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