SYSTEM OF PARTICLES – ALL DERIVATIONS
CENTER OF MASS OF A TWO PARTICLE SYSTEM
Consider a system of two particles P1 and P2 of masses m1 and m2. Let 𝑟⃗1 and 𝑟⃗2 be
their position vectors at any instant t with respect to the origin O, as shown in Fig.
Let m1,m 2 be masses of two particles
let r1,r2 be position vectors of particles
let f1,f2 be external forces on particles
Let v 1,v 2 be velocities of particles
Let F12 ,F21 be internal forces on particles (due to each other)
According’s to Newton’s second law:
  



d
f1  f2  F12  F21  m1v1  m2 v 2 
dt




dr1
dr 2
As v1 
and v 2 
dt
dt


  

d  dr1
dr 2 
f1  f2  F12  F21   m1
 m2

dt 
dt
dt 


But F12   F21 so they will cancel out
 


d2
f1  f2  2  m1r1  m2 r2 
dt
Multiply and divide L.H.S by m1 + m2 we get


 
d2  m1r1  m2 r2 
f1  f2   m1  m2  2
dt  m1  m2 
 
Let f  f1  f2



d2  m1r1  m2 r2 
f   m1  m2  2
dt  m1  m2 

d2 
Comparing this equation with f   m1  m2  2 Rcm , we get
dt



m1r1  m2 r2 

Rcm 
m1  m2 
Centre of mass of an n particle system





m1r1  m2 r2  m3 r3 .........  mn rn 

Rcm 
 m1  m2  m3  ..............mn 
RELATION BETWEEN ROTATIONAL K.E. AND MOMENT OF INERTIA
Relation between rotational kinetic energy and moment of inertia. As shown in Fig.,
consider a rigid body rotating about an axis with uniform angular velocity ω. The
body may be assumed to consist of n particles of masses m1 , m2 , m3 ,..........mn ;
situated at distances r1 , r2 , r3 ,..........rn from the axis of rotation. As the angular
velocity ω of all the n particles is same, so their linear velocities are
v1 = r1 ω , v2 = r2 ω , v3 = r3 ω,.........,vn = rn ω
Hence the total kinetic energy of rotation of the body about the axis of rotation is
Rotational K.E.
1
1
1
m1v12  m2 v 22  ............  mn v n2
2
2
2
1
1
1
 m1r12ω2  m2r22ω2  ............. mnrn2ω2
2
2
2
1
 m1r12  m2r22  ..........mnrn2 ω2
2
1
 Iω2
2
where I  m1r12  m2r22  ..........mnrn2 (moment of inertia)



RADIUS OF GYRATION
The radius of gyration of a body about it axis of rotation may be defined as the
distance from the axis of rotation at which, if the whole mass of the body were
concentrated, its moment of inertia about the given axis would be the same as with
the actual distribution of mass.
Expression for k. Suppose a rigid body consists of n particles of mass m each,
situated at distances r1 , r2 , r3 ,..........rn from the axis of rotation AB.
The moment of inertia of the body about the axis AB is
I  mr12  mr22  mr32  .......mnrn2

I  m r12  r22  r33  .......

Multiplying and dividing RHS by n, we get
I

m  n r12  r22  r33  .......

n
now m × n = M, total mass of the body.
If k is the radius of gyration about the axis AB, then
I = MK2 , therefore
Mk 
2

M r12  r22  r33  .......

n
r12  r22  r33  .......
 k
n
RELATION BETWEEN ANGULAR MOMENTUM AND MOMENT OF INERTIA
Consider a rigid body rotating about a fixed axis with uniform angular velocity ω. The
body consists of n particles of masses m1, m2, m3, ......., mn ;situated at distance r1 ,
r2 , r3 ......, rn from the axis of rotation. The angular velocity ω of all the n particles will
be same but their linear velocities will be different and are given by
v1 = r1 ω, v2 = r2 ω, v3 = r3 ω,......., vn = rn ω
Linear momenta of particles,
p1 = m1r1 = m1r1ω,
p2 = m2r2 = m2r2ω,
p3 = m3r3 = m3r3ω…………
Therefore, the angular momenta of particles
2
L1 = p1r1 = m1r ω
1
2
L2 = p2r2 = m2r ω
2
2
L3 = p3r3 = m3r ω ………
2
The angular momentum of a rigid body about an axis is the sum of moments of linear
momenta of all its particles about that axis. Thus
L = L1 + L2 + L3 + ......,+ Ln
 m1r12ω  m2r22ω  m3r32ω  ...........mnrn2ω


 m1r12  m2r22  m3r32  .......mnrn2 ω
Since
m1r12  m2r22  m3r32  .......mnrn2  I (moment of inertia)
 L  Iω
RELATION BETWEEN ANGULAR MOMENTUM AND TORQUE
As
Torque,
  
τ  r F
  
and Angular Momentum, L  r  p
Differentiating both sides w.r.t. time t, we get



dL d   dr   dp
 (r  p) 
p  r 
dt dt
dt
dt


  
 dp  
 v p  r F
 dt  F



  

0τ
 v  p  v  mv  0

 dL
τ
dt
RELATION BETWEEN TORQUE AND MOMENT OF INERTIA
Consider a particle P of mass m1 at a distance r1 from the axis of rotation. Let its
linear velocity be v1 .
Linear acceleration of the first particle, a1 = r1 α
Moment of force F1 about the axis rotation is
1  r1f1 = m1 r12 α
Similarly, 2  r2 f2 , 3  r3 f3 , ...............n  rn fn ,
Total torque acting on the rigid body is
τ = τ1+ τ2 + τ3 +..........+ τn
τ  m1r12 α  m2r22α  m3r32 α  .......mnrn2 α


 τ  m1r12  m2r22  m3r32  ........mnrn2 α
Since m1r12  m2r22  m3r32  ........mnrn2  I (moment of inertia)
 τ  Iα
EQUATIONS OF ROTATIONAL MOTION
First equation of motion. Consider a rigid body rotating about a fixed axis with
constant angular acceleration α. By definition,
dω
dt
 dω  αdt
α
Integrting both sides within limits
ω2
t
 dω  α dt
ω1
0
 [dω]
ω2
ω1
 α[t]0t
 ω2  ω1  α[t  0]
 ω2  ω1  αt
Second equation of motion. Let ω2 be the angular velocity of a rigid body at any
instant t. By definition,
dθ
dt
dθ  ω2dt
ω2 
 dθ  (ω1  αt)dt
Integrting both sides within limits
θ
t
t
0
0
0
 dθ  ω1  dt  α tdt
t
 t2 
(θ)  ω1(t)  α  
 2 0
θ
0
t
0
 t2

(θ  0)  ω1(t  0)  α   0 
2

θ  ω1t 
1 2
αt
2
Third equation of motion. The angular acceleration α may be expressed as
dω
dt
m u lt ip ly a n d d iv id e b y d θ , w e g e t
α 
dω
dθ

dθ
dt
dω
α  ω
dθ
 αdθ  ω dω
α=
I n t e g r a t in g b o t h s id e s w it h in g iv e n lim it s
θ
ω2
0
ω1
α dθ 

ω dω
ω2
 α(θ )
θ
0
ω2 
 

 2  ω1
 ω 22
ω 12 
 α(θ  0 )  


2 
 2
 2 α θ  ω 22  ω 12
ACCELERATION OF A BODY MOVING DOWN A ROUGH INCLINED PLACE
Consider a body of mass M and radius R rolling down a plane inclined at an angle θ to the
horizontal.
The external forces acting on the body are
I.
II.
III.
The weight Mg of the body acting vertically downwards through the center of mass of the
cylinder.
The normal reaction N of the inclined plane acting perpendicular to the plane at P.
The frictional force f acting upwards and parallel to the inclined plane.
The weight Mg can be resolved into two rectangular components:
I.
II.
Mg cos θ perpendicular to the inclined plane.
Mg sin θ acting down the inclined plane.
As there is no motion in a direction normal to the inclined plane, so
N = Mg cos θ
Applying Newton’s second law to the linear motion of the center of mass, the net force on
the body rolling down the inclined plane is
F = Ma = Mg sin θ – f
It is only the force of friction f which exerts torque τ on the cylinder and makes it rotate with angular
acceleration α. It acts tangentially at point of contact P and has lever arm equal to R.
τ = Force x force arm = fR
Also,
τ = Iα
FR = Iα
f
Or
Iα
R
Putting the value of f in equation, we get
Mgsinθ  f  Ma
 Mgsinθ 
As α 
Iα
 Ma
R
a
R
Ia
 Ma
R2
Ia
Mgsinθ  2  Ma
R
 I

 a  2  M   Mgsinθ
R

 Mgsinθ 
 a
Mgsinθ
 I

 2  M
R

LAW OF CONSERVATION OF ANGULAR MOMENTUM
It states that if external torque acting on a system is zero, the total angular momentum of the
system remains conserved.
Proof:
As we know
τ ext 
if τ ext
dL
dt
0
dL
0
dt
or L  constant
then