PHYSICS OF MANY PARTICLES

1. A body is made of a collection of particles. We would like to think of this body having
a "centre". For two masses the "centre of mass" is defined as: . cm
r m r m r
m m
+
≡
+
􀁇 􀁇 􀁇
1 1 2 2 1 1 2 2
1 2 1 2
In 2 dimensions (i.e. a plane) this is actually two equations:
and . These give the coordinates
of the centre of mass of the two-particl
cm cm
x m x m x y m y m y
m m m m
+ +
= =
+ +
e system.
2. Example: one mass is placed at 2 and a second mass, equal to the first, is placed
at 6 . The cm position lies halfway between the two as you can see from:
x cm
x cm
=
=
1 2 2 6 4 .
2
Note that there is no physical body that is actually located at 4 ! So the centre
of mass can actually be a point where there is no matter. Now supp
cm
cm
x mx mx m m cm
m m m
x cm
+ +
= = =
+
=
1 2
ose that the first
mass is three times bigger than the first:
(3 ) 2(3 ) 6 3
3 4
This shows that the cm lies closer to the heavier body. This is
cm
x m x mx m mcm
m m m
+ +
= = =
+
1 1 2 2
1 2 1
always true.
3. For N masses the obvious generalization of the centre of mass position is the following:
1 .
n N
N N
cm n n
N n
rm r m r m r mr
m m m M
=
=
+ + ⋅ ⋅ ⋅ ⋅ + ⎛ ⎞
= + + ⋅ ⋅ ⋅ ⋅ + = ⎜⎝ ⎟⎠
Σ
􀁇 􀁇 􀁇 􀁇 􀁇
1 2
1 2
In words, this says that the following: choose any origin and draw vectors , ,
that connect to the masses , , . Heavier masses get more importance in the sum.
So suppose that
N
N
r r r
m m m
⋅ ⋅ ⋅
⋅ ⋅ ⋅
􀁇 􀁇 􀁇
2 2
2 2
2
2
is much larger than any of the others. If so, . Hence, the
cm is very close to the position vector of .
4. For symmetrical objects, it is easy to see where the cm position lies:
cm
m rm r r
m
m
≈ =
􀁇 􀁇 􀁇
for a sphere or circle
it lies at the centre; for a cylinder it is on the axis halfway between the two faces, etc.
5. Our definition of the cm allows Newton's Second Law to be written for entire collection
( )
( )
( )
of particles:
v 1 v
a v 1 a
a use 0
cm
cm n n
cm
cm n n
cm n ext int int
ex
dr m
dt M
d m
dt M
M F F F F
F
= =
= =
∴ = = + =
⇒
Σ
Σ
Σ Σ Σ
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
􀁇
12 21
(the sum of external forces is what causes acceleration)
In the above we have used Newton's Third Law as well: 0 etc.
6. Consider rotational motion now for a rigid system of N par
t cm M a
F F
=
+ =
Σ 􀁇
􀁇 􀁇
2 2 2
1 1 2 2 3 3
ticles. Rigid means that
all particles have a fixed distance from the origin. The kinetic energy is,
1 v 1 v 1 v
2 2 2
K= m + m + m + ⋅ ⋅ ⋅ ⋅
( )
2 2 2 2 2 2
1 1 2 2 3 3
2 2
2
2
1 1 1
2 2 2
1
2
Now suppose that we define the "moment of inertia" . Then clearly the
kinetic energy is 12 .
i i
i i
m r m r m r
mr
I mr
K I
ω ω ω
ω
ω
= + + + ⋅ ⋅ ⋅ ⋅
=
≡
=
Σ
Σ
2
1 2
How similar this is to 1 v !
2
7. To familiarize ourselves with I, let us consider the following: Two particles and
are connected by a light rigid rod of length . Neglecting the mass of the
K M
m m
L
=
1
rod, find the
rotational inertia of this system about an axis perpendicular to the rod and at a
distance from .
I
x m
2 ( )2
1 2 Answer: . Of course, this was quite trivial. Now we can ask
a more interesting question: For what is the largest? Now, near a maximum, the
slope of a function is zero.
I mx m L x
x I
= + −
( ) 2
1 2 max
1 2
So calculate and then put it equal to zero:
2 2 0 .
dI
dx
dI m x m L x x m L
dx m m
∴ = − − = ⇒ =+
1 m 2 m
x
L
x
y
z
dφ
F r
r
ds
dW = F ⋅ ds
􀁇 􀁇
8. Although matter is made up of discrete atoms, even if one takes small pieces of any
body, there are billions of atoms within it. So it is useful to think of matter as being
continuously distr
cm
2
cm
ibuted. Since a sum becomes an integral , it is obvious that
the new definitions of and R becom:
and R 1 .
9. A simple application
I
I rdm rdm
M
≡ ≡
Σ ∫
∫ ∫
􀁇
􀁇 􀁇
2 2 2
: suppose there is a hoop with mass distributed uniformly over it.
The moment of inertia is: = .
10. A less trivial application: instead of a hoop as above, now consider a solid plat
I=∫rdm R∫dm=MR
2
0
3 2 2 2
0 0
0
e:
( 2 )
= 2 1( ) 1
2 2
11. You have seen that it is easier to turn things (e.g. a nut, when changi
R
I r dm dm rdr
r dr R R MR
π ρ
π ρ π ρ
= =
= =
∫
∫
ng a car's tyre
after a puncture) when the applied force acts at a greater distance. This is because
the is greater. We define from the magnitude is sin .
Here i
torqueτ τ r F τ rF θ
θ
= × =
􀁇 􀁇 􀁇
s the angle between the radius vector and the force.
12. Remember that when a force acts through a distance it does an amount of work
equal to . Now let us ask how much work is done wh
F dr
F⋅dr
􀁇 􀁇
􀁇 􀁇 en a torque acts through a
certain angle as in the diagram below:
( )( )
( ) 1 1 1 2
The small amount of work done is:
cos cos
Add the contributions coming from from all particles,
cos cos net
dW F d s F ds F rd d
dW F r d F
θ θ φ τ φ
θ φ θ
= ⋅ = = =
= +
􀁇 􀁇
( ) ( ) 2 2 cos n n n r dφ + ⋅ ⋅ ⋅ + F θ r dφ
( )
( ) ( )
1 2
ext ext
2
τ dφ= τ ωdt .....(1)
Now consider the change in the kinetic energy ,
1
2
n
net
d
dW
K
dK d I
τ τ τ φ
ω
= + + ⋅ ⋅ ⋅ ⋅ +
∴ =
= ⎛⎜ ⎞
⎝ ⎠
Σ Σ
( ) ......(2)
By conservation of energy, the change in K must equal the work done, and so:
In words this says that the total torque equals
net ext
I d I dt
dW dK I
ω ω α ω
τ α
⎟= =
= ⇒ Σ =
the moment of inertia times the angular
acceleration. This is just like Newton's second law, but for rotational motion !
13.A comparison between linear and rotational motion quantities and formulae:
LINEAR ROTATIONAL
, ,
v
x M I
dx d
dt dt
φ
φ
= ω=
2 2
a v
a
1 v 1
2 2
d d
dt dt
F M I
K M K I
ω
α
τ α
ω
= =
= =
= =
14. Rotational and translational motion can occur simultaneously. For example a car's
wheel rotates and translates. In this case the total kinetic energy is clearly the s
W=∫Fdx W=∫τdφ
2 2
cm
um of
the energies of the two motions: 1 v 1 .
2 2
15. It will take a little work to prove the following fact that I simply stated above: for a
system of N particles, the total kineti
cm K= M + I ω
cm
c energy divides up neatly into the kinetic energy
of rotation and translation. Start with the expression for kinetic energy and write v +v
where v is the velocity of a particle with r
i
i
′
′
􀁇 􀁇
􀁇
( ) ( )
2
cm cm
espect to the cm frame,
1 v 1 v v
2 2
1 v +v v +v
2
i i i i i
i i i
K m m
m
= = ⋅
= ′ ⋅ ′
Σ Σ
Σ
􀁇 􀁇
􀁇 􀁇 􀁇 􀁇
(2 2)
cm cm i i
1 v +2v .v +v
2 i = Σ m 􀁇 􀁇′ ′
cm i cm cm
2 2
cm i
Now, v v v v v 0. Why? because the total momentum
is xero in the cm frame! So this brings us to our result that,
1 v 1 v
2 2
i ii i
i i
m m p
K m m
⋅ ′= ⋅ ′= ⋅ ′ =
= + ′
Σ 􀁇 􀁇 􀁇 Σ 􀁇 􀁇 Σ 􀁇
2 2 2
cm i
1 v 1 .
2 2 i = M + mr′ ω
Σ Σ
Σ