OSCILLATIONS: I

1. An oscillation is any self-repeating motion. This motion is characterized by:
a) The period , which is the time for completing one full cycle.
b) The frequency 1/ ,which is the number of
T
f= T cycles per second. (Another
frequently used symbol is ).
c) The amplitude A, which is the maximum displacement from equilibrium (or the
size of the oscillation).
2. Why does a syste
ν
m oscillate? It does so because a force is always directed towards
a central equilibrium position. In other words, the force always acts to return the
object to its equilibrium position. So th
restore
e object will oscillate around the equilibrium
position. The restoring force depends on the displacement , where
is the distance away from the equilibrium point, the negative si
F = −kΔx Δx
gn shows that the
force acts towards the equilibrium point, and is a constant that gives the strength of
the restoring force.
k
3. Let us understand these matters in the context of a spring tied
to a mass that can move freely over a frictionless surface. The
force ( ) (or - because the extension will be
ca
F x = −kx kΔx Δx
2
lled for short). In the first diagram is positive, is
negative in the second, and zero in the middle one. The energy
stored in the spring, ( ) 1 , is positive in the first and
2
th
x x x
U x = kx
2
ird diagrams and zero in the middle one. Now we will use Newton's second law to
derive a differential euation that describes the motion of the mass: From ( )
and it follows that
F x kx
ma F m d
= −
=
2
2 2
2 2 , or 0 where . This is
the equation of motion of a simple harmonic oscillator (SHO) and is seen widely in
many different branches of physics. Although we have derived it
x kx dx x k
dt dt m
= − +ω = ω ≡
for the case of a
mass and spring, it occurs again and again. The only difference is that , which is
called the oscillator frequency, is defined differently depending on the situation.
4. In
ω
order to solve the SHO equation, we shall first learn how to differentiate some
elementary trignometric functions. So let us first learn how to calculate d cos
dt
starting from the basic defini
ωt
tion of a derivative:
Compressed
Stretched
( ) ( ) ( )
( ) ( )
cos and cos . Take the difference:
( ) cos cos
sin sin( / 2)
Start : x t t x t t t t
x t t x t t t t
t t t
ω ω
ω ω
ω ω ω
= +Δ = + Δ
+ Δ − = + Δ −
= − Δ + Δ
sin as becomes very small.
cos sin .
(Here you should know that sin for small , easily proved by drawing trian
t t t
d t t
dt
ω ω
ω ω ω
θ θ θ
≈ − Δ Δ
∴ =−
≈ gles.)
You should also derive and remember a second important result:
sin cos .
(Here you should know that cos 1 for small .)
5. What
d t t
dt
ω ω ω
θ θ
=
≈
( )
( )
2
2
2
2
2
2
happens if you differentiate twice?
sin cos sin
cos sin cos .
So twice differentiating either sin
d t d t t
dt dt
d t d t t
dt dt
t
ω ω ω ω ω
ω ω ω ω ω
ω
= =−
= − = −
2
2
2
or cos gives the same function back!
6. Having done all the work above, now you can easily see that any function of the
form ( ) cos sin satisfies . But what do , a, b represe
t
x t a t b t d x x
dt
ω
= ω + ω =−ω ω
( )
nt ?
a) The significance of becomes clear if your replace by 2 in either sin
or cos . You can see that cos 2 cos 2 cos . That is, the
function merely repeats it
t t t
t t t t
π
ω ω
ω
π
ω ω ω π ω
ω
+
⎛⎜ + ⎞⎟= + =
⎝ ⎠
self after a time 2 / . So 2 / is really the period of the
motion , 2 2 . The frequency of the oscillator is the number of
complete vibrations per unit time: 1 1 so 2
2
T T m
k
k
T m
π ω π ω
π
π ν
ω
ν ω πν
π
= =
= = =
[ ] 1
2 .
Sometimes is also called the angular frequency. Note that dim , from
it is clear that the unit of is radian/second.
b) To understand what and mean let us note that fro
k
T m
T
a b
π
ω ω
ω
−
= =
=
m ( ) cos sin it
follows that (0) and that d ( ) sin cos (at 0). Thus,
dt
is the initial position, and is the initial velocity divided by .
x t a t b t
x a xt a t b t b t
a b
ω ω
ω ω ω ω ω
ω
= +
= =− + = =
44
Mg
P
θ C
θ
c) To understand what and mean let us note that from ( ) cos sin it
follows that (0) and that d ( ) sin cos (at 0). Thus,
dt
is the initial position, and
a b xt a t b t
x a xt a t b t b t
a b
ω ω
ω ω ω ω ω
= +
= =− + = =
( )
is the initial velocity divided by .
d) The solution can also be written as: ( ) cos . Since cos and sin never
beome bigger than 1, or less than -1, it follows that . For obviou
m
m m
x t x t
x x x
ω
= ω +φ
− ≤ ≤+ s reason
is called the amplitude of the motion. The frequency of the simple harmonic motion
is independent of the amplitude of the motion.
e) The quantity is called the phase of t
m x
θ =ωt +φ he motion. The constant is called the
A different value of just means that the origin of time has been
chosen differently.
7. . Put 0
phase constant.
φ
φ
Energy of simple harmonic motion φ =
2
for convenience, and so imagine a mass
whose position oscillates like cos . Let us first calculate the potential energy:
1 1
2 2
m
m
x x t
U kx kx
= ω
= = 2 2
2
2 22 2 2 2
cos .
Now calculate the kinetic energy:
1 v 1 1 sin 1 sin
2 2 2 2
The sum of potential + kinetic is:
m m
t
K m m dx m x t kx t
dt
ω
= = ⎛⎜ ⎞⎟ = ω ω = ω
⎝ ⎠
( )
2 2 2 2
2 2 2 2
1 cos 1 sin
2 2
1 cos sin 1 .
2 2
Note that this is independent of time and energy goes from kinetic to potential,
m m
m m
E K U kx t kx t
kx t t kx
ω ω
ω ω
= + = +
= + =
( 2 2)
then
back to kinetic etc.
8. From the above, you can see that v . From this it is clear that
the speed is maximum at 0 and that the speed is zero at .
9. Putting two springs i
m
m
dx k x x
dt m
x xx
= =± −
= =±
1 2
1 2
1 2
n parallel makes it harder to stretch them, and . In
series they are easier to stretch, and . So a mass will oscillate faster in
the first case as compared to the
eff
eff
k k k
k k k
k k
= +
⎛ ⎞
=⎜⎝ + ⎟⎠
second.

EQUILIBRIUM OF RIGID BODIES

1. A rigid body is one where all parts of the body are fixed relative to each other (for
example, a pencil). Fluids and gases are non-rigid.
2. The translational motion of the centre of mass of a rigid body is governed by:
where is the external force.
Similarly, for rotational motion, where is the net external torque.
ext
ext
dP F F F net
dt
dL
dt
τ τ τ
= =
= =
Σ
Σ
􀁇 􀁇 􀁇 􀁇
􀁇
􀁇 􀁇 􀁇
3. A rigid body is in mechanical equilibrium if the linear momentum and angular
momentum have a constant value. i.e., 0 0.
refers to 0 a
P
L dP dL
dt dt
P
= =
=
both
and Static equilibrium
􀁇
􀁇 􀁇 􀁇
􀁇
nd 0.
4. As an example of static equilibrium, consider a beam resting on supports:
L =
􀁇
1 2 We want to find the forces and with which the supports push on the rod
in the upwards direction. First, balance forces in the vertical direction:
y
F F
y
F
( )( ) ( )( ) ( )( ) ( )( )
1 2
1 2
1 2
0
Now demand that the total torque vanishes:
0 / 4 / 2 0
From these two conditions you can solve for and ,
y
F F Mg mg
F F L Mg L mg L
F F
τ
= + − − =
= + − − =
Σ
Σ
( ) ( )
1 2
3 2 2
, and .
4 4
5. Angular momentum and torque depend on where you choose the origin of your
coordinates. However, I shall now prove that for a body in equi
M mg M mg
F F
+ +
= =
1 2 1 1 2 2
librium, the choice of
origin does not matter. Let's start with the origin O and calculate the tor
que about O,
Now, if we
O N N N τ =τ +τ + ⋅ ⋅ ⋅ +τ =r×F+r ×F + ⋅ ⋅ ⋅ +r ×F
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
take a second point P, then all distances will be measured from P and each
L
L / 4 M
m
1 F 2 F
i r􀁇
CG r􀁇
O x
y
z
CG
i m g􀁇
M g􀁇
( ) ( ) ( ) 1 1 2 2
1 1 2 2
vector will be shifted by an amount . Hence the torque about P is,
P
P P P N P N
N N
r
r r F r r F r r F
r F r F r F
τ = − × + − × + ⋅ ⋅ ⋅ ⋅ + − ×
= ⎡ × '2B × +⋅⋅⋅+ ×
􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
( )
( )
1 2
1 2
but 0, for a body in translational equilibrium
P P P N
O P N
O P ext
ext
r F r F r F
r F F F
r F
F
τ
τ
⎣ ⎦⎤ − ⎣⎡ × + × + ⋅ ⋅ ⋅ + × ⎦⎤
= −⎡⎣ × + + ⋅ ⋅ ⋅ + ⎤⎦
= −⎡⎣ × ⎤⎦
= ∴
Σ
Σ
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇
􀁇
.
6. Let us use the equilibrium conditions to do something of definite practical importance.
Consider the balance below which is in equilibrium when two known weights are hung
as shown.
P O τ􀁇 =τ􀁇
1 2 We want to know m in terms of m and m .
( ) ( ) 1 2
1
1 2
2
Taking the torques about the knife edge in the two cases, we have:
and
or .
Remarkably, we do not need the values
mgx m g L x m gx mg L x
m m m mm
m m
= − = −
⇒ = =
of or .
7. The centre of gravity is the average location of the weight of an
object. This is not quite the same as the centre of mass of a body (see lecture 12) but
suppos
x L
Centre of Gravity.
e the gravitational acceleration has the same value at all points of a body.
Then: 1) The weight is equal to , and 2)the centre of gravity coincides with
the centre of mass. Remember tha
g
M g
􀁇
􀁇
t weight is force, so the CG is really the centre of
gravitational force acting on the body. The net force on the whole body = sum of forces
over all individual particles, . If
i ΣF=Σmg g 􀁇 􀁇 􀁇
( ) ( )
has the same value at all points of the body, then
. So the net torque about
the origin is
Hence, . So the torque
due to gravit
i
i i ii
cm cm
F g m Mg
O r m g m r g
Mr g r Mg
τ
τ
= =
= × = ×
= × = ×
Σ Σ
Σ Σ Σ
Σ
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
y about the centre of mass of a body
(i.e. at 0 ) is zero !! cm r􀁇 =
L x
mg 1 m g
L x
2 m g mg
L
W
x
θ φ
+
CM
A
B O
W
C
D
1 R
2 R
1 1 μ R
2 2 R μ
θ
8. In the demonstration I showed, you saw how to find the CG of an irregular object
by simply suspending it on a pivot,
9. Let's solve a problem of static equilibrium: A non-uniform bar of weight W is suspended
at rest in a horizontal position by two light cords. Find the distance from the left-hand
end to the
x
( )
1 2
2 1
2 1 2
center of gravity.
Call the tensions and . Put the forces in
both directions equal to zero,
a) sin sin 0 (horizontal)
b) cos cos 0 (vertical)
sin
Solution : T T
T T
T T W T W
φ θ
φ θ
θ φ
− =
+ − = ⇒ =
+
( ) ( )
( )
2
2
The torque about any point must vanish. Let us choose that point to be one end of the bar,
cos 0 cos cos
sin
10.Here is another problem of the same k
Wx T L x T L L
W
φ φ
φ
θ φ
− + = ⇒ = =
+
1 2 2 2 1 1
ind: find the least angle at which the rod can lean
to the horizontal without slipping.
Considering the translational equilibrium of
the rod, and . This
Solution :
R R R R W
θ
=μ +μ =
2 ( )
1 2
2 22
2 22
2 2 2
gives,
. Now consider rotational equilibrium
1
about the point A:
or, cos cos sin .
2
This gives cos sin from which tan
2
R W
R OB W OD R OA
R AB W AB R AB
R W R
μ μ
μ
θ
θ μ θ
θ μ θ θ
=
+
× = × + ×
× = × + ×
⎛⎜ − ⎞⎟=
⎝ ⎠ ( )
2
2
2 2 1 2
1 2
2
2
2 with .
1
Using this value of , we get tan 1- .
2
10. In the lecture you heard about:
a)Stable equilibrium: object returns to its original positi
R W R W
R
R
μ μ μ
μ μ
θ
μ
−
= =
+
=
Types of Equilibrium.
on if displaced slightly.
b)Unstable equilibrium: object moves farther away from its original position if
displaced slightly.
c)Neutral equilibrium: object stays in its new position if displaced slightly.

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

ANGULAR MOMENTUM

( )
( )
1. Recall the definition of angular momentum: . The magnitude can be written
in several different but equivalent ways,
(a) sin
(b) sin
(c) sin
L r p
L rp
L r p rp
L r p rp
θ
θ
θ
⊥
⊥
= ×
=
= =
= =
􀁇 􀁇 􀁇
2. Let us use this definition to calculate the angular momentum of a projectile thrown from
the ground at an angle . Obviously, init θ ial angular momentum is zero (why?).
( ) ( ) 2
0 0
0
We know what the projectile's coordinates will be at time after launch,
v cos , v sin 1
2
as well as the velocity components,
v v c x
t
x= θt y= θ t − gt
=
( ) ( ) ( )
0
2 2 2
0 0 0
os , v v sin .
Hence, ˆ ˆ vˆ v ˆ v v v ˆ
1 v cos v cos ˆ v cos ˆ.
2 2
In the above, ˆ ˆ ˆ is a unit vector pe
y
x y x y x
gt
L r p xi yj i j m m x y k
m gt gt k mgt k
k i j
θ θ
θ θ θ
= −
= × = + × + = −
= ⎛⎜ − ⎞⎟ = −
⎝ ⎠
= ×
􀁇 􀁇 􀁇
2
rpendicular to the paper. You can see here
that the angular momentum increases as .
2. Momentum changes because a force makes it change. What makes angular momentum
change? Answer: torque. H
t
ere is the definition again: . Now let us establish a
very important relation between torque and rate of change of L.
Begin:
. At a slightly later time,
r F
L r p L L r
τ = ×
= × + Δ = + Δ
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 (􀁇 ) ( )
By subtracting, .
r p p
r p r p r p r p
L r p r p L r p r p r p r p
t t t t
× +Δ
= × + ×Δ + Δ × + Δ ×Δ
Δ ×Δ +Δ × Δ Δ
Δ = ×Δ + Δ × = = × + ×
Δ Δ Δ Δ
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
x
y 0 v
v
x v
y v
O
θ
z
L 􀁇
O
Δφ
ΔL
􀁇
z Lsinθ
θ
( )
0
Now divide by the time difference and then take limit as 0 :
lim
But is v and v ! Also, v v v v
t
t
L dL dL r dp dr p
t dt dt dt dt
dr p m dr p m m
dt dt
Δ →
Δ →
Δ
= ∴ = × + ×
Δ
= × = × = ×
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
∵
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 0. So we arrive
at . Now use Newton's Law, . Hence we get the fundamental
equation , or . So, just as a particle's momentum changes with time
bec
dLrd p Fd p
dt dt dt
dL r F dL
dt dt
τ
=
= × =
= × =
􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
ause of a force, a particle's momentum changes with time because of a torque.
3. As you saw in the lecture, the spinning top is an excelllent application of .
angular
dL
dt
=τ
􀁇
􀁇
Start from where sin . But is perpendicular to and
so it cannot change the magnitude of . Only the direction changes. Since ,
you can see from the
r F F mg Mgr L
L L t
τ τ θ τ
τ
= × = ∴ =
Δ = Δ
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇
diagram that . So the precession speed
sin sin
is: sin . As the top slows down due to friction and
sin sin
decreases, the top precesses faster and faster.
4
P
P
L t
L L
Mgr Mgr L
t L L L
τ
φ ω
θ θ
φ τ θ
ω
θ θ
Δ Δ
Δ = =
Δ
= = = =
Δ
. Now consider the case of many particles. Choose any origin with particles moving with
respect to it. We want to write down the total angular momentum,
L=L
􀁇 􀁇
1 2
1
1 2
1
1
Since , it follows that . Thus the time rate of change of the total
N
N n
n
N
N n
n
N
n
n n
n
L L L
dL dL dL dL dL
dt dt dt dt dt
dL dL
dt dt
τ τ
=
=
=
+ + ⋅ ⋅ ⋅ + =
= + + ⋅ ⋅ ⋅ + =
= =
Σ
Σ
Σ
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇
􀁇 􀁇
angular momentum of a system of particles equals the net torque acting on the system.
I showed earlier that internal forces cancel. So also do internal torques, as we shall see.
5. The torque on a system of particles can come both from external and internal
forces. For example, there could be charged particles which attract/repel each
other while they are all in an exte
int
rnal gravitational field. Mathematically,
. Now, if the forces between two particles not only are equal and
opposite but are also directed along the line joining the two parti
Στ􀁇=Στ􀁇 +Στ􀁇ext
int
int 1 2 1 12 2 21
12
cles, then can easily
show that the total internal torque, 0. Take the case of two particles,
But
r F r F
F F
τ
τ τ τ
=
= + = × + ×
= −
Σ
Σ
􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 ( ) ( ) ( ) 21 12 int 1 2 12 12 12 12 12 ˆ, ˆ ˆ 0
Thus net external torque acting on a system of particles is equal to the time rate of change
of the of the total angular momentum of the sy
=Fr ∴Στ = r−r ×F =r × Fr =F r ×r = 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
stem.
6. It follows from that if no net external torque acts on the system, then the angular
momentum of the system does not change with the time: 0 a constant.
This is simple
dL
dt
dL L
dt
=τ
= ⇒ =
􀁇
􀁇
􀁇 􀁇
2
but extremely important. Let us apply this
to the system shown here. Two stationary discs, each with
1 , fall on top of a rotating disc. The total angular
2
momentum is unchanged so, i
I MR
I
=
2
2
.
Hence, 2 =1 .
2 3 3
7. You shoud be aware of the similarities and differences between the equations for linear
and rotational motion:
i
i f f f i
f
f i i
I I
I
MR
MR
F d p d
dt
ω ω ω ω
ω ω ω
τ
⎛ ⎞
= ⇒ = ⎜⎜ ⎟⎟
⎝ ⎠
⎛ ⎞
= ⎜ × ⎟
⎝ ⎠
= ⇔ =
􀁇 􀁇 􀁇 􀁇 . One big difference is that for momentum,
v, it does not matter where you pick the origin. But definitely depends on the
choice of the origin. So changing to changes to
L
dt
p m L
r c r L
=
+
􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇
( )
( )
:
.
8. Linear and angular acceleration:
a v
L
L r p c r p c p L
d d r d r dr
dt dt dt dt
ω
ω ω
′
′ = ′× = + × = × +
= = × = × + ×
􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
T R
v
So the acceleration has a tangential and radial part, a a a .
=α×r +ω×
= +
􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 􀁇
2
i 2
I = MR
3 2
f 2
I = MR

ROTATIONAL KINEMATICS

1. Any rotation is specified by giving two pieces of information:
a) The point about which the rotation occurs, i.e. the origin.
b) The angle of rotation is denoted by in the diagram.
2. The a
φ
rc length = radius angular displacement, or .
Here is measured in radians. The maximum value of is
2π radians, which corresponds to 360 degrees or one full revolution. From this it
s rφ
φ φ
× =
0
follows
that 1 radian 57.3 or 1 radian 0.159 revolution. Obviously, if = 2π, then 2 ,
which is the total circumference.
3. Suppose that there is a particle located at the tip of the radi
= = φ s= πr
1 2
us vector. Now we wish to
describe the rotational of this particle, i.e. describe its motion as goes around
the circle. So, suppose that the particle moves from angle to in t
kinematics
φ φ 2 1
2 1
2 1
0
ime . Then, the
is defined as, . Suppose that we look at
over a very short time. Then, lim is called the instantaneous angular
t
t t
average angular speed
t t t
d
t dt
φ φ φ
ω ω ω
φ φ
ω
Δ →
−
− Δ
= =
− Δ
Δ
= =
Δ
second
speed.
4. To familarize ourselves with the notion of angular speed, let us compute for a clock
second, minute and hour hands:
2 0.105 / ,
60
rad s
ω
π
ω = =
3
minute
4
hour
2 1.75 10 / ,
60 60
2 1.45 10 / .
60 60 12
5. Just as we defined accelaration for linear motion, we also define acceleration fo
rad s
rad s
π
ω
π
ω
−
−
= = ×
×
= = ×
× ×
2 1
2 1
2
0 2
r
circular motion:
(average angular speed)
Hence, lim becomes (angular acceleration). Let us
see wha
t
t t t
d dd d
t dt dt dt dt
ω ω ω
α
ω ω φ φ
α α
Δ →
− Δ
≡ =
− Δ
Δ
= = = =
Δ
t this means for the speed with which a particle goes around. Now use .
Differentiate with respect to time : . The rate of change of arc length is clearly
s r
tds rd s
dt dt
φ
φ
=
=
A B C
A r C r
what we should call the circular speed, v. So v . Since r is held fixed, it follows that
v . Now define v. Obviously, . Here stands for tangential, i.e.
in the direc
T T
r
d rd a d a r T
dt dt dt
ω
ω
α
=
= = =
tion of increasing .
6. Compare the formulae for constant linear and angular accelerations:
LINEAR ANGULAR
s
( )
0 0
2 2
0 0 0 0
2 2
0 0
v v a
v1a 1
2 2
v v 2a
t t
x x t t t t
x x
ω ω α
φ φ ω α
= + = +
= + + = + +
= + − 2 2 ( )
0 0 2
Why are they almost identical even though they describe two totally different physical
situations. Answer: because the mathematics is identic
ω =ω + α φ −φ
2
al!
7. The angular speed of a car engine is increased from 1170 rev/min to 2880 rev/min in
12.6 s. a)Find the average angular acceleration in rev/min . (b) How many revolutions
does t
2 2
he engine make during this time?
SOLUTION: this is a straightforward application of the formulae in point 5 above.
8140rev/min , 1 425rev.
2
8. W
f i
it t
t
ω ω
α φ ω α
−
= = = + =
heel A of radius 10.0 cm is coupled by a chain B to wheel C of radius 25.0 cm.
Wheel A increases its angular speed from rest at a uniform rate of 1.60 rad/s2. Determine
the time for
A C r= r=
wheel C to reach a rotational speed of 100 rev/min.
SOLUTION: Obviously every part of the chain moves with the same speed and so
v v . Hence . From the definition of acceleration,
0 . From this, 16.4
C C
A C A A C C A
A
A A C C
A
r r r
r
t r
t r
ω
ω ω ω
ω ω ω
α
α α
= = ⇒ =
−
= = = = .
9. Imagine a disc going around. All particles on the disc will have same ' ' and ' '
but different 'v' and 'a' . Clearly' ' and ' ' are simpler choices !!
s
ω α
ω α
2 v
1 v
10. Now consider a particle going around a circle at constant speed. You might think that
constant speed means no acceleration. Bu this is wrong! It is changing its direction and
accelerating. This is called "centripetal acceleration", meaning acceleration directed
towards the centre of the circle. Look at the figure below:
1 2
2 2
0
2
Note that the distance between points and is v . Similarly,
v v v v v . More generally, lim v v . In vector form,
/ v
a v . The negative sign in
t
R
P P r t r
a a
t r r t r
r
r
θ
θ
θ
θ Δ →
Δ = Δ ≈
Δ Δ
Δ ≈ ⇒ = ≈ = = =
Δ Δ
= −
􀁇 􀀃 dicates that the acceleration is towards the centre.
12. Vector Cross Products: The vector crossproduct of two vectors is defined as:
A×B=ABsinθ nˆ where nˆ is a unit vector that is perpendicul
􀁇 􀁇
ar to both and .
Apply this definition to unit vectors in 3-dimensions: , , .
A B
i j k k i j j k i
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
× = × = × =
􀁇 􀁇
( ) ( ) ( )
13. Some key properties of the crossproduct:
0
ˆ ˆ ˆ
( ) ˆ ( x y z y z z y
x y z
A B B A
A A
A B C A B A C
i j k
A B A A A AB AB i
B B B
• × = − ×
• × =
• + × = × + ×
• × = = − +
􀁇 􀁇 􀁇 􀁇
􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 )ˆ ( )ˆ
14. The cross product is only definable in 3 dimensions and has no meaning in 2-d. This is
unlike the dot product which as a meaning in any number of dimensions.
z x x z x y y x A B −A B j+ A B −A B k
1 P
2 P
2 v
C r
r
θ
1 v
2 1 v −v
1 m
2 m
1 r􀁇
2 r􀁇

COLLISIONS

1. Collisions are extremely important to understand because they happen all the time -
electrons collide with atoms, a bat with a ball, cars with trucks, star galaxies with other
galaxies,...In every case, the sum of the initial momenta equals the sum of the final
momenta. This follows directly from Newton's Second Law, as we have already seen.
2. Take the simplest collision: two bodies 1 2 1 2
1 2
1 1 2 2 1 1 2 2
of mass and moving with velocities and .
After the collision they are moving with velocities v and v . From momentum conservation,
v v
m m u u
mu +m u = m +m ⇒ 1 1 1 2 2 2
1 2 1 2 1 2 1 2 1
2 1 1 2 2 2 2 1 1 2 2 2
( v) (v )
This is as far as we can go. There are two unknowns but only one equation. However, if the
collision is elastic then,
v v
m u m u
mu m u m m
− = −
+ = + ⇒ 2 2 1 2 2
21 1 1 2 2 2 2
1 1 2 2 1 2 2 1
( v ) (v ).
Combine the two equations above,
v v v v .
In words, this says that in an elastic collision the relative speed of
m u m u
u u u u
− = −
+ = + ⇒ − = −
1 2
1 2
1
the incoming particles
equals the relative speed of the outgoing particles.
3. One can solve for v and v (please do it!) easily and find that:
v (m m
m
−
= 2
1 2
1 2 1 2
1 2 1
2 1 2
1 2 1 2
1 2 1 2 2 1
) ( 2 )
v ( 2 ) ( )
Notice that if = , then v and v . So this says that after the collision, the
bodies will
u m u
m m m
m u m mu
m m m m
m m u u
+
+ +
−
= +
+ +
= =
2 1 2
just reverse their velocities and move on as before.
4. What if one of the bodies is much heavier than the other body, and the heavier body is at
rest? In this case, m>>m andu = 0. We can immedi 1 1 2 ately see that v and v 0.
This makes a lot of sense: the heavy body continues to stay at rest and the light body just
bounces back with the same speed. In the lecture, you saw a demonstr
= −u =
2 1 2 1 1 2 1
ation of this!
5. And what if the lighter body (rickshaw) is at rest and is hit by the heavier body (truck)? In
this case, and 0. From the above equation we see that v and v 2 .

So the truck's speed is unaffected, but the poor rickshaw is thrust in the direction of the
truck at twice the truck's speed!
6. Sometimes we wish to slow down particles by making them collide with other particles.
In a nuclear reactor, neutrons can be slowed down in this way. let's calculate the
fraction by which the kinetic energy of a neutron of mass decreases in a head-on
collision with an atomic
m
2
2
f f
2
i i
1 2 1 2
f i 2
1 2 1 2
nucleus of mass that is initially at rest:
m M M + m
Solution: 1 1 v
v
For a target at rest: v ( )v 4 .
( )
7. A bullet with mass m, is fired into a b
i f
i
i f
i
m
K K K
K K
m m K K mm
m m K m m
−
= − = −
− −
= ∴ =
+ +
lock of wood with mass M, suspended like a
pendulum and makes a completely inelastic collision with it. After the impact, the
block swings up to a maximum height y. What is the initial speed of the bullet?
Solution:
By conservation of momentum in the direction of the bullet,
v ( ) , v ( )
The block goes up by distance y, and so gains potential energy. Now we can use the
m m MV m MV
m
+
= + ⇒ =
conservation of energy to give, 1 ( ) 2 ( ) , where V is the velocity
2
acquired by the block+bullet in the upward direction just after the bullet strikes. Now
use 2 . So finally, th
m MV m M gy
V gy
+ = +
= e speed of the bullet is: v ( ) 2 .
8. In 2 or 3 dimensions, you must apply conservation of momentum in each direction
separately. The equation looks as if it is one equation, but it i i f
m M gy
m
P P
+
=
=
􀁇 􀁇
s actually 3
separate equations: , , . On the other hand, suppose you had
an elastic collision. In that case you would have only one extra equation coming from
energy co
ix fx iy fy iz fz p =p p =p p =p
nservation, not three.
9. What happens to energy in an inelastic collision? Let's say that one body smashes into
another body and breaks it into 20 pieces. To create 20 pieces requires doing work
against the intermolecular forces, and the initial kinetic energy is used up for this.
φ
r
O
y
x
(x, y)
s

MOMENTUM

1. Momentum is the "quantity of motion" possessed by a body. More precisely, it is defined as:
The dimensions of momentum are
Mass of the body × Velocity of the body.
-1 and the units of momentum are kg-m/s.
2. Momentum is a vector quantity and has both magnitude and direction, v. We can
easily see that Newton's Second Law can be reexpressed in terms
MLT
p=m􀁇
of momentum. When
I wrote it down originally, it was in the form a . But since v a, this can also be
written as (new form). In words, the rate of change of momentum of a bod
m d
dt
d
dt
= =
=
F
p F
􀁇 􀁇 􀁇􀁇
􀁇 􀁇
y equals
the total force acting upon it. Of course, the old and new are exactly the same,
( ) .
3. When there are many particles, then th
d dm md m
dt dt dt
p= v= v= a=F
􀁇 􀁇 􀁇􀁇 􀁇
1 2
1 2
1 2
e total momentum is,
Thi
N
N
N
d d d d
dt dt dt dt
= + +⋅⋅⋅
= + + ⋅ ⋅ ⋅
= + + ⋅ ⋅ ⋅ =
P
P p p p
P p p p
F F F F
􀁇
􀁇 􀁇 􀁇􀁇
􀁇 􀁇 􀁇􀁇
􀁇 􀁇 􀁇􀁇
s shows that when there are several particles, the rate at which the total momentum
changes is equal to the total force. It makes sense!
4. A very important conclusion of the above is that if the sum of the total external forces
vanishes, then the total momentum is conserved, 0 = 0. This is quite
independent of what sort of forces act between the bodies - electric, gravitati
ext

F d
dt
Σ = ⇒ P
􀁇 􀁇
onal, etc. -
or how complicated these are. We shall see why this is so important from the following
examples.
5. Two balls, which can only move along a straight line, collide with each other. The
1 1 2 2 1 1 2 2
initial
momentum is and the final momentum is v v . Obviously one
ball exerts a force on the other when they collide, so its momentum changes. But, from the
fact that th
i f P=mu +mu P =m +m
1 1 2 2 1 1 2 2 ere is no external force acting on the balls, , or v v .
6. A bomb at rest explodes into two fragments. Before the explosion the total momentum is
i f P=P mu +mu =m +m
zero. So obviously it is zero after the explosion as well, . During the time that the
explosion happens, the forces acting upon the pieces are very complicated and changing
rapidly with
Pf =0
1 1 2 2 1 1 2 2
time. But when all is said and done, there are two pieces flying away with a
total zero final momentum v v . Hence v v . In other words, the
fragments fly apart with equal momentum
f P =m +m m =−m
but in opposite directions. The centre-of-mass
stays at rest. So, knowing the velocity of one fragment permits knowing the velocity of
the other fragment.
0
7. If air resistance can be ignored, then we can do some
interesting calculations with what we have learned.
So, suppose a shell is fired from a cannon with a speed
10 m/s at an angle 60 with the horizontal. At the highest
point in its path it explodes into two pieces of equal
masses. One of the pieces retraces its path to the cannon.
Let us find the velocity of the other piec
1 1 2
2 2 2 2
e immediately
after the explosion.
Solution: After the explosion: 5 (why?). But 10cos60
2
5 5 . Now use: v v 15 / .
2 2
8. When yo
x x x x
x x x x
P M P P P M
P M M P M ms
= − + = = ×
⇒ = + = ⇒ =
u hit your thumb with a hammer it hurts, doesn't it? Why? Because a large
amount of momentum has been destroyed in a short amount of time. If you wrap your
thumb with foam, it will hurt less. To understand this better, remember that force is the
rate of change of momentum: . Now define the as:
F dp dp Fdt I
dt
= ⇒ = impulse
force × time over which the
2 2
1 1
If the force changes with time between the limits , then one should define as,
. Since , therefore . In words, the change of momentum
equal
f
i
t t p
f i
t t p
I
I=∫Fdt ∫Fdt=∫dp I=p −p
force acts.
s the impulse, which is equal to the area under the curve of force versus time. Even
if you wrap your thumb in foam, the impulse is the same. But the force is definitely not!
Δt
9. Sometimes we only know the force numerically
(i.e. there is no expression like F=something).
But we still know what the integral means: it is
the area under the curve of force versus time.
The curve here is that of a hammer striking a
table. Before the hammer strikes, the force iszero, reaches a peak, and goes back to zero