# Moment of Inertia

 Science > Rotational Motion > You are Here

Rigid Body:

• A rigid body is one whose geometric shape and size remains unchanged under the action of any external force.

#### Axis of rotation:

• When a rigid body performs the rotational motion, the particles of the body moves in circles. The centres of these circles lie on a straight line called the axis of rotation, which is fixed and perpendicular to the plane of circles.
• The particles on the axis of rotation are stationary.

#### Moment of Inertia:

• The moment of inertia of a rigid body about a given axis is, defined as the sum of the products of the mass of each and every particle of the body and the square of its distance from the given axis.
• S.I. unit of moment of inertia is kg m² and C.G.S. system it is g cm². Dimensions of the moment of inertia are [M1L2T0]

#### Explanation:

• If the system of masses consists of a number of point masses m1, m2, m3, m4, ……..msituated at a distance of r1, r2, r3, r4, ……..rfrom the axis of rotation, then by definition of moment of Inertia we have

• Instead of assuming body to be composed of discrete masses, it can be considered to be composed of continuous matter (mass); then the process of summation should be replaced by integration with proper limits.

#### Physical Significance of Moment of Inertia:

• Newton’s first law of motion is also called a law of inertia indicates that a body is unable to change by itself its state of rest or state of uniform motion along a straight line.  This property of inertness is known as inertia. It is an inherent property of matter. It is due to the inertia, a body opposes any change in its state of rest or of uniform motion in a straight line.
• In the translational motion, the mass of a body is a measure of its inertia.  Greater the mass, larger is the inertia, greater is the force required to produce a given linear acceleration in it.
• In rotational motion, the moment of inertia of a body is a measure of its inertia.  Greater the moment of inertia, larger is the torque required to produce a given angular acceleration in it.
• Thus moment of inertia in the rotational motion is analogous to the mass in translational motion because it plays the same role in rotational motion as the mass plays in translational motion. This is clear from the following table
 Sr.No. Translational motion Rotational motion 1 Linear Momentum p = m v Angular Momentum L = Iω 2 Force F = m a Torque τ =  Iα 3 Kinetic Energy K.E. =  ½mv² Kinetic Energy K.E. =  ¼mv²

Radius of Gyration:

• The radius of gyration of a rigid body about a given axis is defined as the distance from the axis at which the whole mass of the body must be supposed to be concentrated so that this imaginary point mass has the same moment of Inertia as the actual body, about the given axis.
• The radius of gyration is denoted by letter ‘K’.  As it is a distance, its S.I. unit is m and C.G..S. unit is cm.
• Its dimensions are [M0L1T0]
• The expression for moment of inertia in terms of the radius of gyration is I = MK².

Where I Moment of inertia of a body

M = Mass of the body

K = Radiation of gyration of the body

#### Physical Significance of Radius of Gyration:

• Moment of inertia depends on the mass of the body, the distribution of mass about the axis of rotation and the position of the axis of rotation.
• These factors can be separated by expressing the moment of inertia as a product of the mass and the square of a particular distance from the axis of rotation.  This particular distance is called as radius of gyration (K) of the body.
• By definition of radius of gyration

• From the above explanation we can conclude that, the radius of gyration is the measure of distribution of the mass of the body about the given axis

#### Example – 1:

• Four point masses 1kg, 2 kg, 3 kg and 4 kg are located at the corners A, B, C and D of a square ABCD of side 1m. Find the moment of inertia and radius of gyration in each of the following cases when axis of rotation is
• passing through A and perpendicular to the plane of ABCD
• passing through O, the centre of the square  and perpendicular to plane of ABCD
• along the side AB
• along diagonal AC
• Solution:
• Given: m1 = 1 kg, m2 = 2 kg, m3 = 3 kg, m4 = 4 kg, AB = BC = CD = AD = 1 m
• Axis passing through A and perpendicular to the plane of ABCD

 Sr.No. Mass Distance from A 1 m1 = 1 kg r1 = 0 2 m2 = 2 kg r2= 1 m 3 m3 = 3 kg r3= √2 m 4 m4 = 4kg r4 = 1m

IA = ∑ miri² = m1r1² + m2r2² + m3r3² + m4r4²

∴ IA =  (1)(0)² + (2)(1)² + (3)(√2)² + (4)(1)²

∴ IA =  0 + 2 + 6 + 4 = 10 kg m²

M = m1 + m2 + m3 + m4 = 1 + 2 +3 + 4  = 10 kg

By definition of radius of gyration

IA = MKA²

∴  10 = 10 KA²

∴ KA² = 1

KA = 1 m

Ans: Moment of inertia of system about axis passing through A  and perpendicular to the plane of ABCD is 10 kg m² and corresponding radius of gyration is 1 m.

• Axis passing through O, the centre of the square  and perpendicular to plane of ABCD

 Sr.No. Mass Distance from O 1 m1 = 1 kg r1 = √2 /2 m 2 m2 = 2 kg r2= √2 /2 m 3 m3 = 3 kg r3= √2 /2 m 4 m4 = 4kg r4 = √2 /2 m

IO = ∑ miri² = m1r1² + m2r2² + m3r3² + m4r4²

∴ IO =  (1)(√2 /2)² + (2)(√2 /2)² + (3)(√2 /2)² + (4)(√2 /2)²

∴ IO =  (1)(0.5)² + 2(0.5)² + 3(0.5)² + 4(0.5)² = 10(0.5)² = 10 × 0.25 = 2.5 kg m²

M = m1 + m2 + m3 + m4 = 1 + 2 +3 + 4  = 10 kg

By definition of radius of gyration

IA = MKO²

∴  2.5 = 10 KO²

∴ KO² = 0.25

∴  KO = 0.5 m

Ans: Moment of inertia of system about axis passing through O  and perpendicular to the plane of ABCD is 2.5 kg m² and corresponding radius of gyration is 0.5 m.

• Axis passing through AB:

 Sr.No. Mass Distance from AB 1 m1 = 1 kg r1 = 0 m 2 m2 = 2 kg r2= 0 m 3 m3 = 3 kg r3= 1 m 4 m4 = 4kg r4 = 1 m

IAB = ∑ miri² = m1r1² + m2r2² + m3r3² + m4r4²

∴ IAB =  (1)(0)² + (2)(0)² + (3)(1)² + (4)(1)²

∴ IAB =  0 + 0 + 3 + 4  = 7 kg m²

M = m1 + m2 + m3 + m4 = 1 + 2 +3 + 4  = 10 kg

By definition of radius of gyration

IAB = MKAB²

∴  7 = 10 KAB²

∴ KAB² = 0.7

KAB = 0.837 m

Ans: Moment of inertia of system about side AB is 75 kg m² and corresponding radius of gyration is 0.837m.

• Axis passing through Diagonal AC:

 Sr.No. Mass Distance from AC 1 m1 = 1 kg r1 = 0 m 2 m2 = 2 kg r2= √2 /2 m 3 m3 = 3 kg r3= 0 m 4 m4 = 4kg r4 = √2 /2 m

IAC = ∑ miri² = m1r1² + m2r2² + m3r3² + m4r4²

∴ IAC =  (1)(0)² + (2)(√2 /2)² + (3)(0)² + (4)(√2 /2)²

∴ IAC =  0 + (2)(0.5)² + 0 + 4(0.5)²  = 7 kg m²

M = m1 + m2 + m3 + m4 = 1 + 2 +3 + 4  = 10 kg

By definition of radius of gyration

IAC = MKAC²

∴  7 = 10 KAC²

∴ KAC² = 0.7

KAC = 0.837 m

Ans: Moment of inertia of system about diagonal AC is 75 kg m² and corresponding radius of gyration is 0.837m.

#### Example – 2:

• Three point masses 1 kg, 2 kg and 3 kg are located at the vertices A, B and C of an equilateral triangle ABC of side 1m. Find the moment of inertia and radius of gyration in each of the following cases when axis of rotation is
• passing through A and perpendicular to the plane of triangle ABC.
• passing through centroid G of the triangle and perpendicular to the plane of triangle ABC
• along the side AB
• Solution:
• Given: m1 = 1 kg, m2 = 2 kg, m3 = 3 kg, , AB = BC = AC = 1 m
• Axis passing through A and perpendicular to the plane of triangle ABC

 Sr.No. Mass Distance from A 1 m1 = 1 kg r1 = 0 m 2 m2 = 2 kg r2=  1 m 3 m3 = 3 kg r3= 1m

IA = ∑ miri² = m1r1² + m2r2² + m3r3²

∴ IA =  (1)(0)² + (2)(1)² + (3)(1)²

∴ IA =  0 + 2  +3  = 5 kg m²

M = m1 + m2 + m3  = 1 + 2 +3  = 6 kg

By definition of radius of gyration

IA = MKA²

∴  5 = 6 KA²

∴ KA² = 5/6 = 0.8333

KA = 0.913 m

Ans: Moment of inertia of system about A is 5 kg m² and

corresponding radius of gyration is 0.913 m.

• Passing through centroid G of the triangle and perpendicular to the plane of triangle ABC

For equilateral triangle GA = GB = GC = side/√  = 1 / √  = 0.577 m

 Sr.No. Mass Distance from A 1 m1 = 1 kg r1 = 0.577 m 2 m2 = 2 kg r2=  0.577 m 3 m3 = 3 kg r3= 0.577 m

IG = ∑ miri² = m1r1² + m2r2² + m3r3²

∴ IG =  (1)(0.577)² + (2)(0.577)² + (3)(0.577)²

∴ IG =  1/3 + 2/3  +3/3  = 6/3 = 2 kg m²

M = m1 + m2 + m3  = 1 + 2 +3  = 6 kg

By definition of radius of gyration

IG = MKG²

∴  2 = 6 KG²

∴ KG² = 2/6 = 0.3333

KG = 0.577 m

Ans: Moment of inertia of system about G is 2 kg m² and corresponding radius of gyration is 0.577 m.

• Along the side AB

 Sr.No. Mass Distance from A 1 m1 = 1 kg r1 = 0 m 2 m2 = 2 kg r2=  0 m 3 m3 = 3 kg r3= √3 /2=0.866 m

IAB = ∑ miri² = m1r1² + m2r2² + m3r3²

∴ IAB =  (1)(0)² + (2)(0)² + (3)(0.866)²

∴ IAB =  0 + 0  + 1.25  = 1.25 kg m²

M = m1 + m2 + m3  = 1 + 2 +3  = 6 kg

By definition of radius of gyration

IAB = MKAB²

∴  1.25 = 6 KAB²

∴ KAB² = 1.25/6 = 0.2083

KAB = 0.456 m

Ans: Moment of inertia of system about G is 1.25 kg m² and corresponding radius of gyration is 0.456 m.

#### Example – 3:

• A light thin uniform rod of length 2 m has two bodies stuck to its ends. The mass of each body is 100 g. Find the moment of inertia of the system about a transverse axis passing through i) the centre of mass of the rod ii) one end of the rod.
• Case – I: Axis through the centre of mass of the rod

 Mass (m) Distance from axis of rotation (r) m1 r1 = l/2 = 2/2 = 1m m2 r2 = l/2 = 2/2 = 1m

• Case – II: Axis through the centre of mass of the rod

 Mass (m) Distance from axis of rotation (r) m1 r1 = l = 2 m m2 r2 = 0 m

Ans: The M.I. of the system about the axis passing through the centre is 0.2 kg m2

and about the axis passing through the end is 0.4 kg m2.

#### Kinetic Energy of a Rotating Body:

• Consider a rigid body rotating about axis passing through point O and perpendicular to the plane of paper in anticlockwise sense as shown. Consider infinitesimal element at P of mass dm in the plane of the paper.  Let distance of point P from the axis of rotation be r.

The moment of inertia of the rigid body is given by

• As the body is a rigid body, the element at P will start moving in a circular motion. Let v be its tangential or linear velocity.

Then by concepts of circular motion

v  =  r ω      ……….. ..  (2)

Due to this velocity, the element at P will possess kinetic energy which is given by

From equation (2) and (3)

• Similarly, we can find kinetic energy of each and every infinitesimal element in the body.  Total kinetic energy of body can be found by integrating both sides of above equation.

This is an expression for the kinetic energy of a rotating body about a given axis.

#### Relation between kinetic energy of a rotating body and the frequency of  its rotation

Substituting ω = 2πn  in equation (5) we get

The quantities in the bracket are constant.

∴  E α n²

Thus, the kinetic energy of rotating body about a given axis is directly proportional to the square of the frequency of the body.

#### Relation between kinetic energy of rotating a body and its angular momentum

• The kinetic energy of  rotating body about a given axis is given by

#### Torque Acting on Rotating Body:

• Consider a rigid body rotating about an axis passing through point O and perpendicular to the plane of the paper in an anticlockwise sense as shown.  Let us consider infinitesimal element at P of mass dm in the plane of the paper.  Let distance of this element from the axis of rotation be r.

The moment of Inertia of the rigid body is given by

• Let τ be the external torque acting on the body.  Under the influence of this torque let the body rotates with angular acceleration α. Let dF  be  the magnitude  of external  force acting on the element, in the plane of page and at right angles to  position vector of point P, then by Newton’s  Second Law of motion,

dF = dm . a  ………..    (2)

where ‘a’  is the magnitude of linear acceleration (tangential acceleration) of the element.

By concept of circular motion, we know that

a  = r  α    ………….   (3)

From Equations (2) and (3)

dF = dm . r. α       ………  (4)

Let dτ be the torque acting on the element, then magnitude of the torque is given by

dτ = r , dF   ……………… (5)

From equations (4) and   (5)

dτ = r , dm . r. α

dτ = r ² dm α

• Similarly, we can find torque on each and every infinitesimal element in the body. The total torque acting on the body can be found by integrating both sides of above equation

• As body is rigid body angular acceleration (α) of all the elements is the same and it is taken out of integration sign

From equations (1) and (6)

τ =  Iα

This is an expression for torque acting on rotating body.

#### Example – 3:

• A disc of mass 50 Kg and radius 0.4 m is capable of rotation about an axis passing through its centre and at right angles to its plane. If a constant torque of 12 Nm acts on it, find the angular acceleration caused in the disc.
• Solution:
• Given: Mass of disc = M = 50 kg, radius of disc = R = 0.4 m, Torque = τ = 12 Nm
• To Find: Angular acceleration = α = ?

The M.I. of a disc about an axis passing through its centre and at right angles to its plane is given by

I = ½MR² = ½ × 50 × 0.4² = 25 × 0.16 = 4 kg m²

We have torque, τ = I α

∴  α  = τ / I = 12/4 = 3 rad/s²

Ans: Angular acceleration = 3 rad/s²

#### Example – 4:

• A constant couple of 25 Nm acts on a flywheel of mass 100 kg and radius of gyration 25 cm. What is the resulting angular acceleration?
• Solution:
• Given: Mass of flywheel (disc) = M = 100 kg, radius of gyration = K = 25 cm = 0.25 m,  Torque = τ = 25 Nm
• To Find: Angular acceleration = α = ?

The M.I. of a body in terms of radius of gyration is given by

I = MK² = 100 × 0.25² = 100 × 0.0625 = 6.25 kg m²

We have torque, τ = I α

∴  α  = τ / I = 25/6.25 = 4 rad/s²

Ans: Angular acceleration = 4 rad/s²

#### Example – 5:

• A constant torque of magnitude 5000 Nm acting on a body increases its angular velocity from 4 rad/s to 20 rad/s in 8 s. Calculate the M.I. of the body about the axis of rotation.
• Solution:
• Given: Initial angular speed = ω1 = 4 rad/s, Final angular speed = ω2 = 20 rad/s, Time taken = t = 8 s,  Torque = τ = 5000 Nm
• To Find: Moment of Inertia = I  = ?

α  = (ω– ω1)/t  = (20 – 4)/8 = 2 rad/s²

We have torque, τ = I α

∴  I = τ/ α  = 5000/2 = 2500 kg m²

Ans: Moment of inertia of body is 2500 kg m²

#### Example – 6:

• The speed of rotation of the body increases from 60 rpm too 90 rpm in 1 minute. Calculate the torque acting on the body, if its M.I. is 500 kgm².
• Solution:
• Given: Initial angular speed = N1 = 60 r.pm., Final angular speed = N2 = 90 r.pm., Time taken = t = 1 min = 60 s,  Moment of Inertia = I = kgm²
• To Find:  Torque = τ = ?

We have torque, τ = I α

∴  τ = 500 × 0.0524  = 26.18 Nm

Ans: Torque acting is 26.18 Nm

#### Example – 7:

• A solid sphere of radius 25 cm and mass 25 kg rotates about a fixed axis passing through its centre. If its angular velocity changes from 2π rad/s to 12π rad/s in 5 s, calculate the torque applied.
• Solution:
• Given: Radius of sphere = R = 25 cm = 0.25 m, Mass of sphere = 25 kg, Initial angular speed = ω1 = 2π rad/s, Final angular speed = ω2 = 12π rad/s, Time taken = t = 5 s,
• To Find: Torque = τ = ?

M.I. of sphere about a fixed axis passing through its centre is given by

α  = (ω– ω1)/t  = (12π – 2π)/5 = 2π  = 2 × 3.142 = 6.284 rad/s²

We have torque, τ = I α

∴  τ = 0.625 × 6.284 = 3.928 Nm

Ans: Torque required is 3.928 Nm

#### Example – 8:

• The angular velocity of a disc rotating in its plane changes from 2 rad/s to 10 rad/s in one minute when a constant torque of 2 Nm applied. What is the M. I. of the disc?
• Solution:
• Given: Initial angular speed = ω1 = 2 rad/s, Final angular speed = ω2 = 10 rad/s, Time taken = t = 1 min = 60 s,  Torque = τ = 2 Nm
• To Find: Moment of inertia of the disc = I = ?

α  = (ω– ω1)/t  = (10 – 2)/60 = 8/60  = 0.1333 rad/s²

We have torque, τ = I α

∴ I = τ / α = 2/0.1333  = 15 kg m²

Ans: Moment of inertia of the disc is 15 kg m²

#### Example – 9:

• A flywheel in the form of a disc is initially at rest. When a constant torque acts upon it for one minute, it attains a speed of 300 r.p.m. Calculate the torque if the mass and radius of the disc are 10 kg and 0.4 m respectively.
• Solution:
• Given: Radius of disc = R = 0.4 m Mass of disc = 10 kg, Initial angular speed = N1 = 0 rad/s, Final angular speed = N2 = 300 r.p.m., Time taken = t = 1 min = 60 s,
• To Find: Torque = τ = ?

M.I. of disc about an axis passing through its centre and perpendicular to plane is given by

I = ½MR² = ½ × 10×0.4² = 0.8 kg m²

ω= 2πN2/ 60 = (2× π × 300)/60 = 10π rad/s

α  = (ω– ω1)/t  = (10π – 0)/60 = π/6  = 0.5237 rad/s²

We have torque, τ = I α

∴  τ = 0.8 × 0.5237 = 0.419 Nm

Ans: Torque required is 0.419 Nm

#### Statement :

• The moment of inertia of a rigid body about any axis is equal to the sum of its moment of inertia about a parallel axis through its centre of mass and the product of the mass of the body and the square of the distance between the two axes.

#### Explanation :

• Consider a rigid body of mass M rotating about an axis passing through point O and perpendicular to the plane of the paper. Let IO be the moment of inertia of the body about the axis passing through O and perpendicular to the plane of the paper. Let IG be the moment of inertia of the body about the axis passing through the centre of mass of the body (G) and parallel to given axis passing through O.
• Let ‘h’ be the distance between the two axes i.e. (OG) =  h. Then by principle of parallel axes’

IO =  IG +  Mh²

This is the mathematical statement of the principle of parallel axes.

#### Proof :

• Consider the infinitesimal element of the body of mass   ‘dm’   at P, in the plane of the paper.  Join OP, OG and  GP.  Draw a perpendicular from P on extended line OG, such that PD ⊥ OG.

By definition Of moment of inertia

Applying Pythagoros theorem to Δ OPD

OP² =   OD² +  PD²

∴ OP²2  =   (OG + GD)² + PD²

∴ OP² = OG²+2OG..GD + GD² + PD² …(3)

In DPGD, by Pythagoros theorem,

PG²  = GD² + PD²  ……..   (4)

From-equations (3) and (4)

OP² = OG² + 2 OG..GD + PG²

Multiplying both sides of above equation by dm

dm.OP² = dm.OG² + 2 dm.OG.GD + dm.PG²

integrating both sides

IO =  IG +  Mh² (Proved)

#### Statement:

• Moment of inertia of a rigid plane lamina about an axis perpendicular to its plane is equal to the sum of its moment of inertia about any two mutually perpendicular axes in its plane and meeting in the point where the perpendicular axis cuts the lamina.

#### Explanation:

• Consider a thin rigid plate or layer or lamina (in this case thickness of the body is less compared to its surface area). Let OX and OY be two mutually perpendicular axes in the plane of a rigid lamina, intersecting at O. Let OZ be, other axes which is passing through point O and is perpendicular to the plane of the lamina. Let lx , ly and Iz be the moments of Inertia of the lamina about OX, OY and OZ axes respectively.
• Then by principle of perpendicular axes,

lz    =    lx  +   ly

This is the mathematical statement of the principle of perpendicular axes.

#### Proof:

• Consider infinitesimal element in the plane of lamina having mass dm situated at point P. Let the coordinates of point P be (x, y). Join OP and draw PM and PN perpendicular on OX and OY, respectively.

Then OM  =  NP  =  x   and  MP = ON  =  y, Let OP  =  r

Moment of inertia of the lamina about Z- axis is given by

Similarly, M.I. of the lamina about X-axis is given by

Similarly Moment of Inertia of the lamina about Y-axis la given by

Applying Pythagoros theorem to triangle OPM

OP²  =  OM² + PM²

∴ z²  =  x² + y²

Multiplying both sides of equation by dm and integrating

From equation (1), (2), (3) and (4)

lz    =    lx  +   ly     (Proved)

 Science > Rotational Motion > You are Here

### 2 Comments

1. Sanket Talwekar

How did u find distance in every case?

• Hemant More

Each side of the square is 1 m. Hence by Pythagoras theorem diagonal is √2 m. and all the distances should be measured from the point through which the axis of rotation passes.