MOMENT

OF

INERTIA

BSc I (UNIT IV)

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Contents

1) Moment of Inertia 3

Radius of Gyration 3

2) Theorem of Perpendicular Axes 4

Theorem of Parallel Axes 4

3) Moment of Inertia of Geometrical Bodies

Circular Disc 5

Solid Cylinder 6

Annular Ring 7

Hollow Cylinder 7

Solid Sphere 8

Thin Spherical Shell 9

Thick Spherical Shell 10

Rectangular Lamina 11

Rectangular Bar 11

Solid Cone 12

4) Kinetic Energy of a Body Rolling on a Horizontal Plane 13

5) Acceleration of a Body Rolling Down on an Inclined Plane 14

6) Products of Moment of Inertia 16

7) Principal Moments 17

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For translator motion, value of inertia depends only on the mass of the body. The kinetic energy in such

motion depends on the mass and linear velocity of the body.

When a body rotates about an axis, the kinetic energy of rotation is determined not only by its mass and

angular velocity but also upon the position of the axis about which it rotates and distribution of mass

about this axis.

Let us consider a body of mass rotating about an axis with angular velocity . All its particles

have the same angular velocity but as they are at different distances from the axis of rotation, their linear

velocities are different. Hence, we get

where, is the moment of inertia of the body.

Radius of Gyration

If the entire mass of the body is supposed to be concentrated at a point such that the kinetic energy of

rotation is the same as that of the body itself, then the distance of that point from the axis of rotation is

called the radius of gyration of the body about that axis.

where, is the number of particles each of mass into which the given mass is divided.

The dimension of moment of inertia is and its unit is .

The dimension of radius of gyration is and its unit is .

Moment of inertia is scalar because its value about a given axis remains unchanged by reversing its

direction of rotation about that axis.

Greater the moment of inertia of a body, greater is the couple required to produce a given angular

acceleration.

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The moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to

the sum of the moment of inertias of the lamina about the two axes at right angles to each other in its own

plane intersecting each other at the point where the perpendicular axis passes through it.

Let and be two perpendicular axes in the plane of the lamina. Let be the mass of a particle at

point at distance from an axis through origin perpendicular to plane .

Moment of inertia of the particle about - axis =

Moment of inertia of the particle about - axis =

If we divide the whole lamina into a number of particles of masses at distances

from the axis. Hence, the moment of inertia about - axis and - axis are given as,

Moment of inertia of the lamina about a perpendicular axis through

origin is given as,

=

=

The moment of inertia of a body about any axis is equal to

the sum of its moment of inertias about a parallel axis

through its centre of gravity and the product of its mass and

the square of the distance between the two axes.

Let be an axis in the plane of paper and a parallel axis

through , the centre of mass of the body. The perpendicular

distance between the two axes is . Let be the mass of the

body and , the mass of the element at at a distance

from .

Moment of inertia of the element having mass about axis

Moment of inertia of the body about axis

Let be the moment of inertia of the body about , an axis through . Hence, .

Hence, moment of inertia of the body is given as

is sum of the moments of all the particles about passing through , the centre of gravity.

Since the body is balanced about the centre of mass , algebraic sum of all the moments about is zero.

Therefore, . Hence,

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Circular Disc

a) About an axis through the centre of the disc perpendicular to its plane

Let us consider a circular disc of radius . Let us consider an elementary ring of radius and

width . The area of the disc is given as . Mass per unit area of the circular disc is

. Hence, mass of the elementary ring is given as,

Moment of inertia of the element about an axis through

its centre perpendicular to its plane is given as,

Hence, the moment of inertia is given as,

b) About the diameter of the circular disc

Let us consider to be the diameter. Let and be the moment of inertia about the end points

of the diameter. Therefore,

By the symmetry of the figure, . Hence,

c) About the tangent of the circular disc

Let be a tangent at .

By the theorem of parallel axes,

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Solid Cylinder

a) About its own axis of symmetry

Let us consider a solid cylinder of mass , length and radius . The volume of the cylinder is

given as, . Hence, the mass density of the cylinder is . Now let us consider a coaxial

cylinder of width at distance from the axis of symmetry . Hence, the volume of the

coaxial cylinder is . Hence, the mass of the coaxial cylinder is given as,

Moment of inertia of the coaxial cylinder is,

Hence, the moment of inertia is given as,

b) About the axis passing through the centre and perpendicular to its own axis of symmetry

Let be the axis of symmetry and be the axis perpendicular to . Let us consider a

circular disc of width at a distance from axis. Mass per unit length of the cylinder is

. Hence the mass of the disc is

. Moment of inertia of this disc about the diameter of the

rod is,

Moment of inertia of the disc about axis is given by

parallel axes theorem is,

Hence, the moment of inertia of the cylinder is given as,

For a thin rod, . Hence moment of inertia is given as,

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Annular Ring

a) About an axis passing through the origin and perpendicular to its plane

Let us consider a ring having inner radius and outer radius having mass . Area of the face

of the ring is . Mass per unit area of the ring is given as, . Let us now

consider a ring having radius and . Face area of this ring is . Mass of this ring is,

Moment of inertia of this ring is given as,

Hence, the moment of inertia is given as,

b) About its diameter

Let us consider to be the diameter. Let and be the moment of inertia about the end points

of the diameter. Therefore,

By the symmetry of the figure, . Hence,

Hollow Cylinder

a) About its own axis of symmetry

Let us consider a hollow cylinder of mass , length and inner radius and outer radius . The

volume of the cylinder is given as, . Hence, the mass density of the cylinder is

. Now let us consider a coaxial cylinder of width at distance from the axis of

symmetry. Hence, the volume of the coaxial cylinder is . Hence, the mass of the coaxial

cylinder is given as,

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Moment of inertia of the coaxial cylinder is,

Hence, the moment of inertia is given as,

b) About the axis passing through the centre and perpendicular to its own axis of symmetry

Let be the axis of symmetry and be the axis perpendicular to . Let us consider a

circular disc of width at a distance from axis. Mass per unit length of the cylinder is

. Hence the mass of the disc is

. Moment of inertia of this disc about the diameter of the

rod is,

Moment of inertia of the disc about axis is given

by parallel axes theorem is,

Hence, the moment of inertia of the cylinder is given as,

Solid Sphere

a) About its diameter

Let us consider a solid sphere of radius and mass . Consider a thin circular slice of radius,

. The volume of the slice is . Let be the mass per unit volume of

the sphere. Hence, mass of the slice is given as, . Moment of inertia of this slice

about a diameter is given as,

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Hence, the moment of inertia of the disc about its diameter is,

But mass of the sphere is

. Hence, the above

expression can be written as,

b) About a tangent

Let be a tangent at .

By the theorem of parallel axes,

Thin Spherical Shell

About its diameter

Let us consider a thin spherical shell of radius and mass . The mass per unit area of the shell

is given as, . Let us consider a thin element of shell bounded by two parallel planes

& at and . Let the radius of the shell be

and its thickness be .

The area of the thin element

By the geometry of the figure, and .

Differentiating we get,

And,

Area of the thin element

Mass of the thin element

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Hence, the moment of inertia about the diameter,

The moment of inertia about the diameter of the spherical shell is given as,

Thick Spherical Shell

About its diameter

Let us consider a thick spherical shell of inner radius , outer radius and mass . The total

volume of the shell is

. Hence, the mass per unit volume of the shell is

. Let us consider a concentric shell of thickness at a distance from the

centre. Hence, the mass of this concentric shell is,

Moment of inertia of this thin shell about the diameter

is given as,

Moment of inertia of the hollow sphere about its diameter is,

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Rectangular Lamina

About an axis perpendicular to its plane and passing through its centre of gravity

Let us consider a rectangular lamina of mass having length and breadth . The surface area

of the lamina is . Hence, the mass per unit area of the lamina is . Let us consider a thin

strip of width at a distance from the centre of gravity . The area of the strip is ,

The mass of the strip is given by,

The moment of inertia of the strip about an

axis parallel to side is given by,

The moment of inertia of the lamina is given as,

The moment of inertia of the lamina about an axis parallel to side is given by,

Applying perpendicular axes theorem, we find the moment of inertia of the lamina about an axis

perpendicular to its plane and passing through its centre of gravity as,

Rectangular Bar

a) About an axis perpendicular to its plane and passing through its centre of gravity

Let us consider a bar having length and breadth having mass . Let us consider the bar to be

made up of laminas placed one above the other. Hence the moment of inertia of the bar about an

axis perpendicular to its plane and passing through its centre of gravity as,

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b) About an axis perpendicular to its length and passing through one of its edges

The moment of inertia of the bar through its centre of gravity is . Let us find the moment of

inertia at the edge . By the geometry of the figure,

By using parallel axes theorem we can find the

moment of inertia of the bar about an axis

perpendicular to its length and passing through

one of its edges as,

For a square bar, . Hence, the moment of inertia is given as,

Solid Cone

a) About its vertical axis

Let us consider a solid cone having mass , base radius and height . The mass per unit

volume of the cone is given as, . Let be the semi vertical angle of the cone. Let

us consider a small circular disc of radius at a distance from the vertex. Let the disc have a

thickness . Hence, volume of the disc is,

According to the geometry of the figure we have, and

Hence, mass of the disc is given by,

Hence, moment of inertia of the disc about the vertical axis

perpendicular to its plane is given as,

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Hence, the moment of inertia of the solid cone about its vertical axis is given by,

b) About an axis through its vertex and parallel to the base

Let us consider an axis through the vertex parallel to the base. Let us consider a small circular

disc of radius at a distance from the vertex. Let the disc have a thickness . Hence, the

moment of inertia of this disc about its diameter is given as,

By parallel axis theorem, the moment of inertia of the disc about the axis parallel through the

vertex and parallel to the diameter of the disc is given as,

Hence, the moment of inertia of the solid cone about the axis parallel through the vertex and

parallel to the base is given as,

Let us consider a body of mass having radius rolling on a horizontal plane. Let be the linear

velocity and be the angular velocity of the body. The linear and rotational kinetic energy are given as,

Linear kinetic energy

Rotational kinetic energy

Hence, the total kinetic energy is given as,

Total energy

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Special Cases

a) Spherical Ball

For a spherical ball moment of inertia about its diameter is

. Hence,

Therefore, the total kinetic energy of the rolling spherical ball is given by,

Total Kinetic Energy

Total Kinetic Energy

b) Circular Disc

For a circular disc moment of inertia about an axis through its centre and perpendicular to its

plane is

. Hence,

Therefore, the total kinetic energy of the rolling circular disc is given by,

Total Kinetic Energy

Total Kinetic Energy

Let us consider a body of mass having radius rolling on a inclined plane. Let be the linear velocity

and be the angular velocity of the body. In one revolution, it will move a distance of . The total

kinetic energy of the moving body is,

. In moving a distance , the vertical distance

travelled is . Hence the change in potential energy is .

Since, there is no slipping, no energy is dissipated. Hence,

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Differentiating the above equation with respect to , we get

Now,

is the velocity of the body and

is the acceleration of the body. Hence,

Special Cases

a) Solid Cylinder

For a solid cylinder, moment of inertia about an axis through its centre is

. Hence,

Therefore, the total acceleration of the rolling solid cylinder is given by,

Acceleration,

Acceleration,

b) Solid Sphere

For a solid sphere, moment of inertia about its diameter is

. Hence,

Therefore, the total acceleration of the rolling solid sphere is given by,

Acceleration,

Acceleration,

c) Hollow Sphere

For a hollow sphere, moment of inertia about its diameter is

. Hence,

Therefore, the total acceleration of the rolling hollow sphere is given by,

Acceleration,

Acceleration,

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Let us consider a rigid body of mass composing of particles. For any particle, angular

momentum is given by,

But, . Hence,

Hence, total angular momentum of the rigid body is given by,

But, . Hence, the above equation becomes,

The - component of the above expression is,

Similarly equations for and can be written as,

Since,

, we get

Hence, the equations for angular momentum can be written as,

This can be written as,

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where,

The diagonal elements of the matrix are called the Coefficient of Moment of Inertia.

The other elements of the matrix are called the Products of Moment of Inertia.

Principal moments can be solved by finding the eigen values of the above matrix,

One can find a set of cartesian axes for which the inertia tensor will be a diagonal tensor. The axes are

called as the Principal axes and the corresponding diagonal components are called Principal moments

of inertia.

The angular momentum is given by,

We know that, torque applied to a body is the rate change of angular momentum. Hence,

Now we know that,

Hence,

can be treated as an operator. Hence, we get

If we select the axes of rotation in the body where all the products of moment of inertia vanish, i.e.

These axes where the products of moment of inertia vanish are called the Principal Axes. Hence,

where, , and are called the principal moments of inertia.

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