Download I L - IBPhysicsLund

Topic
2.2 Extended
FYI: For all you brainiacs,
the genuine
official definitions are...
Angular
momentum
FYI: The direction H
of L–is the
same as the
direction for , and given by
the right hand rule.
v
If you recall, linear momentum p was
the mass m times the velocity v.
Just as torque  was the rotational
r
equivalent of force, and defined as
( = rF)
 = rF sin 
we define angular momentum L as the
rotational equivalent of linear
momentum:
L = rp sin 
L = rmv sin  (L = rp)
The effect of the "sin " term of course is to take
into account the fact that only the perpendicular
component of v can cause the mass to move in the
circle.
If  = 90° then v = vt, the tangential velocity.
Thus L = rmvt. Since vt = r we have L = rm(r)
so that L = mr2.
L = I
Angular
Momentum
Topic 2.2 Extended
H – Angular momentum
Now we can derive the following:
L = I
L = I
Given
Provided I is constant
L = I  Why?
t
t
L = I
Why?
t
L = 
Newton's 2nd
t
(angular momentum)
Compare with Newton's 2nd law (linear momentum):
P = F
t
Newton's 2nd
(linear momentum)
FYI: You may recall Topic
doing the same
when we looked at linear
2.2thing
Extended
0
momentum...
HP– =Angular
Fint + Fext momentum
t
CONSERVATION OF ANGULAR MOMENTUM
We can divide the torque  into internal and external
torques, so that
L =  0+ 
int
ext
t
From Newton's 3rd, for every force there is an equal
and opposite reaction. Thus all the internal forces
sum to zero:
In an analogous way, all the internal torques also
sum to zero: ...Thus
L = 
ext
t
Then if all of the external torques sum to zero, we
have
L = 0
t
which implies that L = a constant, or
Conservation of
L0 = Lf
Angular Momentum
I00 = Iff
Provided ext = 0
Topic 2.2 Extended
H – Angular momentum
CONSERVATION OF ANGULAR MOMENTUM
A thin rod of mass M and
length L having two point
masses m on its ends is
spinning about its central
axis at an angular speed ωi
as shown:
M
m
m
L
m
m
ωi
If an internal force pulls
each point mass to half
its original radius, what
is the new angular speed
ωf?
m
ωf
m
Topic 2.2 Extended
H – Angular momentum
CONSERVATION OF ANGULAR MOMENTUM
Since there are no external
torques, angular momentum is
conserved:
Li = Lf
Iiωi = Ifωf
Ii
ωf =
ω
If i
M
m
L
m
Ipoint,i
=
Ii =
m L
2
1
ML2
12
2
+
× 2
1
2
mL2
mL2
= 2
m
ωi
The total rotational inertia is
given by the sum of the
individual ones:
Irod,i = Irod,f =
m
m
m
ωf
1
ML2
12
Ipoint,f
=
If =
m L
4
1
ML2
12
2
+
× 2
1
8
mL2
mL2
= 8
Topic 2.2 Extended
H – Angular momentum
CONSERVATION OF ANGULAR MOMENTUM
Our final results:
ωf =
1
ML2
12
+
1
2
mL2
+
1
8
mL2
1
2
ML
12
m
ωi
Which reduces to
ωf =
M
2M + 12m
ωi
2M + 3m
m
L
m
m
ωi
m
ωf
m
Note: The numerator is bigger than the denominator showing that the
angular speed increases as the masses approach the center.
If M = 3 kg, m = 2 kg, and ωi = 10 rad/s, what is
the final angular speed?
2·3 + 12·2
ωf =
10
2·3 + 3·2
= 25 rad/s
Topic 2.2 Extended
H – Angular momentum
CONSERVATION OF ANGULAR MOMENTUM
Suppose we send a satellite into space whose
orientation has to be fine-tuned after being
placed in orbit. There are basically two ways to
do this:
(1)Fire small external rockets.
(2)Rotate an internal flywheel.
Method (1) is very hard to manage because you
also have to fire rockets to stop the motion.
Method (2) is very easy to manage because you
can simply run an electric motor until the craft
is aimed the right way.
Topic 2.2 Extended
H – Angular momentum
CONSERVATION OF ANGULAR MOMENTUM
Here’s how Method (2) works:
If you want to turn the craft clockwise, you
rotate the flywheel counterclockwise:
Conservation of angular momentum does the rest.
L0 = Lf
0 = Icraftωcraft - Iflyωfly
Icraftωcraft = Iflyωfly
Δθfly
Δθcraft
Icraft
= Ifly
Δt
Δt
Why are the Δts
unsubscripted?
Icraft Δθcraft = Ifly Δθfly
Δθfly
Icraft
=
Δθcraft
Ifly
Rotational Motion and Equilibrium
8-5 Angular Momentum
CONSERVATION OF ANGULAR MOMENTUM
Suppose Icraft is 3.75 kg·m2, and Ifly is .45 kg·m2.
If you want to turn the craft clockwise through
90° then we must turn the flywheel
Δθfly =
Icraft
Ifly
Δθfly =
3.75
0.45
Δθcraft
·90° = 750°
Question: Why can we use degrees if we want?