7-asystem-of-practicles-and-rotational-motion.ppt

SYSTEMS OF PARTICLES AND ROTAIONAL MOTION – PART-1
1. Rigid body; Pure Translation: Example-1, Example-2
2. Translation Plus Rotation: Example-1, Example-2
3. Rotation: Examples
4. A rigid body rotation about Y-axis; A rigid body rotation about Z-axis
5. Precision; Centre of Mass – 2-Particle System, 3-Particle System, n-Particle system
6. Vector Notation of Centre of Mass
7. Centre of Mass of a Rigid Body – CM of Thin Rod, Ring, Disc, Sphere & Cylinder
8. Motion of Centre of Mass
9. Example for Motion of Centre of Mass independent of Internal Forces
10. Linear Momentum of a System of Particles
11. Cross or Vector Product – Properties of Cross Product
12. Angular Velocity and its Relation with Linear Velocity; Angular Acceleration
13. Torque, Angular Momentum; Relation between Torque and Angular Momentum
14. Torque and Angular Momentum for a System of Particles
15. Conservation of Angular Momentum
Created by C. Mani, Assistant Commissioner, KVS RO Silchar Next

Rigid body
Ideally a rigid body is a body with a perfectly definite and unchanging shape.
The distances between all pairs of particles of such a body do not change.
Note: No real body is truly rigid, since real bodies deform under the
influence of forces. But in many situations the deformations are negligible
and hence can be considered as rigid bodies.
Types of motion a rigid body can have:
Pure translation
In pure translation all particles of the
body have the same velocity at any
instant of time.
P1
P2
P1
P2
P1
P2
Example-1: Motion of a block on an inclined plane
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Example-2: Motion of a vase on a parabolic path (projectile)
In this example, the body is restricted to move only with translation.
All particles of the body have the same velocity at any instant of time.
The direction of velocities of the various particles will be parallel (tangential)
to the trajectory at any instant.
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Translation plus Rotation
A wooden or metallic cylinder rolling
down the inclined plane undergoes
both translational and rotatory
motion.
P1
P2
P4
The velocities of different particles
P1, P2, P3 and P4 at the same instant
are different depending on their
locations in the body.
The velocity at P4 is zero (velocity of
any particle coming into contact with
the surface of the inclined plane is
momentarily zero, if the body is
rolling without slipping).
P3
Example-1: Rolling of a cylinder on an inclined plane
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Example-2: Translation as well as rotation of a vase as a projectile
In this example, the body is moving with translation as well as rotation.
The particles of the body have different velocities at any given instant of time.
The direction of velocities of the various particles will be different w.r.t. the
trajectory at any given instant.
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Rotation
In rotation of a rigid body about a fixed axis, every particle of the body moves
in a circle, which lies in a plane perpendicular to the axis and has its centre
on the axis.
Axis of rotation
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A spin rotating about its axis (vertical) A pot rotating about its vertical axis
Z A table fan rotating about its axis (┴ to screen)
Examples
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The particles describe circular motion
with the planes of the circles lying in XZ plane. i.e. perpendicular to Y-axis.
X
Y
Z
The particles describe circular motion
with the planes of the circles lying in XY plane. i.e. perpendicular to Z-axis.
X
Y
Z
A rigid body rotation about Y-axis:
A rigid body rotation about Z-axis:
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Precision
In precision of a rigid body about a fixed point, the axis of rotation itself
rotates about an another axis passing though that fixed point.
Play Up(↑) and Down(↓) arrow keys alternately to repeat the precision!!
Click to see precision of top! Click to see precision of table fan!
Y
O
(fixed)
Z
Y
O
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In precision, the axis of rotation is not fixed but the point of contact is fixed.
At any instant, this axis of rotation passes through the point of contact.
In case of spinning top, the point of contact ‘O’ with the ground is fixed but
the axis of rotation passing through this point is rotating about Y-axis
(vertical).
Interestingly, the axis of rotation describes a cone with its vertex at ‘O’.
In case of oscillating table fan or pedestal fan, the blades are rotating about
the horizontal axis (here it is Z-axis and in the plane XY) but the arm carrying
the blades rotates about the vertical axis (Y-axis).
The meeting point of both horizontal and vertical axes is ‘O’ and is fixed.
Thus, in precision, the axis (one line) is not fixed but one point on the rigid
body is fixed.
IMPORTANT OBSERVATIONS:
The motion of a rigid body which is not pivoted or fixed in some way is either
a pure translation or a combination of translation and rotation.
The motion of a rigid body which is pivoted or fixed in some way is rotation.
For our study, we shall consider the simpler cases involving only rotation of
rigid bodies where one line (axis) is fixed. Home Next Previous

CENTRE OF MASS
Consider a two-particle system along the x- axis with mass m1 and m2 at the
distances x1 and x2 respectively from the origin ‘O’.
x
y
O
m1 m2
x1
x2
C
XCM
The centre of mass of the system is
that point C which is at a distance XCM
from O, where XCM is given by
XCM can be regarded as the mass-
weighted mean of x1 and x2.
If the two particles have the same
mass m1 = m2 = m, then
2m
mx1 + mx2
XCM =
m1 + m2
m1x1 + m2x2
XCM =
2
x1 + x2
XCM =
Thus, for two particles of equal
masses the centre of mass lies
exactly midway between them.
Two-particle system
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Consider a n-particle system along the x- axis with mass m1, m2, m3 …….. mn
at the distances x1, x2, x3 …….. xn respectively from the origin ‘O’.
The centre of mass of the system is that point C which is at a distance XCM
from O, where XCM is given by
m1 + m2 + ………. + mn
m1x1 + m2x2 + ………. + mnxn
XCM =
∑ mi is the sum of mass of all the
particles and hence is the total mass M
of the body.
XCM =
∑ mixi
i=1
n
∑ mi
i=1
n
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x
y
O
(x1, y1)
m1
m2
C
m3
(x2, y2)
(x3, y3)
XCM
YCM
Consider a 3-particle system in the x-y
plane with mass m1, m2 and m3 at the
positions (x1, y1), (x2, y2) and (x3, y3)
respectively.
The centre of mass C of the system
located at (XCM, YCM) is given by
For the particles of same mass,
m1 = m2 = m3 = m.
m1 + m2 + m3
m1x1 + m2x2 + m3x3
XCM =
m1 + m2 + m3
m1y1 + m2y2 + m3y3
YCM =
Thus, for three particles of
equal masses the centre of
mass lies at the centroid of
the triangle formed by
them.
3
x1 + x2 + x3
XCM =
3
y1 + y2 + y3
YCM = m m
m
C
Three-particle system
Then,
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x
O
y
z
(x1, y1, z1)
m1
m2
m3
mn
Consider a n-particle system in space
with mass m1, m2, …….. mn at the
positions (x1, y1), (x2, y2), ………, (xn, yn)
respectively.
The centre of mass C of the system
located at (XCM, YCM, ZCM) is given by
m1 + m2 + ………. + mn
m1x1 + m2x2 + ………. + mnxn
XCM =
m1 + m2 + ………. + mn
m1y1 + m2y2 + ………. + mnyn
YCM =
m1 + m2 + ………. + mn
m1z1 + m2z2 + ………. + mnzn
ZCM =
or
XCM =
∑ mixi
i=1
n
∑ mi
i=1
n YCM =
∑ miyi
i=1
n
∑ mi
i=1
n ZCM =
∑ mizi
i=1
n
∑ mi
i=1
n
, and
∑ mi is the sum of mass of all the particles and is the total mass M
of the system.
(x2, y2, z2)
(x3, y3, z3)
(xn, yn, zn)
n-particle system
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i
= xi
ri j
+ yi k
+ zi
If ri is the position vector of the ith
particle and R is the position vector of the
centre of mass of the system, then
The above equations written explicitly for 3 different axes can be expressed
in a single equation using vector notation.
and i
= Xi
R j
+ Yi k
+ Zi
R =
∑ miri
i=1
n
∑ mi
i=1
n
Vector notation of centre of mass
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A rigid body is a system of closely packed particles.
Therefore, the equations discussed so far are applicable to a rigid body.
The number of particles such a body is so large that it is impossible to carry
out the summations over individual particles in these equations.
Centre of mass of a rigid body
Since the spacing of the particles is small, the body can be treated as a
continuous distribution of mass.
We subdivide the body into n small elements of mass; Δm1, Δm2... Δmn;
the ith
element Δmi is taken to be located at the point (xi, yi, zi).
The coordinates of the centre of mass are approximately given by
XCM =
∑ (Δmi)xi
i=1
n
∑ Δmi
i=1
n
, and
YCM =
∑ (Δmi)yi
i=1
n
∑ Δmi
i=1
n ZCM =
∑ (Δmi)zi
i=1
n
∑ Δmi
i=1
n
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If n is made bigger and bigger and each Δmi smaller and smaller, these
expressions become exact.
In that case, we denote the sums over i by integrals.
Thus,
Σ (Δmi) xi becomes ∫ x dm
Σ Δmi becomes ∫ dm = M,
Σ (Δmi) yi becomes ∫ y dm
Σ (Δmi) zi becomes ∫ z dm
The coordinates of the centre of mass now are
The vector expression equivalent to these three scalar expressions is
, and
XCM =
M
1
∫ x dm YCM =
M
1
∫ y dm ZCM =
M
1
∫ z dm
RCM =
M
1
∫ r dm
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Consider a thin rod, whose radius (in case of cylindrical rod) is much smaller
than its length.
Taking the origin ‘O’ to be at the geometric centre of the rod and x-axis to be
along the length of the rod, we can say that on account of reflection
symmetry, for every element dm of the rod at x, there is an element of the
same mass dm located at –x.
O x dm
dm -x
Centre of mass of a thin rod
The net contribution of every such pair to the integral and hence the integral
∫ x dm itself is zero.
The point for which the integral itself is zero, is the centre of mass.
Thus, the centre of mass of a homogenous thin rod coincides with its
geometric centre.
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The same symmetry argument will apply to homogeneous rings, discs,
spheres, or even thick rods of circular or rectangular cross section.
For all such bodies, for every element dm at a point (x,y,z ) one can always
take an element of the same mass at the point (-x,-y,-z ).
(In other words, the origin is a point of reflection symmetry for these bodies.)
As a result, all the integrals are zero. This means that for all the above
bodies, their centre of mass coincides with their geometric centre.
Centre of mass of a ring, disc, sphere and thick rod
CM CM CM
CM
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MOTION OF CENTRE OF MASS
Let us discuss the physical importance of centre of mass for a system of
particles.
We have
R =
∑ miri
i=1
n
M
or MR = ∑ miri
i=1
n
or MR = m1r1 + m2r2 + ………….. + mnrn
Differentiating the two sides of the equation with respect to time, we get
dr1
dt
= m1
dR
dt
M
dr2
dt
+ m2 + …………..
drn
dt
+ mn
or MV = m1v1 + m2v2 + ………….. + mnvn
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MV = m1v1 + m2v2 + ………….. + mnvn
Differentiating the two sides of the equation with respect to time, we get
dv1
dt
= m1
dV
dt
M
dv2
dt
+ m2 + …………..
dvn
dt
+ mn
or MA = m1a1 + m2a2 + ………….. + mnan
Now, from Newton’s second law, the force acting on the first particle is
given by F1 = m1 a1.
The force acting on the second particle is given by F2 = m2 a2 and so on.
MA = F1 + F2 + ………….. + Fn
Thus, the total mass of a system of particles times the acceleration of its
centre of mass is the vector sum of all the forces acting on the system of
particles.
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When we talk of the force F1 on the first particle, it is not a single force, but
the vector sum of all the forces on the first particle; likewise for the second
particle etc.
Among these forces on each particle there will be external forces exerted
by bodies outside the system and also internal forces exerted by the
particles on one another.
We know from Newton’s third law that these internal forces occur in equal
and opposite pairs and in the sum of forces, their contribution is zero.
Only the external forces contribute to the equation.
M A = Fext
where Fext represents the sum of all external forces acting on
the particles of the system.
The above equation states that the centre of mass of a system of particles
moves as if all the mass of the system was concentrated at the centre of
mass and all the external forces were applied at that point.
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The system may be a collection of particles in which there may be all kinds
of internal motions, or it may be a rigid body which has either pure
translational motion or a combination of translational and rotational motion.
Whatever is the system and the motion of its individual particles, the centre
of mass moves according to the above equation.
Instead of treating extended bodies as single particles as we have done in
earlier chapters, we can now treat them as systems of particles.
We can obtain the translational component of their motion, i.e. the motion
of centre of mass of the system, by taking the mass of the whole system
to be concentrated at the centre of mass and all the external forces on the
system to be acting at the centre of mass.
M A = Fext
To obtain we did not need to specify the nature of the system
of particles.
To determine the motion of the centre of mass no knowledge of internal
forces of the system of particles is required; for this purpose we need to
know only the external forces.
NOTE:
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Example for motion of centre of mass independent of internal forces
X
Y
O
j
a = -g
CM
CM
A projectile, following the usual parabolic trajectory, explodes into fragments
midway in air. The forces leading to the explosion are internal forces. They
contribute nothing to the motion of the centre of mass.
The total external force, namely, the force of gravity acting on the body, is the
same before and after the explosion.
The centre of mass under the influence of the external force continues,
therefore, along the same parabolic trajectory as it would have followed if
there were no explosion. Home Next Previous

LINEAR MOMENTUM OF A SYSTEM OF PARTICLES
Linear momentum of a particle is defined as p = m v
Newton’s second law written in symbolic form for a single particle is
dP
dt
F =
where F is the force on the particle.
For the system of n particles, the linear momentum of the system is
defined to be the vector sum of all individual particles of the system,
Consider a system of n particles with masses m1, m2, ….., mn respectively
and velocities v1, v2, …….., vn respectively.
The particles may be interacting and have external forces acting on them.
The linear momentum of the first particle is m1 v1 , of the second particle is
m2 v2 and so on.
But MV = m1v1 + m2v2 + ………….. + mnvn
= m1v1 + m2v2 + ………….. + mnvn
P = M V
Thus, the total momentum of a system of particles is equal to the product
of the total mass of the system and the velocity of its centre of mass.
P = p1 + p2 + …….…. + pn
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P = M V
Differentiating w.r.t. time,
dP
dt
=
dV
dt
M
But, M A = Fext
dP
dt
= MA
dP
dt
= Fext
This is the statement of Newton’s second law extended to a system of
particles.
If the sum of external forces acting on a system of particles is zero,
then
dP
dt
= 0
or P = constant
This is the statement of law of conservation of linear momentum of a
system of particles. Home Next Previous

Cross Product or Vector Product
In vector product, the symbol of multiplication between the vectors
is represented by ‘ x ‘. The result of the product is a vector.
Eg. Torque is defined as the product of force and moment arm.
Here force, moment arm and torque are all vectors.
Consider two vectors A and B making an angle θ
with each other.
The cross or vector product is given by
and
A x B = | A | | B | sin θn
By Right Hand Thumb Rule or Maxwell’s Cork Screw Rule, the direction
of the resultant in this case is perpendicular and into the plane of the
diagram.
A
P
O
B
θ
Q
A x B means the vector A is rotated towards B and its effect is taken on B.
n
where is the unit vector
along a direction which is
perpendicular to plane
containing A and B.
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sin θ =
A x B
| A | | B |
| |
B x A means the vector B is rotated towards A and its effect is taken on A.
The resultant in this case is perpendicular and emerging out of the plane of
the diagram.
Closing and opening a tap is the best example.

Properties of Cross Product
1. Vector product results in a vector.
2. Vector product or Cross product is not commutative.
i i
x j j
x
= = k k
x = 0
3.
A x B = - B x A
i j
x = k
4.
j k
x = i
k i
x = j
j i
x = - k
5.
k j
x = - i
i k
x = - j
6. If i
= ax
a j
+ ay k
+ az
i
= bx
b j
+ by k
+ bz
a x b = i j k
ax ay az
bx by bz
= (ay bz – az by) i – (ax bz – az bx) j + (ax by – aybx) k
| a x b | = (ay bz – az by)2
+ (ax bz – az bx)2
+ (ax by – aybx)2
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r
Angular velocity and its relation with linear velocity
X
Y
Z
v
P
P’
C
∆θ
∆s
The particle at P describes a circle with
the centre C on the axis and radius
r, the perpendicular distance of the point
P from the axis.
We also show the linear velocity vector v
of the particle at P. It is along the tangent
at P to the circle.
The average angular velocity
of the particle over the interval Δt is
 =
Δθ
Δt
lim
Δt→0
∆θ
∆t
 =
Let P′ be the position of the particle after
an interval of time (Δt). The angle PCP′
describes the angular displacement Δθ of
the particle in time Δt.
As Δt tends to zero (i.e. takes smaller and smaller values), the ratio Δθ/Δt
approaches a limit which is the instantaneous angular velocity and is given
by
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The linear velocity is the rate of change of linear displacement.
, r and v are mutually perpendicular to each other and  is perpendicular
to the plane containing r and v.
v =  x r
v =
Δs
Δt
lim
Δt→0
But Δs = Δθ x r
P
r
O
v
 v =
Δt
lim
Δt→0
Δθ x r
Here, v and r are shown on the plane of the screen and hence  is
perpendicular to the plane and emerges out of the plane. (RHSR)
Directions of r, v and 
or v =
Δθ
Δt
lim
Δt→0
x r
For rotation about a fixed axis, the direction of the vector  does not
change with time. Its magnitude may, however, change from instant to
instant.
For the more general rotation, both the magnitude and the direction of 
may change from instant to instant.

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Angular acceleration
Angular acceleration α is the time rate of change of angular velocity.
Thus, d
dt
α =
TORQUE OR MOMENT OF FORCE
The rotational analogue of force is moment
of force. It is also referred to as torque.
X
Y
Z
r
r sin θ
O
F
τ θ
θ
the moment of the force acting on the
particle with respect to the origin O is
defined as the vector product
If a force acts on a single particle at a
point P whose position with respect to
the origin O is given by the position
vector r, then
P
= r × F
τ
The moment of force (or torque) is a vector quantity.
The magnitude of τ is given by τ = r F sinθ
where r is the magnitude of the position vector r, i.e. the length OP, F is
the magnitude of force F and θ is the angle between r and F as shown.
Its direction is given by the right handed screw rule and is ┴ to the plane
containing r and F. In this case, it emerges out of the plane. Home Next Previous

In the previous treatment position vector r was resolved and perpendicular
component r sin θ was used for torque.
Similarly, the force vector F can be resolved and perpendicular component
F sin θ can be used for torque.
Therefore, torque can be interpreted in
two ways viz.
i) τ = (r sin θ) F
or τ = r┴ F
ii) τ = r (F sin θ)
or τ = r F┴
Note that τ = 0 if r = 0, F = 0 or θ = 0° or 180°.
Thus, the moment of a force vanishes if
either the magnitude of the force is zero, or
if the line of action of the force passes
through the origin.
If the direction of F is reversed, the direction of the moment of force is reversed.
If directions of both r and F are reversed, the direction of the moment of force
remains the same.
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X
Y
Z
r
O
F
τ θ
P
F sin θ

ANGULAR MOMENTUM OR MOMENT OF MOMENTUM
The rotational analogue of momentum is
moment of momentum or angular
momentum.
the angular momentum of the particle
with respect to the origin O is defined
as the vector product
= r × p
L
The angular momentum is a vector quantity.
The magnitude of L is given by L = r p sinθ
Its direction is given by the right handed screw rule and is ┴ to the plane
containing r and p. In this case, it emerges out of the plane.
If a single particle at a point P (whose
position with respect to the origin O is
given by the position vector r )
experiences linear momentum p, then
where r is the magnitude of the position vector r, i.e. the length OP, p is
the magnitude of momentum p and θ is the angle between r and p as
shown.
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X
Y
Z
r
r sin θ
O
p
L θ
θ
P

In the previous treatment position vector r was resolved and perpendicular
component r sin θ was used for angular momentum.
Similarly, the momentum vector p can be resolved and perpendicular
component p sin θ can be used for angular momentum.
Therefore, angular momentum can be
interpreted in two ways viz.
i) L = (r sin θ) p
or L = r┴ p
ii) L = r (p sin θ)
or L = r p┴
Note that L = 0 if r = 0, p = 0 or θ = 0° or 180°.
Thus, the angular momentum vanishes if
either the magnitude of the momentum is
zero, or if the line of action of the
momentum passes through the origin.
If the direction of p is reversed, the direction of the angular momentum is
reversed. If directions of both r and p are reversed, the direction of the angular
momentum remains the same.
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X
Y
Z
r
O
p
L θ
P
p sin θ

Relation between Torque and Angular Momentum
It is the rotational analogue of the relation between force and linear momentum.
= r × p
L
Angular momentum is given by
Differentiating w.r.t. time,
dL
dt
=
d
dt
( r × p )
p
dr
dt
x +
dp
dt
r x
=
mv
x +
dp
dt
r x
= v (since
dr
dt
v
= mv
=
p
and )
+
dp
dt
r x
v )
x
= m ( v
+
dp
dt
r x
0
= v )
x =
( v 0
(since )
= r × F (since
dp
dt
F
= )
= r × F
τ
(since )
τ
=
dL
dt
Thus, the time rate of change of
the angular momentum of a
particle is equal to the torque
acting on it. This is the rotational
analogue of the equation F = dp/dt,
which expresses Newton’s second
law for the translational motion of
a single particle.
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Torque and angular momentum for a system of particles
For a system of n particles,
L = L1 + L2 + …….…. + Ln
∑ Li
i=1
n
L =
The angular momentum of the ith
particle is given by Li = ri × pi
 ∑ ri x pi
i=1
n
L =
This is a generalisation of the definition of angular momentum for a single
particle to a system of particles.
Differentiating ,
∑ Li
i=1
n
L =
dL
dt
=
d
dt
∑ Li
i=1
n
or
dL
dt
=
d
dt
∑
i=1
n
Li or
dL
dt
= ∑
i=1
n
τi
where ri is the position vector of the ith
particle with respect to a
given origin and pi = (mivi) is the linear momentum of the particle.
where τi is the torque of the ith
particle and = ri × Fi
τi
or τ = ∑
i=1
n
τi
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The force Fi on the ith
particle is the vector sum of external forces Fi
ext
acting
on the particle and the internal forces Fi
int
exerted on it by the other particles
of the system.
 τ = ∑
i=1
n
τi
can be written as
τ = τext + τint where = ri × Fi
ext
τext
and = ri × Fi
int
τint
The forces between any two particles are not only equal and opposite but
also are directed along the line joining the two particles.
In this case the contribution of the internal forces to the total torque on the
system is zero, since the torque resulting from each action-reaction pair of
forces is zero.
Thus,
= 0
τint
 τ = τext

dL
dt
= τext
Thus, the time rate of the total angular
momentum of a system of particles
about a point is equal to the sum of the
external torques acting on the system
taken about the same point.
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Conservation of angular momentum
dL
dt
= τext
dL
dt
= 0
If τext = 0, then
or L = constant
Thus, if the total external torque on a system of particles is zero, then the
total angular momentum of the system is conserved.
The above equation is equivalent to three scalar equations,
Lx = K1, Ly = K2 and Lz = K3
where K1, K2 and K3 are constants; Lx, Ly and Lz are the components of the
total angular momentum vector L along the x, y and z axes respectively.
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Acknowledgement
1. Physics Part I for Class XI by NCERT
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