Rotational Motion
Angular Quantities
Angular Kinematics
Rolling Motion
Chapter 8
Rotational Motion
We consider only rigid (non-squishy) bodies in this
chapter.
We have already studied translational motion in
considerable detail. We now know that the translational
motion we have been describing is the motion of the
center of mass.
We begin this chapter by considering purely rotational
motion (the center of mass does not change its xyz
coordinates)… but soon we will consider objects which
both rotate and translate.
8.1 Angular Quantities
For rotational motion, we specify angles in radians
instead of degrees.
Look at a point P somewhere on a rotating circular disc.
P is a distance r from the center
of the disc.
Choose the x-axis to be
horizontal. Then the line from
the center of the disc to P makes
an angle  with the x-axis.
=ℓ/r, where ℓ is the length of the
arc from the x-axis to P.
P
O
r

ℓ
x
If the angle  is 360 degrees, the
arc length is 2r, so 2 radians
are equal to 360 degrees. This is
how I remember the conversion
factor between degrees and
radians.
P
O

ℓ
x
Strictly speaking, radians are not
a unit (angles are unitless).
When you studied kinematics, you defined an object’s
motion by specifying its position, velocity, and
acceleration.
Now we are about to study angular kinematics. We will
define an object’s angular motion by specifying its angle
(rotational position relative to some axis), angular
velocity, and angular acceleration.
If we call the z axis the axis of rotation, angular velocity
is defined by avg=/t, where  is the angular
displacement of a rotating object in the time t.
Instantaneous angular velocity is =d/dt.
“Wait a minute! Velocity is a vector! What is the
direction of ?”
Consider our rotating wheel.
Here’s a point and its
instantaneous linear velocity.
Another. And another.
How do we define a unique
direction for the angular
velocity?
If we put the x and y axes in the
plane of the wheel, then the wheel
is rotating about an axis
perpendicular to this plane.
We usually call this the z-axis.
z
The direction of  is along the zaxis, perpendicular to the wheel,
and is given by the right hand rule.
To remind you that angular velocity has a direction, I’ll
write z,avg=/t and z=d/dt.
Our rules for vectors apply. You get to choose the
direction of the z-axis. Whether z is positive or negative
depends on the direction of rotation.
this symbol indicates an arrow coming out of the screen
If this business about direction is confusing, don’t worry.
It is not particularly important this semester.
You’ll learn to love the right hand rule if you are lucky
enough to take the second semester of algebra-based
physics!
Note that all points on a rigid body rotate with the same
angular velocity (they go around once in the same time).
What about the tangential
velocities of different points?
We’ll get to that question in a
minute.
Angular velocity z,avg=/t and
z=d/dt.
You can probably figure out how
angular acceleration is defined.
P

z

ℓ
z,avg=z/t, where z is the
change in angular velocity of a
rotating object in the time t.
Instantaneous angular acceleration is z=dz/dt.
The subscript z on  and  emphasizes the rotation is
relative to some axis, which we typically label “z.”
I should really put a subscript z on  because it is also
measured relative to the z axis. To be consistent with
Physics 23, I will leave it off.
x
Because points on a rotating
object also have an
instantaneous linear motion,
linear and angular motion must
be connected.
vtang
z
As your book shows, vtangential=rz
and atangential=rz. Note that z and
z are the same for all points on a
rotating rigid body, but vtang and
atang are not.
Notation: vtangential=vtang=v.
, , v and a are all magnitudes of vectors!
z and z are vector components!
r
.
Acceleration has both tangential and radial components.
We know from before that ar=v2/r. Total acceleration is
a=atangential+ar.
We often use frequency and period of rotation: f=/2
and T=1/f.
Example 8.3. What are the linear speed and
acceleration of a child seated 1.2 m from the center of a
steadily rotating merry-go-round that makes one
complete revolution in 4.0 s?
z = 2/4 s-1
vtang = rz = (1.2 m) (/2 s-1)

vtang = 0.6  m/s
z = 0 (the angular velocity is not changing)
vtang
z
r
atang = rz = 0
aradial = ar = (0.6 
m/s)2
/ (1.2 m)

vtang
z
r
ar = 2.96 m/s2
The total acceleration is the vector sum of the radial and
tangential accelerations:
a = 2.96 m/s2, towards center of merry-go-round
New OSE’s introduced in this section:
z,avg=/t
z,avg=z/t
vtang=rz
atang=rz
You can also use aradial = aR=Rz2
8.2 Kinematic Equations for Uniformly
Accelerated Rotational Motion
Remember our equations of kinematics from chapter 2?
Analogous equations hold for rotational motion.
linear
angular
vx=v0x+axt
z=0z+zt
x=x0+v0xt+½axt2
θ=θ0+0zt+½zt2
vx2=v0x2+2ax(x-x0)
z2=z2+2z(θ-θ0)
new OSE’s
Example 8.6. Through how many revolutions did the
centrifuge motor of example 8.5 turn during its
acceleration period? Assume constant angular
acceleration.
In example 8.5, a centrifuge motor goes from rest to
20,000 rpm in 5 minutes.
First we need to calculate , as in example 8.5.
z,avg=z/t
0
0
z,avg=(f-i) / (tf-ti)
f=(20000rev/min)(min/60s)(2 radians/rev)
tf=(5 min)(60 s/min)
z,avg=7.0 rad/s2
Next calculate the angle through which the centrifuge
motor turns.
0 0
θ=θ0+0zt+½zt2
θ=½ (7.0 rad/s2) (300 s)2
θ=3.15x105 radians
There are 2 radians in each revolution so the number of
revolutions, N, is
N=(3.15x105 radians)(revolution/2 radians)
N=5x104 revolutions).
I chose to do the calculations numerically rather than
symbolically for practice with angular conversions.
8.3 Rolling Motion
Many rotational motion situations involve rolling objects.
Rolling without slipping
involves both rotation and
translation.

Friction between the rolling
object and the surface it rolls
on is static, because the
rolling object’s contact point
this point on the wheel is
with the surface is always
instantaneously at rest if the
instantaneously at rest.
wheel does not slip (slide)
the illustration of  in this diagram is misleading;
the direction of  would actually be into the screen
Your text discusses the rolling motion of a bicycle wheel
by changing reference frames; the center of mass of the
wheel is taken to be at rest with the ground moving past
it.
We skipped the sections on relative motion, so I
won’t expect you to perform a detailed analysis of
rolling motion by switching reference frames.
However, you should be able to calculate how
many revolutions the bicycle wheel makes in
example 8.7.
Here are some things you need to know about rolling
without slipping.
vtop=2vcm
The point on the rolling
vcm
object in contact with the
vcm
“ground” is instantaneously
at rest.
vcm
The center of the wheel
moves with the speed of the
center of mass.
vbot=0
The point at the “top” of the rolling wheel moves with a
speed twice the center of mass.
The two side points at the level of the center of mass
move vertically with the speed of the center of mass.
What is important for us is
that if an object rolls without
slipping, we can use our
OSE’s for z and z, using the
translational speed of the
object.
vCM=rz
vcm
vtop=2vcm
vcm
vcm
vbot=0
aCM=rz
These relationships will be valuable when we study the
kinetic energy of rotating objects.