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Rotational Motion
We are now going to study one of the most common types of motion, that of a body constrained
to move on a circular path.
is the relation between ω and the speed at which
the body travels along the circular path.
direction of
motion
Description of the motion
Let’s first consider only the description of rotational motion, without worrying about the cause
of the motion. This means that what we are now
doing is mathematics and not physics. We will
consider the physics later, when we talk about the
nature of forces which give rise to circular motion, and discuss some examples.
Suppose the radius of the circle is r, and that the
angle θ is changing at a rate (so many radians per
second) symbolized by ω (the Greek letter
“omega”):
dθ
=ω .
dt
This is called the angular velocity. It may or may
not be constant. If the body is speeding up or
slowing down as it goes around, then ω will
change in time.
One of the first important things to know is what
θ
r
rθ
The distance covered along the circumference is
rθ. The rate of change of this distance is
v=
d
dθ
(r θ ) = r
,
dt
dt
where we used the fact that r is constant to
perform the last step. Hence, the speed is
v = rω
.
It turns out that there’s another way to obtain this
result, using an expression for the velocity in
polar coordinates in two dimensions derived in
an earlier section on vectors. Because the circular path lies in a single plane, these are the
appropriate coordinates to use. In order to get
some practice with polar coordinates, let’s see
how it works.
In the general case where both coordinates r and
θ are changing in an arbitrary way, we showed
that the velocity vector is
vP(t) =
dr
dθ ˆ
rˆ + r
θ .
dt
dt
In the present case, the radial coordinate r is
constant, so dr/dt=0. Also, dθ/dt=ω. Therefore,
the velocity is
vP = r ωθˆ
,
where the unit vector θˆ points around the circle
in the direction of increasing θ. Taking the
magnitude of both sides, we find that the speed is
v = rω ,
which agrees with the above.
How large is the acceleration? To answer this,
we will use another general expression derived in
the section on vectors:
äåå d 2 r
ëì
1 d 2
ár ω é
aP(t) = rˆ ååå 2 − rω2 ìììì + θˆ
r dt
ã dt
í
In our particular case, the radial coordinate r is
constant, so we get
aP = – r ω 2 rˆ + r α θˆ
centripetal tangential ,
where we have introduced a new symbol
α=
dω
d2θ
=
dt
dt2
denoting the angular acceleration (in the same
sense that ω=dθ/dt is the angular velocity).
The first term points towards the center of the
circle (in the − rˆ direction). For this reason, it is
called centripetal acceleration. (In Latin, the
word petere means “to go toward, to seek”.) The
second term, the tangential acceleration, points
along the circumference of the circle, and is proportional to the angular acceleration.
In order to better understand centripetal acceleration, let’s consider for a moment the special
case where the body is moving with constant
speed along the circumference of the circle.
Then its angular acceleration is zero, and the
body’s entire acceleration is centripetal. Let’s try
to see why the acceleration points towards the
center of the circle.
The next figure shows the velocity vector at a
particular time t , and then at a later time t+∆t .
The difference vP(t + ∆ t) −vP(t), shown in red, points
inward although not directly through the center
of the circle.
Pv(t + ∆ t)
In the limit as ∆ t goes to zero, the difference
points directly towards the center. The
acceleration is defined as the limit of this
difference, divided by ∆t:
P(t) =
a
Pv(t)
Pv(t + ∆ t) – Pv(t)
dvP(t)
vP(t + ∆t) − vP(t)
= lim
.
dt
∆t6 0
∆t
Hence, the acceleration points directly towards
the center of the circle, for a body moving with
constant speed along the circle.
In the general case, the body is also changing
speed as it travels along the circle. In that case,
the acceleration has a tangential component as
well.
The angular velocity vector
As ∆t is made smaller and smaller, the difference
points more and more towards the center:
Pv(t + ∆ t)
Pv(t)
Pv(t + ∆ t) – Pv(t)
Until now, we haven’t been concerned about the
orientation of the circular path in space. We have
only been looking at the plane containing the
motion. But sometimes we would like to keep
track of both the angular velocity and the
orientation of the circle. To do this in a concise
way, it is useful to define a vector called the
angular velocity vector
ω
P = ω nˆ
,
where nˆ is a unit vector perpendicular to the
plane of the circle. Its direction is given by the
nˆ
The magnitude of the angular velocity vector is
the angular velocity ω.
There is a nice expression for the velocity in
terms of the angular velocity vector:
vP = ω
P H Pr
,
as shown in the following figure:
ω
P
ω
P HPr
Pr
The angular velocity vector is particularly useful
when we are talking about the rotation of a rigid
body. For example, consider this blob-shaped
plate:
P
ω
0101010101010101
01010101010101010101010101010101
01010101010101010101010101010101
01010101010101010101010101010101
01010101010101010101010101010101
01010101010101010101010101010101
10010101010101010101010101010101
right-hand rule: you curl the fingers of your right
hand around in the direction of the body’s motion
and stick out your thumb. Your thumb points in
the direction of nˆ :
Each piece of the plate has a different velocity,
but all have the same angular velocity. The
angular velocity vector is a property of the plate
as a whole. Once we know the angular velocity
vector of the plate, we can get the velocity of any
piece of the plate just by knowing its position,
thanks to the last formula.
The same goes for the rotation of a threedimensional rigid body, of course. The above
formula for the velocity in terms of the angular
velocity is still valid, except now there is one
small subtlety involving the position vector of a
given piece of the body. Notice that we don’t
r from the center of the
really have to measure P
circle. We can measure it from any point on the
axis of the circle:
ω
P
ω
P HPr
Then the velocity of that piece of the body
located at position i is
010101010101010101010101
100101010101010101010101
vPi = ω
P H Pr i .
We will make use of this formula when we talk
about the inertia tensor in a later section.
Pr
Cause of the motion
We can do this because any part of Pr which is
P does not contribute to the cross
parallel to ω
product ω
P H Pr .
This is helpful because we usually want to refer
all the position vectors of the pieces of the rigid
body to the same point within the body. We can
choose any point on the axis of rotation of the
body:
P
ω
Pv1
0101010101010101
Pr1
Pr2
Pv2
Pr3
Pv3
Suppose we are now concerned with how to
constrain a body to move on a circular path. We
would like to know what force is necessary to
achieve this.
Why is a force necessary, in the first place? Well,
we have just learned that a body moving on a
circular path is always accelerating, even if its
speed is constant. But, according to Newton’s
second law, if a body accelerates with respect to
an inertial frame, a force must be acting.
What force is causing this acceleration? Whatever is constraining the body to move on its
circular path must be exerting a force on the
body. This force of constraint is just the body’s
mass times its acceleration. Using what we
learned above about the acceleration, we deduce
that the force splits into two pieces. The piece
which is always present even if the speed is
constant is called the centripetal force:
P
F
= − m ω 2 r rˆ
centripetal
,
and the extra piece which is present if the speed
along the circle is changing is the tangential
force:
P
F
= m r α θˆ
tangential
.
(Recall that α is the angular acceleration.)
The following diagram shows the centripetal
force
Pv
0101
10010101
P
F
centripetal
another force called the centrifugal force, which
we will discuss later. All you have to know right
now is that it is not the same as the centripetal
force. We will explain the distinction in more
detail later, but for now just remember that the
two forces point in opposite directions. The centrifugal force points outwards from the center of
the circle. In Latin, the word fugere means “to
run away from, to flee” - as in the English word
“fugitive”.)
Example: a rock on a string
Suppose you have tied a rock to a piece of string,
and are twirling it around at constant speed in a
circle. Then the above analysis applies, and a
centripetal force must be acting on the rock. In
this case, the centripetal force is the tension in
the string. Its magnitude is just mω2 r , where m is
the rock’s mass, ω is the angular frequency of
rotation, and r is the radius of the rock’s orbit.
Example: circular orbit around earth
For example, if we attach a rock to a string and
twirl it around, this is the force we must exert in
order to keep the rock on a circular path with
constant speed. We exert this force by pulling on
the string.
(Here’s a short note about terminology: there’s
The gravitational force on a body of mass m orbiting earth at a distance r from earth’s center is
P = − GMm rˆ ,
F
r2
where G is Newton’s constant and M is earth’s
mass. Because this force is in the negative radial
direction, it can give rise to circular motion
around earth’s center (although the motion need
not be circular, as we will see later.)
and earth’s mass is
We’ll suppose that we know the frequency with
which the body orbits earth, and calculate the
radius of its orbit. To do this, we use the fact that
the centripetal force on the body is the gravitational force:
Inserting these values, we find
− m ω2 r rˆ = −
GMm
rˆ .
r2
Solving for the radius, we find
ä GM ë
r = ååå 2 ììì
ã ω í
1 /3
.
M = 5.98 H 10 24 kg .
r . 4.2 H 10 7 m = 42,000 km .
For comparison, earth’s radius is about 6,400 km.
Example: charged particle in magnetic field
Suppose an electrically charged particle moves in
P . If the
a region containing a magnetic field B
P and its electric charge is q,
particle’s velocity is v
then the force on the particle is
P .
P = q vP H B
F
Suppose a satellite orbits earth in such a way that
it remains stationary above a particular spot on
the equator. What is the radius of its orbit?
Suppose now that the particle moves in a region
where the magnetic field is constant in time and
uniform in space. Then it turns out to be possible
for the particle to be executing uniform circular
P , as we will
motion in a plane perpendicular to B
now see.
The angular frequency is the same as that of the
earth:
Let the magnetic field point in the positive z
direction:
Sub-example: geosynchronous orbit
ω=
2π
2π
=
= 7.27 H 10 − 5 s − 1
24 hours
86,400 s
The value of the gravitational constant is
G = 6.67 H 10 − 11 m 3 kg − 1 s− 2
P = Bzˆ , B > 0 .
B
A particle undergoing uniform circular motion
with radius r in the plane perpendicular to the z
direction has velocity
vP = r ω θˆ ,
as we showed before. Hence, the force on such a
particle is
q á rωθˆ é H á Bzˆ é = qrωB rˆ .
If the product qω is negative, this force is
directed radially inward. It is then a centripetal
force. To find the frequency of rotation, we
equate the magnetic force to our previous expression for the centripetal force
− mω2 r rˆ = qBrωrˆ .
Solving for the frequency, we find
ω= −
qB
m
.
The magnitude of ω is called the cyclotron
frequency. Notice that it increases with increasing magnetic field strength, and that it does not
depend on the particle’s speed.
The radius of the circular orbit is r=v/|ω|, or
mv
.
qB
As in the gravitational case, circular motion turns
out not to be the most general kind of motion in a
uniform magnetic field. It turns out that the most
general path in this situation is a spiral, or helix.
The projection of the helix onto a plane perpendicular to the magnetic field is the circular
path just described. The component of the velocity parallel to the magnetic field is constant
(since the component of the magnetic force
P in that direction is zero).
qvP H B
Here is a plot of such a path:
If q is positive, then the particle goes around the
magnetic field in the negative (i.e. left-handed)
sense.
r=
The radius increases with increasing particle
speed v, and decreases with increasing field
strength. If the field goes to zero, then the radius
is infinite (a straight line).
B
vo
Exercise: what is the sign of the electric charge
of the above particle?
We will treat the more general case in which both
electric and magnetic fields are present, in
another section.