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Internal And External Forces: Every body of finite size is made of large number of particles and
it is not possible to describe the position and velocity of the individual particles. It is much easier
to describe the motion of the body if we can find some relation between their velocities. For this
we introduce the concept of system in rotational mechanics. System is a collection of any
number of particles interacting with one another are said to form a system. When we say
interaction it means they are exerting force on each other, these forces enable them to form a
system.
Internal Forces: the forces which particles of the system exert of each other are called
internal forces. Internal forces are always mutual and form action reaction pair. Thus, these
forces always cancel out and can’t produce change in momentum of the body.
To change the momentum of the body we have to apply force from outside this system.
This force exerted by some source outside the system is called external force.
Kinds of Motion of rigid Body: A rigid body can have three kinds of motion
[a] Translational motion
[b] Rotational Motion
[c] Rolling Motion
[a] Translational Motion: In translational motion of the body all points of the rigid body will
have the same velocity i.e. all the points on the body will be moving parallel to each other with
the same speed.
[b] Rotational Motion: In rotational motion the body rotates about a fixed axis. Every particle in
the body will move in a circle when the body rotates but radius
of circular path
can be different for different particles. As all the particles will be
completing
their circular path in the same time the angular velocity of all
points in the
body is same. The linear velocity of particles will be different.
Larger
the
distance of the particle from the axis of rotation larger will be the
linear velocity
whereas point on the axis of rotation is at rest.
If we consider the rotational of disc about an axis passing through the center and
perpendicular to the plane of the disc with angular velocity ω, then linear velocity will be
maximum for points lying on the circumference of the disc. The diametrically opposite points
will have same speed but opposite direction of motion.
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[c] Rolling Motion: Rolling motion is the resultant of translational motion and the rotational
motion of the body. Thus, the velocity at different points of the body is sum of velocity due to
translational and rotational motion.
As shown in the figure, the velocity at top most point of the body is 2v whereas the velocity of
the point of contact with the surface is zero.
Concept Of Center of Mass: Center of mass for a body is that point where whole mass of the
body is supposed to be concentrated and if external force is applied at the center of mass, the
body will move in the same way as when force is applied on it.
Center of mass of the geometrical body is the center of the body for e.g. for sphere or
disc it the center of sphere or disc respectively. But center of mass can lie either on the body or
outside it.
Center of Mass of Two particle system: Consider two masses m1 and m2 having position vector




r 1 and r 2 respectively. We assume that let F 1 and F 2 be the external force which acts on the


two masses m1 and m2 respectively. Also F 12 and F 21 denotes internal forces which acts on m1


and m2 respectively. Thus, the total force which acts on m1 is F 1 + F 12 , using Newton’s second
law this force is equal to the change in linear momentum of m1.



d (m1v1 )
F 1 + F 12 =
dt
Similarly, we can also write
Adding [1] and [2]



d ( m2 v 2 )
F 2 + F 21=
dt
[1]
m
[2]






d (m1v1 ) d (m2 v2 )
[ F 1 + F 12] + [ F 2 + F 21]=
+
dt
dt


 

As internal forces form action and reaction pair, therefore F12  F21  0 . Also, F1  F2  F , is the
net external force which acts on the system of two masses. Thus,
 d (m1v1 ) d (m2 v2 )
F
+
dt
dt
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 d


F  m1v1  m2 v 2 
dt

 d  dr1
dr2 
F   m1
 m2

dt 
dt
dt 
 d2


F  2 m1 r1  m2 r2 
dt


2

d  m1r1  m2 r2 

F  m1  m2  2 
[3]
dt  m1  m2 
When whole mass i.e. m1 +m2 is concentrated at the center of mass its equation can be written
as


d 2R
[4]
F  m1  m2  2
dt
Comparing the two equations [3] and [4] , we get
 m1r1  m2 r2
R
m1  m2
Conclusions: [a] The center of mass of the two particles having same mass m1=m2=m, is
 r1  r2
R
2
i.e. the center of mass of two particles of equal mass lie midway between the two masses.
[b] It two particles are of different masses , then center of mass always lies near the heavy mass.
[c] If center of mass of two particle system lies at the center then we can substitute R=0


m1r1  m2 r2
Center of mass of n particle system:
Consider n masses placed in a system having masses m1, m2…….mn and having position vector
  

r 1, r 2, r 3….. r n respectively. Then the position vector of CM of the system of n particles is given
by
n

mi ri
 
R  i 1n
 mi
i 1
As, sum of all the masses is total mass M of the system, thus we can write this equation as
n

mi ri
 
R  i 1
M
In terms of position co ordinates, the position vector of any mass can be written as

ri  xi iˆ  yi ˆj  z i kˆ
If X, Y and Z be the coordinates of center of mass of the system
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1 n
X iˆ  Y ˆj  Z kˆ 
 mi xi iˆ  yi ˆj  z i kˆ
M i 1
Equating x, y and z components in the equation we get
1 n
X=
 mi x i
M i 1
1 n
Y=
 mi y i
M i 1
1 n
Z=
 mi z i
M i 1
In case of continuous body, we can write the above equation as
1
xdm
X=
M 
1
Y =  ydm
M
1
zdm
Z=
M
The vector expression for continuous body is
 1 
R
r dm
M
Conclusions: The position of the center of mass of the system is independent of the choice of
coordinate system.
[b] The position of center of mass depends on the size and shape of the body and the
distribution of mass. Hence center of mass may lie inside or outside the body.
Motion of center of mass:
Linear Momentum of system of particles:

Let us consider a system of n particles with masses m1, m2 …………mn moving with velocities v1 ,


v 2…… v n respectively. The particles may be interacting and only external forces acting on the
particles. The total linear momentum of the system of particles is sum of momentum of n
particles.




P  m 1 v 1 +m2 v 2+……………….mn v n
If M is the total mass of the system and V is the velocity of center of mass of the system, then





P  MV  m1 v 1 +m2 v 2+……………….mn v n
Thus, total linear momentum of the system of particles is product of total mass of the body and
velocity of center of mass of the body.
Differentiating equation for linear momentum and using Newton’s second law
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


dP
dV
Fext 
M
dt
dt
Thus, for an isolated system on which no external force is acting the linear momentum of center
of mass or velocity of center of mass is constant.
If= Fext =0

P  cons tan t

V  cons tan t
i.e. if no external force acts on the system the velocity of the center of mass remains constant.
For e.g. if diwali rocket explodes in mid air, different parts moves in different direction in such a
way that the velocity of center of mass remains constant i.e. the center of mass keeps on moving
along the parabolic path as it was moving before the explosion.
Similarly consider the case of uranium nucleus which disintegrates at rest emitting an
alpha particle, as the velocity of CM before radium decay is zero, therefore daughter nucleus and
alpha particle will move in such a way that the momentum of alpha particle and daughter
nucleus is equal and opposite and net momentum is zero.
Expression for torque in co-ordinate form:
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Equilibrium of rigid bodies
A body in general can be in equilibrium if it has uniform motion either translational or rotational.
The two conditions for equilibrium of rigid bodies are
[a] Translational equilibrium: A body is said to be in translational equilibrium if the net external
force on the body is zero. Such a body moves with uniform velocity i.e. its momentum remains
same both in magnitude and direction Thus for translational equilibrium

Fext  0
This basically implies that net force in all directions x, y and z is zero. For such a body of constant
mass, the velocity will remain constant or its linear acceleration is zero. If the body is at rest and
net external force is zero then it is called static equilibrium. If the body is in uniform motion then
the equilibrium is called dynamic equilibrium.
Equilibrium can also be explained in terms of potential energy, as for conservative forces
acting on the body, the force and potential energy are related as
dU
F= 
dr
dU
Thus for translational equilibrium, 
=0. There can be three cases:
dr
[a] Stable equilibrium: If a body is displaced from the equilibrium position an applying some
force and it returns back to the same position once deforming force is removed then the
equilibrium is said to be stable equilibrium. IN stable equilibrium potential energy of the body is
minimum. For e.g. the pendulum bob at its lowest point is in stable equilibrium.
[b] Unstable Equilibrium: IN unstable equilibrium once the body is displaced from its equilibrium
position it will never return to the equilibrium position on its own. The potential energy of the
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body in unstable equilibrium is maximum. For e.g. o wooden log standing on one edge is in
unstable equilibrium
[c] Neutral Equilibrium: When the body can stay in equilibrium even after it is displaced from its
initial equilibrium position then it is called neutral equilibrium, in such cases the potential energy
of the body is constant. These bodies are said to be in neutral equilibrium and there potential
energy doesn’t change with time.
Rotational Equilibrium: A body is said to be in rotational equilibrium if the net external torque
acting on the rigid body is zero. Such a body either doesn’t rotate or rotate with constant
angular velocity.

i.e.  ext  0
for e.g. as shown in the figure the rod is pivoted at point O as shown in the figure and two
forces F1 and F2 acts on it in downward direction. The system will be in rotational equilibrium if
the moment of forces about point O is zero
F1 x l1 = F2 x l2
This also basically implies that net torque is zero.
As net torque on the body is zero, the angular momentum of the body will be constant, thus
angular acceleration is also zero if MI is constant or angular velocity is constant.
Partial equilibrium: Partial equilibrium means either it is in rotational equilibrium or it is in
translational equilibrium but both conditions are not satisfied.
As shown in the figure, both the forces F acts downwards on the rod and at equal
distance from the point where rod is supported. Thus, torque due to both the forces will be
equal in magnitude and opposite in direction and net torque is zero, whereas the net force on
the rod is 2F.
Similarly in the second figure torque due to both the forces will be in same direction but the
forces are oppositely directed, thus this body is in translational equilibrium and not the
rotational equilibrium.
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Couple: It is a system of two equal and opposite forces with their line of action different is called
couple. Whenever couple is their, the net force on the body is there and torque acts due to
which the body rotates. For e.g. the magnetic compass experience equal and opposite force on
north pole and south pole of the compass due to which the compass needle rotates.
Principles of Moments
: According to principle of moments, a body will be in rotational equilibrium if the algebraic sum
of all moments of all forces acting on the body about a fixed point is zero.
To use this principle consider an ideal lever of negligible mass pivoted at point O called
the fulcrum of the lever. Two forces F1 and F2 act at a distance d1 and d2 from the fulcrum in
downward direction. Let R be the reaction force exerted in the lever at the point of contact.
For translational equilibrium
R = F1 + F 2
For rotational equilibrium the moment of force about O should be zero
F1 x d1= F2 x d2
In case ofl lever F1 is some weight to be lifted and called load and d1 is called the load arm
whereas F2 is called the effort and d2 is called effort arm.
d
load
Mechanical Advantage =
 2
effort d1
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Center of gravity: The center of gravity is the point where whole weight of the body acts and
total gravitational torque is zero.
To determine the center of gravity of an irregularly shaped body we take body like
cardboard and try to balance it on the tip of a pencil as shown in the figure. By hit and trial
method we can find the point where the body can remain balanced and the normal reaction
exerted by pencil tip on the cardboard balances the weight of the cardboard i.e. R=mg
Also, the cardboard is in rotational equilibrium else it would tilt on one side due to torque

and falls. Force of gravity act on all the particles having mass m1 m2 …mn. If r i is the position
vector of mass mi relative to CG, then torque acting on the mass about CG is



 i = ri x mi g
As CG of the cardboard is located that the total torque on it due to forces of gravity is zero,
therefore

 n 
   ri  mi g
i 1
As g is non zero and is same for all points on the body, we can write
n

 mi ri  0
as
i 1
This, is same condition as if CM lies at the origin, thus we can conclude that the CM and CG
coincides if the value of acceleration due to gravity is constant.
There is another way of determining the CG of a body of irregular shape. Suspend the body
from some point like A and draw the vertical line AA1 it will pass through CG. Similarly repeat the
experiment twice by suspending from two points B and C such that BB1 and CC1 are the two
vertical lines passing through CG. The point of intersection of three lines is called center of
gravity.
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Moment Of Inertia
It is term analogous to the mass in linear motion, in translatory motion mass is the quantity that
determines the minimum force required to move the body, similarly in rotational motion
moment of inertia determine the minimum torque required to rotates the body. Larger the
moment of inertia of the body larger will be the torque required to rotate the body.
Kinetic Energy of rotating body
Consider an arbitrary shaped body rotating about some fixed axis with angular velocity ω. We
can assume the body as system of particles having mass m1, m2………………………mn at distance r1,
r2…………….rn from the axis of rotation of the body. The total kinetic energy of the rotating body
is sum of kinetic energy of individual particles.
E = E1+E2+……………………………En
1
1
1
E = m1v12 + m2 v 22 +…………………….. mn v n2
2
2
2
In rigid body rotation the angular velocity of all the particles is same whereas the linear velocity
is different for different particles. Thus, v1=r1ω, v2=r2ω.......... .................vn=rnω. Substituting these
values we get
1
E = m1 r12  m2 r22  ................mn rn2  2
2
1
E = I 2
2
2
2
2
Where I = m1r1  m2 r2  ................mn rn is called moment of inertia of the body. Thus,
mathematically moment of inertia can be defined as sum of product of masses and square of
their distance from the axis of rotation.
I=
n
m r
i 1
2
i i
Thus, moment of inertia depends on
[a] Axis of rotation selected its orientation
[b] the distribution of mass about the axis of rotation which in turn depends on shape and size
of the body.
For a continuous body the MI can be written as
I =  r 2 dm
Also, if ω = 1 , then I= 2[Kinetic energy]
Thus MI of the body is twice the kinetic energy of the body rotating with unit angular velocity.
Units of MI: In SI the units of moment of inertia are kg – m2 and in cgs the units are gm cm2
Radius of Gyration
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It is that point from the axis of rotation where if whole mass of the body in concentrated it will
have the same MI as that of the body with actual distribution of mass. It is generally denoted by
K. If M is mass of the body and K is radius of gyration for any body then the moment of inertia is
given by
I = MK2
[1]
Consider an arbitrary shaped body rotating about some fixed axis, if m1, m2…….mn be n particles
forming the body at a distance of r1, r2…….rn from the axis of rotation. Then, the MI of the body
is also given by
I = m1r12  m2 r22  ................mn rn2
If the body is divided into n parts of equal mass, then
I = m [ r12  r22  .......................rn2 ]
Multiply and divide the equation by n the total number of particles then
r 2  r22  .......................rn2
I=mxn[ 1
]
n
r 2  r22  .......................rn2
I = M[ 1
]
n
Comparing it with equation [1] we get
r12  r22  .......................rn2
n
Thus, radius of gyration can also be defined as the root mean square distance from the axis of
rotation of the body.
K=
Torque And Moment Of Inertia of the body
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