Ch 9 Rotation
9.1 Rotational Kinematics: Angular Velocity and Angular Acceleration
Q: What is angular velocity? Angular speed? What symbols are used to
denote each? What units are used?
Q: What is linear velocity? Linear Speed? What symbols are used to
denote each? What units are used?
Q: What is tangential velocity? Tangential Speed? What symbols are used to
denote each? What units are used?
Q: How do these concepts relate to circular geometry?
Q: Along the same disk, as you go farther out, which value(s) increase(s)?
Which stay the same? Why?
Consider a disk rotating about a fixed axis though its center, where ri = the
distance from the center to its “ith” particle, and i = the angle measured
counterclockwise from a fixed reference line in space to a line representing the
radius from the center to the particle. As the disk rotates, the angle changes.
You denote this with d . As the disk rotates, the particle also moves through
an arc length, ds such that ds = ri |d| as shown in Fig 9-2 below.
Figure 9-1
Figure 9-2
Q: How would you measure d?
A: In radians
Those respective distances, ds and ri vary from particle to particle, but their
respective inter-related ratio, d, or , called the angular displacement, is the
same for each particle on the disk.
Q: Why?
A: As you change the radius of the particle under consideration, you
automatically proportionately change the arc length swept out by that particle.
Q: What is the arc length, in radians, of a full circle?
A: 2r
So s = 2r and  = 2ri / ri = 2 radians = 3600 = 1 revolution
Q: What do you call the time rate of change of the angle? Does it vary from
one particle to the next?
A: . Nope, it’s identical for every particle. In fact, that’s what angular
velocity means and that’s how you symbolize the meaning of angular velocity.
NOTE:  is accepted as increasing in the counterclockwise direction (as that is
how we normally sweep out an angle – schmoes on a mountain say “ugh!”)
Q: Is angular velocity a vector? What do you call the magnitude of angular
velocity?
A: Yep. It’s defined as pointing along the axis of rotation. Its magnitude is
called angular speed. Normally, the units are rev / min = RPM’s in the real
world, and rad/sec in SI.
To convert RPM’s to radians to degrees, use 1 rev = 2 rad = 3600
Q: What would you call the time rate of change of angular velocity? What is
its symbol?
A: Angular acceleration = radians / sec / sec in SI. Its symbol is .
If  is increasing,  is positive; if  is decreasing,  is negative.
Angular displacement is to linear displacement as
Angular velocity is to ________________________ and
Angular acceleration is to ____________________
Q: What do you get when you integrate eq 9-3 above, assuming  is constant?
A: f = i + t where i (or 0) is the constant of integration = initial angular
velocity! This is the rotational analog to vf = vi + at
Integrate it again to get f = 0 + 0t + ½ t2 which is the rotational analog of
xf = x0 + v0t + ½ at2
If you take the top two equations, solve them for t (using the quadratic
formula), and set them equal to each other, you get f2 = 02 + 2 ( - 0). One
morr comparison – this is the rotational analog of vf2 = v02 + 2a(x – x0)
So, if you exchange the word rads for meters, all of your rotational units will
agree!
See Ex 9-1, p 269-270.
A compact disk rotates from rest to 500 rev/min in 5.5 sec.
a) What is its angular acceleration, assuming it is constant?
b) How many revolutions does the disk make in 5.5 sec?
c) How far does a point on the rim 6 cm from the center of the disk travel
during the 5.5 s it takes to get to 500 rev/min?
a) = 0 + t
 = 0 + t
Solve for 
 =  = 500 rev/min x 2  rad x 1 min
t
5.5 sec
1 rev
60 s
2
= 9.52 rad/s
b) Solve for  (in radians)
f = 0 + 0t + ½ t2
= 0 + 0t + ½ t2
= ½ (9.52 rad/s2)(5.5 s)2 = 144 rad
In revolutions, it would be 144 rad x 1 rev/2 rad = 22.9 rev
c) s = r   = 0.06 m  144 rad = 8.64 m
Q: Is there a relationship between the tangential and rotational velocity and /
or tangential and centripetal accelerations?
A:
vtan = ri 
atan = ri 
acentrip = ri 2
9.2 Rotational Kinetic Energy
Q: What is KE again? How do you calculate KE?
Q: What is inertia? Can an object have inertia?
Q: Will a “rolling ball” cause morr damage than a ball that simply
“translates” along?
Q: Is inertia the same thing as mass?
Q: What is the moment of inertia? On which two things does the moment of
inertia depend?
A: Mass and the distribution of that mass around the axis of rotation.
See Ex 9-2, p 272.
An object consists of 4 point particles, each of mass m, that
are connected by rigid massless rods to form a rectangle of
sides 2a and 2b. The system rotates with angular speed 
about an axis in the plane of the figure through the center
as shown.
a) Find the KE of the entire object.
b) Check your results individually by calculating the
KE of each particle, and then finding their sum.
So, rotational KE depends on I and  while translational KE depends on m and
v. In other words, translational KE is still ½ mv2, where v is the tangential
velocity. (Translational and tangential both start with a “t.”) This is
important to know for the problems later on (#’s 18 and 19).
9.3 Calculating the Moment of Inertia
As stated previously, the moment of inertia is dependent upon the mass and the
distribution of that mass around the axis of rotation. Additionally, an object’s
rotational inertia depends on its moment of inertia and its angular speed.
Whenever it is easiest to change the rotational inertia of an object, we say that
it has less rotational inertia. Thus, comparatively, the formula used to
calculate the moment of inertia coincident to an object’s rotational inertia, will
have a smaller value.
Q: Considering two disks of equal mass, which will reach the bottom of a
ramp first – a hollow ring or a solid disk? Why?
EX’s: Turning a meter stick with a mass attached, turning a dumbbell with
the masses extended versus centralized, turning disks of equivalent mass – one
hollowed out in the center and one solid.
Systems of Discrete Particles
Just like you’d expect, you can use eq 9-10 to find the moment of inertia for a
system of discrete particles about an axis of rotation.
Q: Within that defined system, if you change the axis of rotation, will you
affect the moment of inertia? Why?
A: Yep, in just the same way as you affect I when you adjest the weights along
a dumbbell.
Continuous Objects
Hmmm…I think I sense a calculus application…
If an object is continuous, then instead of using the sum of all of the miri’s, we
just use:  r2 dm, where r = radial distance from the axis to the element dm.
See Ex 9.3, P 273.
Find the moment of inertia of a uniform rod of length L and mass M about an
axis perpendicular to the rod and through one end. Assume the rod has
negligible thickness.
Since the total mass, M, is uniformly distributed along the
length, L, the mass per unit length (linear mass density) is 
= M/L.
The moment of inertia is given by:
L
I=
 x dm
2
0
Relating dm to dx gives us:
dm =  dx = M dx
L
So, then,
3
M
M
M  1  M L3 1
2
dx 
x
dx

 ML2
Iy =  x dm   x
 

L
L 0
L  3 L 3 3
0
L
2
L
2
0
Hoop about a Perpendicular Axis Through Its Center – Optional Section
This is the base-line I, off of which all others are defined. For any hoop
(hollowed out center), the moment of inertia is defined as:
I = r2 dm =  R2 dm = R2 dm = MR2
Q: How would I change if the mass were distributed throughout, as in a solid
disk?
I = ½ MR2
Q: Why?
A: All things considered, the moment (excuse the pun ) you distribute the
mass evenly throughout a disk, the average distance that the mass is from the
axis is cut in half, thus, so is the formula! The actual “mathy” derivation is on
p 274, under the optional material Uniform Disk About A Perpendicular Axis.
See and know TABLE 9-1, p 274
Q: The formula for a solid cylinder rotating about a centralized axis is the
same as that of a disk. Why?
A: A solid cylinder is just a phat, err.., fat solid disk. 
The Parallel Axis Theorem
This theorem relates I about an axis through the center of mass of an object to
its I about a second parallel axis, where Icm = the moment of inertia about an
axis through an object’s center of mass, M, and I is the second moment of
inertia about that second parallel axis that is a distance h away.
Really, this is just like applying the moment of inertia twice; once for the actual
com, and once for the relative center of mass!
Try Ex’s 9-4 – 9-7 now. 
9.4 N2 For Rotation
Q: What is N2? What does it mean? What does it imply?
Q: What is torque? Is it a force? Why / why not?
A: If you want to get an object to twist, turn, or spin, you have to apply a force
in that direction. If the object is a disk, for example, nailed to the floor
through its center, then to get it to turn, the force you apply is really a
tangential force. That force, which causes an object to twist, turn, or spin,
creates a torque. To calculate the magnitude of the torque, you multiply Ft
and radial distance.
The magnitude of the torque can also be calculated using Newton’s 2 nd Law
(N2) for rotation. Normally, N2 says that a net force will cause a mass to
accelerate. We write this as F = ma, as it allows us to solve for any missing
values. In terms of rotation, the force becomes the torque (causing the
rotation), the mass becomes mass-and-position of mass, or moment of
inertia, I, and the acceleration becomes angular acceleration, .
Calculating Torques
Figure 9-19
Figure 9-20
The figures at the left show
that the torque is
maximized, as you apply
tangential force, at the
largest r, or distance from
the axis of rotation.
In Fig 9-19 above, a force acts on a restrained object, causing it to rotate about
a fixed point, O, that is  the page. If  is the angle between the radial
direction, and the radial force, and l is the line of action ( the line parallel to
the force and passing through the point of application) = the lever arm, then
we get the equation for equivalent expressions for torque. When  = 900, then
sin  = 1, which just gives us either Ftr or Fl, as shown below. (Always assume
 to be the smaller of the two angles described by the two rays, either F and r,
or F and l.
NOTE: The calculations for torque in this chapter 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
will rotate in strictly the x-y, or y-z, or x-z plane, thus the direction of our
torque, which is always  to the plane of rotation for the circular object, will
be obvious. Next chapter, you will learn to use a method called the “cross
product” to help you find the actual vector for torque, even if the object is not
rotating strictly the x-y, or y-z, or x-z plane. 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. Morr with this later! 
Torque Due to Gravity
Q: Why does this formula make sense for torque due to gravity?
A: The force is just the force due to gravity = weight = mg, and the radial
distance is just the distance to the center of mass.
Q: As a car falls off of a cliff, why does it roll forward?
As an object is exposed to forces of gravity, each point of the object undergoes
a gravitational pull about relative axes of rotation. If you think of an object as
a collection of microscopic point particles, then there is a microscopic
gravitational force on each separate particle.