1. ROTATION OF RIGID BODIES
Linear inertia
Is the tendencyof a body to resist change in its linear velocity.
In other words; objects do not change their state of linear motion unless acted upon by some not
external force.
Rotational inertia
Is the tendency of a body to resist change in its angular velocity.
- In other words, objects do not change their rotational motion unless acted upon by some not
external torque.
- It is also calledmoment of inertia.
CONCEPT OF MOMENT OF INERTIA
Moment of inertia of a body about an axis is a measure of the difficulty in starting,
stopping or changing rotationof the body about that axis.
- It is denotedby I
- The greater the difficulty in starting or stopping , the greater is the moment of inertia of the
body about that axis and vice-versa.
- A body rotates under the actionof a net external torque.
The Greater the moment of inertia of a body about an axis of rotation, the greater is the torque
required to rotate or stop or change rotation of the body that axis and vice-versa.
MOMENT OF INERTIA OF A RIGID BODY
Consider a rigid body rotating about the axis yy-1 with an angular speed ω as shown in
figure 1 below.
2. Figure. 1
Suppose the body is made up of a large number (n) of small particles of masses m1,m2,m3
………..mn situated at perpendicular distances …………… respectively from the
axis of rotationyy'.
As the body rotates, eachparticle of the body follows a circular path around the axis.
Although each particle of the body has the same angular speed ω, the linear velocity (v) of
each particle depends upon particles distance from the axis of rotation.
Thus particle of mass m1 follows a circular path of radius r1. The linear velocityof this
particle is v1
Rotational kinetic energyof the particles of mass m1
The total kinetic energy Kr of the rotating body is the sum of the kinetic energies of all the
particles of which the body is composed.
3. Hence the total K.E of rotatingbody
Moment of inertiaof a rigid body about a given axis of rotation
Is the sum of the products of the masses of its particles and the squares of their respective
perpendicular distance from the axis of rotation.
The moment of inertiaof a body about an axis of rotationis directlyproportional to
the total mass of the body.
The more massive the body, the more difficult will be to start its rotational motion
or stop it from rotating.
For a given mass, the moment of inertia of a body depends upon the distribution of
the mass from the axis of rotation. The larger the distance of the mass from the axis
rotation the larger will be its moment of inertia.
The moment of inertia plays the same role in rotational motion as mass plays in
translational motion.
4. RADIUS OF GYRATION
Radius of gyration Is the distance from the given axis of rotation at which if whole mass of
the object were supposed to be concentrated the moment of inertia would be the same as with
the actual distributionof mass.
The radius of gyration is denotedby the symbol K.
T The moment of inertia of a body of mass M and radius of gyration K is given
by,
For example, the moment of inertia of a thin rod of mass M and length L about an axis
through its centre and perpendicular to its length is
The radius of gyration is a measure of the distribution of mass of a body relative to a
given axis of rotation.
A large radius of gyration means that, on the average, the mass is relatively far from
the given axis of rotation.
The SI unit of radius of gyrationis the metre (m)
Suppose a body consist of n particles each of mass m, then total mass of the body is
M.
5. Therefore radius of gyration is the root mean square distance of the various particles of
the body from the axis of rotation.
EQUATIONS OR UNIFORMLY ACCELERATED ROTATIONAL MOTION
Consider a rigidbody rotatingabout a given axis with uniform angular acceleration.
Let
Initial angular velocity.
Final angular velocityafter time t.
Uniform angular acceleration.
Angular displacement after time t.
i) To derive
From
At
6. For linear motion
v = u + at
ii) To derive
From
At
For the linear motion
iii) To derive
From
Also
7. When
For linear motion
v2 = u2 + 2as
MOTION OF A RIGID BODY AND MOMENT OF INERTIA
Consider a rigid body rotating about the axis with an angular speed as shown in
figure.2 below
Figure. 2
Suppose the body is made up of a large number of particles of masses m1, m2,m3 situated at
perpendicular distances ……… respectivelyfrom the axis of rotation.
8. As the body rotates, eachparticle within the body follows a circular patharound the axis.
Although each particle within the body has the same angular speed , the velocity of each
particle depends upon particles positionwithrespect to axis of rotation.
1. Relationbetweenrotational kineticenergyand moment ofinertia.
When a rigid body rotates about an axis, possesses K.E called the kinetic energy of a
rotatingbody, and its rotational kinetic energydenotedby ( ).
Let us now find the K.E of a rotatingbody.
The particle of mass m1 follows acircular path of radius .
The magnitude of the linear or tangential velocityof the particle onthis circle is v1
Similarly, the rotation kinetic energy of particles of masses
The rotational kinetic energy of the body is equal to the sum of rotational kinetic
energies of all particles
.
But, is the moment of inertiaof the body about the given axis of rotation
Thus the rotational K.E of a body is equal to half of the product of the moment of
inertia of the body and the square of the angular velocity of the body about a given
axis of rotation.
We also know that kinetic energyK of a body for linear motionis,
9. i) Note that both expressions for K.E are one half the products of property of body
and square of velocityterm.
We know that linear velocity v is an analogue of angular velocity ω in
rotational motion
Therefore, mass (m) of the body is an analogue of moment of inertia I of the
body in rotationmotion.
Hence we arrive at a very important conclusion that moment of inertia plays
the same role inrotational as the mass (m) plays in linear motion.
ii) If
From
Thus, the moment of inertia of a rigid body about a given axis of rotation is
numerically equal to twice the rotational K.E of the body when rotating with
unit angular velocityabout that axis.
2. Relationbetweentorque andmoment of inertia
Suppose a body rotates about an axis under the actionof a constant to torque τ .
Let the constant angular accelerationproducedbythe body be ∝.
The angular acceleration (∝) of all the particles of the body will be the same but the
linear acceleration (a) of each particle will depend upon the particles position with
respect to the axis of rotation.
The particle of mass m1 follows a circular path of radius r1. The magnitude of the
linear accelerationof this particle a1 = r1 ∝ .
If F1 is the net external force actingonthis particle
The magnitude of torque due to this force onthis particle
10. Similarly the magnitude of torque on the particles of masses m2,m3 ..... ... ..
are m2r2
2∝, m3 r3
2∝............ respectively.
By right hand rule, the torques onall the particles act inthe same direction.
The magnitude of the total torque on the body is just the sum of individual
torques onthe particles.
But is the moment of inertiaof the body about the axis of rotation
This is basic relationfor rotational motion
It is analogous to Newton’s secondlaw of linear motion
The torque is analogous to force F, the moment of inertia I is analogous to mass m and the
angular acceleration ∝ is analogous to the linear acceleration
If
11. Hence the moment of inertia of a body about a given axis is equal to the torque required to
produce unit angular accelerationinthe body about that axis.
3. Relationbetweenangular momentum and moment ofinertia
The particle of mass m1 follows a circular path of radius about the axis of
rotation.
Therefore, the magnitude v1 of linear velocityof this particle is
The magnitude of the angular momentum of this particle about the axis of
rotation
= linear momentum x radius r1
= m1 v1 x r1
Similarly, the magnitude of angular momentum of particles of masses m2,m3 ..... ... ..
are m2r2
2ω, m3r3
2 ω ................. respectively.
By right hand rule, the angular momentum of all the particles L of the rigid
body point in the same direction, parallel to the angular velocityvector.
Therefore, the magnitude of total angular momentum L of the body about the
axis of rotationis just the sum of individual momentum of the particles.
L = m1r1
2ω + m2r2
2ω + m31r3
2ω + .....
= (m1r1
2 + m2r2
2 + m3r32)ω
But is the moment of the body about the axis of rotation
The SI units of L is kgm2s-1
4 .Relationbetweentorque andangular momentum.
The magnitude of angular momentum of a body about an axis is
12. Differentiatingbothsides with respect to t we get
But Torque
Then
Thus the torque acting on a body is equal to time rate of change of angular
momentum of the body.
Power inRotational Motion
Consider a rigid body rotatingabout an axis due to a constant applied .
If the body rotates through a small change angle in small time then small work
done onthe body is
Thus power of a rotatingbody is equal to the product of torque and angular velocity.
Work-energytheorem inrotational motion
In linear motion, the work-energy theorem states that the net work done on a body by the
external force is equal to the change in body’s linear kinetic energy.
13. Similar, in rotational motion, the work done by a torque in rotating the body about an axis is
equal to the change in body’s rotationkinetic energy.
Therefore, work-energy theorem in rotational motion may be stated as under “The net
work-done by the external torque in rotating a rigid body about an axis is equal to the
change in body’s rotational K.E”.
LAW OF CONSERVATION OF ANGULAR MOMENTUM
If not net external torque acts on a body or system, its angular momentum remains constant
in magnitude and direction.
It is very important to keep in mind that it is - the product of moment of inertia and
angular velocitythat remains constant and not the angular velocityω.
MOMENT OF INERTIA AND RADIUS OF GYRATION OF DIFFERENT BODIES
WITH REGULAR GEOMETRIC SHAPES
1. Moment of inertiaof a uniform rod
(a) About an axis through its centre
Consider a uniform rod of total mass M and length L
Imagine, a small infinitesimally element of mass “ ” and length “ ” at a
distance “ ” from the axis of rotationpassingthrough the middle of the uniform rod.
14. Figure. 3
By similarity
The moment of inertia of the imaginary small infinitesimally element a
distance from the axis of rotationis given by
The total moment of inertia for the whole uniform rod is obtained by
integratingthe above expression
Total moment of inertiafor the whole rodis
15. Its Radius of gyration
But is the moment of inertiaof a uniform rod about an axis through its centre
Then
(b) Moment of inertiaabout an axis at one end
17. 2. Moment of inertiaof hollowSolidcylinder (ring)
Consider a hollow solid cylinder of uniform density is related about an axis passing
through the center alongthe length L of the cylinder.
Consider a small infinitesimal element (shell)at a distance r and having thickness,
then dV = area x thickness
The mass of shell is given by
The moment of inertiafor a shell
18. But the mass of the cylinder is given
Then
Hence for hollowcylinder (ring) the moment of inertiais given by
But if the thickness of the cylinder is very small
I = MR2
isthe momentof inertiaforthinhoop.
Its radius ofgyration
From
MR2 = MK2
19. K2 = R2
Therefore K= R
3. Moment inertiaof a solidcylinder (Disc)
Figure. 6
Area at any radius from the centre of the circle
Differentiate
By similarity
21. 4. Moment of inertiaof a sphere
Consider a spherical rigidbody and a particle at a distance from the centre
Figure. 7
Volume of a small element
Mass of a sphere
22. Moment of inertiaof a shell
PARALLEL AXES THEOREM
The parallel axes theorem state that moment of inertia of a rigid body about any given axis is equal to its moment of inertia about a parallel axis
through its centre ofmass plus theproduct ofthe mass of the body andthe square of the perpendiculardistance between the two parallel axes.
I= IG + Mh
2
This is a very simple anduseful theoremforobjects of regular shapes e.g. rod, sphere, disc.
Applicationof parallel axes theorem
1. For a uniform rod the moment of inertia about an axis through the centre
is
Therefore the moment of inertia about an axis through one endcanbe obtained
23. From
Moment ofinertia of a sphere ofradius R andmass M about an axis through the point onits circumference canbe obtained.
24. Figure 9
IG = 2MR2
5
From parallel axes theorem
I = IG + Mh
2
h = R
2. 3 . Moment of inertia of a disc of radius r and mass m about an axis through a point on the circumference can be obtained.
25. IG = 2MR
2
5
From parallel axes theorem
I = IG + Mh
2
I = 1MR
2
+ MR
2
But h = R
2
I = MR
2
+ 2MR
2
2
I = 3MR
2
2
PERPENDICULAR AXES THEOREM
The perpendicular axis theorem states that “the moment of inertia of plane body about an axis perpendicular to its plane is equal tothe
sum of the moments of inertia of the plane body about any two mutually perpendicular axes in the plane of the body which intersect
the first axis”
COMBINED ROTATIONAL AND TRANSLATIONAL MOTION
26. Consider a wheel rolling on a flat surface without slipping. Each particle of the wheel is undergoing two types of motion at the same
time.
Thecentre of thewheel, which is wheel’s centre of mass, is movinghorizontally withspeedVG.
Figure 10 (a)
At the same time,the wheel is rotatingabout its centre ofmass with angular speedω.
Figure 10 (b)
Thus rollingis the superposition of two motions
(i) Translationof objects centre of mass at velocity VG alonga straight line
(ii) Rotationabout the center of mass at angular velocity ω.
Therefore the wheel possesses bothtranslational androtational K.E.
27. M = Mass of the object
VG = Velocity of object’s centre of mass
I = Moment of inertia ofthe object about an axis through thecentre of mass.
The total K.E of an object undergoing both translational and rotational motion about its centre of mass axis is equal to the sum of
translationK.E ofthe centre ofmass andthe rotational K.Eabout the centre ofmass.
ROLLING OBJECTS
Consider a rigidbody rollingwithout slippingdown theplane inclinedat anangle θ with the horizontal.
Required to determinethe velocity, accelerationandminimumrequiredcoefficient ofstatic friction (μs) forrolling.
Figure. 11
By the conservationof energy we have
Total energyat A = total energy at B E .
EA = EB
K.EA +
P.EA =
28. 1. Rolling a Hollow Sphere
I = MR
2
From
But v
2
= R
2
ω
2
Then
MgH = 1 MR
2
ω
2
+ 1 MR
2
ω
2
2 2
MgH = MR
2
ω
2
gH = MR2
v
2
= gH
Is the velocityof a hollowsphere
From the figure
For Acceleration
From the 3
rd
equationof motion
v
2
= u
2
+ 2aL
Assumingthe body startedfrom rest
u = 0
31. v
2
= u2 + 2aL
But u = 0
v
2
= 2aL
10 gH = 2aL
7
a = 5 . gH
7 L
3. ROLLINGOFA SOLID CYLINDER
From
I = 1 MR
2
2
v
2
= R
2
ω
2
32. Is the velocityof solidcylinder.
For acceleration
From,
v
2
= u
2
+ 2aL
u = 0
v
2
= 2aL
But,
33. Is a accelerationof solidcylinder.
ALTERNATIVELY
Consider a solidcylinder of radius R andmass M rollingdown a plane inclinedat an angle θ to the horizontal.
For rotational motionabout thecentre of mass, the only forcethat produce the torque is.
Torque on cylinder
If I is the moment of inertia andα is the angularacceleration about the axis of rotation, then torque on the cylinderis FS
Equatingequation (ii) and(iii)
Since the cylinder rolls without slipping
Then
Puttingthe value of FSinto equation (i)
34. For a solidcylinder, the moment of inertia about its symmetryaxis is
All uniform rollingsolidcylinders have the samelinearacceleration down the inclineirrespective oftheir masses andradii.
The linear acceleration is just 2/3 as large as it wouldbe if the cylindercouldslide without friction down the slope
Minimum Required for Rolling
From
35. 1.The frictional force is static, the cylinder rolls without slipping andthere is no relative motion between the cylinder andthe inclined
plane at the point ofcontact.
2. If there were no frictionbetweenthe cylinderandtheincline, thecylinder wouldhave slippedinsteadof rolled.
MOTION OFA MASS TIED TO THE STRING WOUNDON A CYLINDER
Consider a solidcylinder of radius R andmass M capable of rotatingfreelywithout friction about its symmetryaxis.
Suppose a stringof negligible mass is wrappedaroundthe cylinderanda heavypoint mass m is suspendedfromthe free end.
Figure 12 (a)
When the point mass m is released, the cylinder rotates about its axis and at the same time the mass falls down due to gravit y and unwinds the
stringwoundon the cylinder
36. Figure 12 (b)
Linear Acceleration of Point Mass (m)
We first examinethe forces ofthe point mass m as shown below
37. Lookingat thecylinder end; we see that tensionin the stringexerts a torque on the cylinder
If I is the moment of inertia of the cylinder about the axis ofrotationandα is the angular accelerationproducedin the cylinder.
..........................(iii)
By equatingequations (ii) and(iii)
38. Since the string unwinds without slippingthe linear acceleration a of the point mass m and the angular acceleration α of the cylinder
are relatedas
Puttingthe value of T intoequation (i)
39. 1. The tensionT in thestringis always less than mg.
2. As moment ofinertiaincreases T approaches mg.
NEWTON’S LAWS OFROTATIONAL MOTION
1. First law
States that "everybody continues in its state of rest or of uniform rotational motion about an axis until it is compelled by some external
torque to change that state".
2. Second law
States that "the rate of change of angular momentum about an axis is directly proportional to the impressed external torque and the
change angular momentumtakes place in the directionof theappliedtorque".
where k = 1.
3. Third law
To everyexternal torque applied, there is an equal andopposite restoringtorque.
SIMPLEPENDULUM
40. Figure. 13
It consists of a stringoflength l suspendedfrom a fixedpoint O andcarryinga bobof mass m at its freeend.
When the bob is slightly displaced from the equilibrium position and then released, it executes rotational motion about an axis through
the suspension point O.
Suppose at anytime t, the bob is at point P making an angle θ with the vertical let be the angular velocity and α be the angular
acceleration at that instant.
The only force which exerts a torque on the bob is that due to the weight of the bob.
The negativesign is used since this is a restoringtorque i.e. a torque which force thebobtoward θ = 0°.
The angular momentum ofthe bobabout the axis of rotationat the consideredinstant (at point P)
L = mvr
v = ωr
L = mωr.r
L = mωr
2
But r = lsinθ where θ = 90º
Therefore L = mωl2
41. From equation (i)and(ii) we have
dω = α
dt
Now, the angular acceleration of the bob at time t
This governs the oscillationof simple pendulum,inthe vertical plane.
(i) When θ = 0º
Therefore, thebobhas zero angular accelerationat the equilibrium position.
(ii) The negative sign shows that the tangential accelerationis directedtowardthe mean position.
(iii) If θ is small, Sin θ= θ
