notice that in all cases Moment of inertia can be expressed as I = Mk² form, where k has the dimension of length. ‘ k’ is a geometric property of the body and axis of rotation.

We know that if a point mass M is at a distance k from the axis of rotation, its moment of inertia, I = Mk². From this, we can define radius of gyration.
k = \(\sqrt{\frac{I}{M}}\) ∴ I = Mk²

Definition: If the whole mass of a body is assumed to be concentrated at a point such that the moment of inertia of the whole body equals the moment of inertia of that point, then the radial distance of the point from the axis of rotation is called the radius of gyration.

Example:
Moment of inertia of a solid sphere about its diameter is, I = \(\frac{2}{5}\)Mr². So, its radius of gyration with respect to its diameter, k = \(\sqrt{\frac{\frac{2}{5} M r^2}{M}}\) = \(\sqrt{\frac{2}{5}} r\)