Ch 10: Conservation of Angular Momentum
Reeeeecap…….
Q: What does angular velocity mean? What is its symbol?
A: The rate of change of angular displacement. ω.
Q: What does angular acceleration mean? What is its symbol?
A: The rate of change of angular velocity. α.
Q: What does angular momentum mean? What is its equation for calculation?
A: How much angular “oomph” exists due to a rotation. People often refer to
it as whip, due to mass / inertia, placement of the mass, and angular velocity.
Iω
ω.
Q: In what way(s) is mass like the moment of inertia? In what way(s) is it
different?
A: They both deal with mass, therefore inertia*, but the moment of inertia is
not only the mass, but also how the mass is distributed with respect to the axis
of rotation.
Q: What is the equation we use to show conservation of linear
momentum? What about that used to show conservation of angular
momentum?
vm ω I
A: m
=
v
I
=
ω
Q: What do you call a net force acting over time? How do you calculate it?
What does it cause? How do you calculate that part?
A: Impulse. F∆
∆t. A change in linear momentum (specifically, the v part).
ma∆
∆t
m ∆v. If you divide both sides by ∆t, you get the rate of change of the linear
momentum. This is how we get from the impulse-momentum theorem to N2.
Q: What do you call a net torque acting over time? How do you calculate it?
What does it cause? How do you calculate that part?
A: Technically, angular impulse. τ∆t. A change in angular momentum
(specifically the ω part). If you divide both sides by ∆t, you get the rate of
change of the angular momentum. This gives you N2 for rotation.
10.1 The Vector Nature of Rotation
So far, for all of our previous rotation discussions, we’ve assumed the axis of
rotation to be fixed in space, and have then subsequently assigned a sign to the
direction of ω. The convention used is called the right-hand rule, which says
when the fingers of the right hand curl in the direction of the rotation, the
thumb will point in the direction of ω. Switch the direction of rotation, and
you switch the direction of ω respectively.
Torque works the same way, in that if you imagine a rotating disk, the torque
causing the rotation:
• is ⊥ to the plane formed by F and r
• has a magnitude = Fr sin φ where φ is the angle between F and r (or the
angle between the directions of F and r).
So what this really means is that in whatever plane F and r exist, τ will be ⊥to
that plane.
Torque is defined as the cross product of r x F where:
τ = r x F and r x F represents a magnitude equal in area to the parallelogram
formed by r and F, and whose direction is ⊥ to the plane formed by F and r.
Sometimes, instead of using r and F, we use the vectors A and B. Really, it’s all
same-same. ☺ The vector n is used to signify a unit vector in the same
direction as would be ⊥ to the plane formed by the two vectors involved.
If A and B are parallel, A x B = 0, as they form no parallelogram.
This also implies that A x A = 0 for the same reason.
A x B can also be written as – B x A. This matters because of the direction of
the resultant vector.
AxB=-BxA
Some other useful properties:
A x (B x C) = (A x B) + (A x C)
If A and B are functions of time, then the product rule for derivatives is
relevant, and you get:
d/dt (A x B) = (A x dB/dt) + (dA/dt x B)
Because the unit vectors i, j, and k, are mutually ⊥,
ixj=k
jxk=i
kxi=j
And, lastly,
i x i = j x j = k x k = 0, as no parallelogram can be formed.
10.2 Torque and Angular Momentum
Q: What symbol is used for linear momentum? What about for angular
momentum?
A: p; L, of course! ☺
The angular momentum of a particle relative to the origin O is the cross
product of r and p.
For Fig 10-8 below, if r and p are in the xy-plane, and L is || the z-axis, then the
cross product, L = r x p = rmv sin φ k. Just like torque, angular momentum is
defined relative to a point. For Fig 10.9 below, a particle of mass m travels
around a circular disk in the xy-plane with its center at O. The disk spins with
angular speed ω. The angular speed and tangential speed are related by the
formula v = rω
ω.
Q: How would you calculate L for Fig 10-9.
A: L = r x p = rmv sin 900 k = rmv k = mr2 ω k = mr2 ω
Fig 10-8
Fig 10-9
Q: What does mr2 signify?
A: Moment of inertia for a single particle about the x-axis!
NOTE: This result does not hold true for angular momentum about a general
point on the z-axis. What would you need to use in this case?
Figure 10-10 shows the angular momentum
vector L’ for the same particle attached to
the same disk but with L’ computed about a
point on the z-axis that is not at the center of
the circle. In this case, the angular
momentum is not parallel to the angular
velocity vector ω, which is parallel to the zaxis.
For any system of particles that rotates about a symmetry axis, the total
angular momentum is parallel to the angular velocity and is given by:
See Example 10-1, p 312
Find the angular momentum about the origin for the following situations.
(a) A car of mass 1200 kg moves in a circle of radius 20 m with a speed of 15
m/s. The circle is in the xy plane, centered at the origin. When viewed at a
point from the z-axis, the car moves counterclockwise.
For (a) use
.
r and p are perpendicular. Really, that means r and v are perpendicular, as the
velocity is the tangential velocity, and linear momentum is the product of mass
times that tangential velocity. Also, r x p is along the z-axis (Figure 10-12). For
all circular motion, r ⊥ p because p = mv; v is the tangential v, ∴ it will always
be ⊥ to the radius at the point of contact. Furthermorr, sin θ = sin 900 = 1.
= rmvsin(90°°) k
= (20 m)(1200 kg)(15 m/s)(1) k
= 3.6 x 105 kg·m2 /s k
This is moment of inertia / time
b) The same car moves in the x-y plane with v = - 15 m/s i. In this part, they
introduce unit vectors, so guess what? You should use ‘em, as well. ☺
Also, since we’re not talking about motion about a symmetry axis, we use
L = r x p, just like in part a).
1st, define
r = xi + yj = xi + y0j
nd
2 , define
p = mv = - mvi
rd
3 , define r x p = (xi + yj) x (- mvi)
= - x mv( i x i) – y0 mv(j x i)
= 0 - y0 mv (-k)
= y0 mv k
= (20 m)(1200 kg)(-15 m/s) k
= 3.6 x 105 kg••m2 / s k
c) For this part, since there is a symmetry axis, use L = I ω
L=Iω
= I ωk = ½ mR2ω k = ½ (1200 kg)(20 m)2(0.75 rad / s)k
= 1.8 x 105 kg••m2 / s k
Q: Why is the moment of inertia of a 1200-kg disk of radius 20 m less than
that of a 1200-kg car at 20 m from the axis?
A: The disk has a lot morr mass closer to the axis of rotation. ☺
Do you remember these notes from last chapter?
NOTE: The calculations for torque in chapter 9 only give us the
magnitude of the torque. Torque is actually a vector quantity. We assume
the direction to always be ⊥ to the plane of contact. In this chapter, our objects
can now rotate in strictly the x-y, or y-z, or x-z plane, or, in some combination
of all three. The direction of our torque, which is still always ⊥ to the plane
of rotation for the circular object, might need to be calculated by calculating
the cross product, if not obvious, as in last chapter. The magnitude of that
torque can then either be calculated with the above formulas, or by applying
the Pythagorean Theorem to the resultant torque vector.
A few morr ditties regarding angular motion
NOTE: The τnet
system.
ext
is the sum of all of the external torques acting on the
In other words, torque is the rate of change of the angular momentum wrt
time, which is also a vector, as it should be.
Q: What do you get, then, if you integrate both sides of the equation wrt time?
A:
So, just like with linear impulse and linear momentum, angular impulse causes
a change in angular momentum.
Furthermorr, the area under the torque-time curve will equal the change in
angular momentum.
For situations where it becomes useful to split the total angular momentum of
a system about some arbitrary point O into its constituent parts (orbital
angular momentum and spin angular momentum), we can use:
For example, see Fig 10-15, which shows the angular momentum of the earth
about the center of the sun as the sum of the orbital and spin angular
momenta.
Figure 10-15
Actually, you can rewrite the above equation as L = rcm x Mvcm, as indicated by
the diagram above.
Q: What’s the diff between the torques we calculate here, and the ones we
calculated in Ch 9?
A: Here, we calculate about points, not just axes.
What this really means is that you can translate the information to pertain to
points or axes. Consider the calculation of the torque computed about point O
compared to the torque computed about the z-axis, for example, as shown in
Fig 10-16 below. This gives you:
Figure 10-16
The components that comprise eq 10-14 are simply the vector components of τ,
r, and F. You can also translate this to describe angular momentum about the
z-axis as:
Pxy just means the projection of the momentum in the xy-plane.
If you look back to equation 10-10:
And take just the z-components of both sides, you get τnet ext, z = dLsys, z / dt
If the object doing the rotating is a rigid object about the z-axis, then
Lsys, z = Iz ω, where Iz is the moment of inertia about the z-axis.
If you substitute this back into eq 10-16, you get:
τnet ext, z = dLsys, z / dt = d/dt (Iz ω) = Iz α
What the last part of that equation means is that the vector α = d ω / dt
Try Ex 10-2, p 315 now. Make sure to read the remarks before you try the
problem to gain some perspective on setting it up, then give it a whirl. ☺
The Gyroscope
Q: What is a gyroscope? How does it work? What does precession mean?
See the following links! The first is a short explanation of
torque and angular momentum. The 2nd is Dr. Lewin’s lecture.
http://hyperphysics.phy-astr.gsu.edu/hbase/rotv.html#rvec3
http://www.youtube.com/watch?v=zLy0IQT8ssk
Bicycle Wheel Gyroscope – Torque and Angular Momentum
τ = rhandle x Fg
L = rwheel x p
rwheel
p
rhandle
Fg
rhandle MUST point from the hang point to
the center of the wheel
Fg is the only force acting on the wheel
once in motion
The torque created by the spinning wheel counteracts the torque due to gravity, and
“lifts” the wheel. Because angular momentum “chases” torque, the wheel precesses
around the hanging rope.
Q: Which way will the wheel precess?
Q: If you sit on a rotating stool, hold the wheel by the handles on either end, and tilt the
wheel, you will start to spin. Why?
A: A rotating bicycle wheel has angular momentum, which is a property involving the
speed of rotation, the mass of the wheel, and how the mass is distributed. For example,
most of a bicycle wheel's mass is concentrated along the wheel's rim, rather than at the
center, and this causes a larger angular momentum at a given speed. Angular momentum
is characterized by both size and direction.
The bicycle wheel, you, and the chair comprise a system that obeys the principle of
conservation of angular momentum. This means that any change in angular
momentum within the system must be accompanied by an equal and opposite
change, so the net effect is zero.
Suppose you are now sitting on the stool with the bicycle wheel spinning. One way to
change the angular momentum of the bicycle wheel is to change its direction. To do this,
you must exert a twisting force, called a torque, on the wheel. The bicycle wheel will then
exert an equal and opposite torque on you. (That's because for every action there is an
equal and opposite reaction.) Thus, when you twist the bicycle wheel in space, the bicycle
wheel will twist you the opposite way. If you are sitting on a low friction pivot, the twisting
force of the bicycle wheel will cause you to turn. The change in angular momentum of
the wheel is compensated for by your own change in angular momentum. The system as
a whole ends up obeying the principle of conservation of angular momentum.
10-3 Conservation of Angular Momentum
This section consists most entirely of examples and proofs. Before you
endeavor to go through it, you must be familiar with the following bits of
information:
If dLsys / dt = 0, then the net external force on an object remains 0, and viceversa.
Q: What does this really mean?
A: Lsys = constant
Q: What does it imply?
The KE, or just K, of a system, in terms of I and ω, can be written as
K = ½ Iω
ω2 = (Iω
ω)2 / 2I. If L = Iω
ω, then K = L2 / 2I
Read the rest of this information, p 319, before you attempt Ex 10-3, p 318-319.
The proofs for all of the equations in this chapter for which you are
responsible are given on p 324 – 325. You should read through these before
you attempt relevant examples to which they pertain.
The last little twist (funny!) comes with the information that the sum of all of
the torques, internal and external may defined as the following:
Your class work / homework is to go through every example in this chapter! ☺
BIG Idea – L chases T (angular momentum chases torque!)
See Dr. Lewin, MIT Open Courseware
Assignment: When are each of the following quantities zero? What does it
mean? Make a chart!
p, ∆v, τ, L, I, ∆ω, Work, Power
Whiteboard Assignment: Summarize how a gyroscope works. ☺