Transcript Document

Rotational Motion
We consider only rigid (non-squishy) bodies in this
section.
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 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.
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
Right hand rule: curl the fingers of your right hand in the
direction of rotation. Your thumb points in the direction
of the angular velocity vector.
You’ll learn to love the right hand rule next week!
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: 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, =d/dt
z=dz/dt
vtang=rz
atang=rz
You can also use aradial = aR=Rz2
Kinematic Equations for Uniformly Accelerated
Rotational Motion
Remember our equations of linear kinematics?
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: A centrifuge motor goes from rest to 20,000
rpm in 5 minutes. Through how many revolutions did the
centrifuge motor of example 8.5 turn during its
acceleration period? Assume constant angular
acceleration.
First we need to calculate :
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.
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
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 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.
Torque
We began this course with a section on kinematics, the
description of motion without asking about its causes.
We then found that forces cause motion, and used
Newton’s laws to study dynamics, the study of forces and
motion.
We have had a deceptively brief section on angular
kinematics. It is brief because we already learned how to
solve such problems on the first day of class.
What’s next?
kinematics  dynamics
rotational kinematics  rotational dynamics
The rotational analog of force is torque.
Consider two equal and opposite forces acting at the
center of mass of a stationary meter stick.
F
F
Does the meter stick move?
Fext = macm = 0 so acm = 0.
Consider two equal and opposite forces acting on a
stationary meter stick.
F
F
Does the meter stick move?
Fext = macm = 0 so acm = 0.
The center of mass of the meter stick does not
accelerate, so it does not undergo translational motion.
However, the meter stick would begin to rotate about its
center of mass.
A torque is produced by a force
acting on an extended (not
pointlike) object.
The torque depends on how strong
the force is, and where it acts on
the object.
F
O
You must always specify your reference axis for
calculation of torque. By convention, we indicate that
axis with the letter “O” and a dot.
Torques cause changes in rotational motion.
Torque is a vector. It is not a force,* but is related to
force.
*So never set a force equal to a torque!
Let’s apply a force to a rod
and see how we get a torque.
First apply the force.
You need to choose an axis of
rotation. Usually there will be a
“smart” choice. Label it with a
point (or line) and an “O.”
+
O
Choose the direction of rotation
that you want to correspond to
positive torque.*
Label the positive direction with a
curved arrow and a “+” sign! Do this
around the point labeled “O.”
F
*Traditionally, the counterclockwise direction is chosen to be
positive. You are free to choose otherwise.
Draw a vector from the origin
to the tail of the force vector.
Give it a name (typically R).
Label the angle between R
and F (you may have to
slide the vectors around
to see this angle).
The symbol for torque is the
Greek letter tau (). The
magnitude of the torque due
to F is RF sinθ.
+
O
R
θ
F
Look at your diagram and determine if the torque would
cause a + or a – rotation (according to your choice of +).
In this case, the rotation would be -, so z=-RF sinθ.
Important: θ is the angle
between R and F.
“Slide” R and F around until
their tails touch. θ is the
angle between them.
θ
F
+
O
R
R
θ
“Between” means go from R to F.
Don’t be fooled by a problem which
gives an angle θ not “between” the
vectors! (Example coming soon.)
F
This θ is the correct
“angle between.”
Watch out for diagrams
containing some other
angle!
In this diagram, which is the angle between R and F?
F
θ?
R
NO!
There are two choices for the angle between R and F.
Because sin(θ)=-sin(-θ), either choice will give you the
correct answer (switch direction of + rotation and
switch sign on sine gives no net switch in sign).
There are other ways to find the
torque.
Often it is easy to visualize F,
the component of F which is
perpendicular to R.
The magnitude of the torque
due to F is RF, and in this case
z=-RF. (Note F=F sinθ.)
Sometimes it is easier to
visualize R, the component of R
which is perpendicular to F.
+
R
O
R
F

θ
F
The magnitude of the torque due to F is RF, and in this
case z=-RF. (Note R=R sinθ.)
Summarizing:
OSE:
z = RF = RF = RF sinθ
+
R is called the lever arm or
moment arm. The line along
which F is directed is its line of
action.
The z axis passes through the
point O and is perpendicular to
the plane of the paper.
R
O
R
F
θ
F
To find the direction of the torque, curl your fingers
around the direction of rotation from R into F. The thumb
of your right hand points in the direction of the torque.
Important reminder: label the point O about which your
torques are calculated and draw a curved arrow around it
with a + sign to show what you have chosen for a
positive sense of direction.
Draw the curved arrow around
the point O, not somewhere else!
A torque producing a + rotation
is +. A torque producing a rotation is -.
R
O
F
+
Example: The biceps muscle exerts a
vertical force of 700 N on the lower
arm, as shown in the figure. Calculate
the torque about the axis of rotation
through the elbow joint.
F
30°
There is no new litany for torques.
You should adapt the litany for force
problems.
When you work with torques, the
first thing you need to do is draw
an extended free-body diagram.
Before that, we need to have a
diagram of the “thing” we are
investigating.
r=5 cm
F
r
We are not interested in the upper arm!

We have our diagram. Now we
must do a free-body diagram. For
rotational motion, we must do an
extended free-body diagram, which
shows where the forces are
applied.
F
O +
r
θ
F
r

Label the rotation axis.
Choose a + direction for
rotation.
How about this for an OSE?
z = RF sinθ
No! No! No!
From the extended free-body
diagram, I see that the angle
between r and F is 90+θ, so
z=RF sin(90+θ) would work.
I think it is better to look for r or
F. In this case, r is easy to see.
F
r
+
O
θ
r
From the diagram r=r cosθ.
OSE: z = RF = RF cosθ. Done! (Except for plugging in
numbers.)
“That was a lot of work for something that takes 2
lines in the textbook!”
No. I showed you a general approach to torque
problems. The text just solved one simple problem.
If more than one torque acts on an object, the net
torque is the algebraic sum of the two torques
(“algebraic” means there may be signs involved).
Example: Calculate the net torque on
the compound wheel shown in the
drawing.
F2
θ
The diagram will serve as an extended
free-body diagram. No need for a
separate one.
z,net = z = z,F1 + z,F2
z,net = +r1F1 + r2(-F2cosθ)
z,net = r1F1 - r2F2cosθ
r2
r1
F1
Rotational Dynamics;
Torque and Rotational Inertia
We saw in our study of dynamics that forces cause
acceleration:
F = ma.
Torques produce angular acceleration, and the rotational
equivalent of mass is the moment of inertia, I:
OSE:
z = Iz.
This is really a vector equation, but our problems will all
have a unique axis of rotation, which is “like” a onedimensional problem, so that the only vestiges of the
vector nature of z will be the sign.
What is this moment of inertia, I?
It is the rotational analog of mass.
I depends on the mass of the object. It also depends on
how the mass is distributed relative to the axis of
rotation.*
The figure on the next slide gives I for various objects of
uniform composition. You should use this figure (or its
equivalent, or appropriate portions of it) in homework
solutions.
Solid cylinder, mass M, radius R
I=½MR2
It doesn’t matter how thick the cylinder is!
*This means a single object can have different I’s for
different axes of rotation!