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Rotational Motion
When we talk about things that rotate (spin on an axis) we need to define how we will
measure how far it has rotated.
A revolution is when an object spins about its axis one time.
In previous mathematics classes, we would have said that one revolution represents a
rotation of 360o. Therefore 1o (one degree) is one 360th of a revolution.
However, we could have said that one revolution represents a rotation of 400 grad.
Therefore 1 grad is one 400th of a revolution.
Since the circumference of a circle is 2pr, we can create a measuring system for
rotation based on this idea as well. We define 1 radian to be 1/2p of a revolution.
Said another way, during one revolution we would rotate through 2p rad.
We will often use radians as our way of measuring the “angle” of revolution in
problems. You should familiarize yourself with the relationship between radian
measure and degree measure for the angle of revolution ().
To start with:
one revolution = 360o = 2p rad
Therefore we can work out some common angles:
 = 1/2 of a revolution = 180o = p rad
 = 1/3 of a revolution = 120o = (2/3)p rad
 = 1/4 of a revolution = 90o = p/2 rad
 = 1/6 of a revolution = 60o = p/3 rad
 = 1/8 of a revolution = 45o = p/4 rad
 = 1/12 of a revolution = 30o = p/6 rad
(see the pie chart on page 198)
Since 360o = 2p rad we can create conversion factors to convert from degrees to
radians and vice versa.
1o = p rad/180o
How many radians is 110o?
and
1 rad = 180o/p rad
How many degrees is 0.785398 rad?
The angle of revolution () is used in problems to describe how much an object
revolves or rotates. The change in the angle of revolution is called the angular
displacement (D).
The angle of revolution () has units of radians, but radians are really just a ratio
between the distance (d) a point on the surface of an object moved and the radius (r) of
the object ( = d/r).
The distance a point on a rotating object moves (linear displacement) can be
determined by the equation:
d = r
Therefore, when we are doing problems where we are looking for d, the units for d will
be determined by the units for r and not by the radian unit. Note that if you enter the
angle of revolution in units of degree, the answer would be incorrect unless you
completed some additional complex mathematical manipulations.
By convention,  = 0 radians is also 0o (starting point is on the +x axis) and a positive
value for  is a rotation in the counterclockwise direction. A negative value for  is a
rotation in the clockwise direction (also starting from the +x axis).
Angular velocity () is a way of expressing how fast an object is spinning.
Average Angular Velocity () is defined as the change in angular displacement (D)
divided by the time it takes to complete that portion of the rotation (Dt).
D
=
Dt
units will be rad/s
If the Angular Velocity is constant over the time interval, then the average angular
velocity will be equal to the instantaneous angular velocity.
If the angular velocity is not constant during a time interval, then the Instantaneous
Angular Velocity would be found by the slope of the graph of the angular position
versus time. (This is just like average velocity and instantaneous velocity from
Chapter 2).
Note that the angular velocity tells us how fast the object is rotating. It does not tell us
how fast a point on the surface is moving. If we want the velocity of a point on the
surface (linear velocity) we need to multiply the angular velocity by the radius as show
below:
 = r
How fast are you moving (due to Earth’s rotation) if you are standing still on the
surface of the Earth?
Using what we know about the Earth: D = 2p rad (one rotation in 24 h)
Dt = 24 h = 86,400 s
D
So  =
rE = 6.38X106 m
Click to see solution
Dt
=
2π rad
=
86,400 s
7.2722X10-5 rad/s
 = r = (6.38X106 m)(7.2722X10-5 rad/s) = 464 m/s
Angular acceleration (a) is a way of expressing how fast a spinning object is
accelerating.
Angular acceleration (a) is defined as the change in angular velocity (D) divided by
the time it takes to make that change (Dt).
D
a=
Dt
units will be rad/s2
Note that the angular acceleration tells us how a rotating object is accelerating. It does
not tell us how fast a point on the surface is accelerating. If we want the acceleration
of a point on the surface (linear acceleration) we need to multiply the angular
acceleration by the radius as show below where r is the radius of the point on the
surface to the center of rotation:
a = ra
See table 8-1 on page 199 for a summary of the relationships between linear and
angular measures.
What are the angular displacements of each of the hands on a clock during a time
period of 3 hours?
a) The Second Hand: the second hand will rotate one complete revolution (-2p rad)
each minute and 3 hours is 180 minutes so 180 revolutions.
Therefore: D = (180)(-2p rad) = -360p rad or -1,130 rad.
b) The Minute Hand: the minute hand will rotate one complete revolution (-2p rad)
each hour so 3 hours is 3 revolutions.
Therefore: D = (3)(-2p rad) = -6p rad or -18.8 rad.
c) The Hour Hand: the hour hand will rotate one twelfth of a revolution (-(1/12)p
rad) each hour so 3 hours is 3(1/12) or 1/4 of a revolution.
Therefore: D = (1/4)(-2p rad) = -(1/2)p rad or -1.57 rad.
What are the linear displacements of the very tip of each of the hands on a clock
during a time period of 3 hours if the second hand is 6.5 cm long, the minute hand
is 5.8 cm long and the hour hand is 3.4 cm long?
See practice problems on page 200
Notice that the angular displacement (D) can be more than one complete
revolution. When we work with angle an use units like degrees, we sometimes
seem to think that 360o is the most that we can use.
One complete revolution is 360o or 2p radians (6.283185… rad). How many
revolutions take place if we have an angular displacement of 57.98 rad?
Dividing 57.98 rad by 2p will give us the number of revolutions.
57.98 rad/2p = 9.228 revolutions
If we know how much time it takes to complete the rotation, then we can know
the angular velocity ().
D
=
Dt
If we divide the angular velocity by 2p rad (amount of radians in one revolution)
then we get the angular frequency: the number of revolutions that occur in 1 s.

f = 2π
so
 = 2pf
Rotational Dynamics
A lever arm is the perpendicular distance from the axis of rotation to the point
where a force is applied.
Imagine that you put a wrench onto a both that you want to turn. The axis of rotation
is the center of the bolt. The point where you apply the force is on the handle of the
wrench. Therefore the level arm is the wrench. Note that when you push (or pull)
on the handle of the wrench, you will be pushing perpendicular to the handle of the
wrench. The length of the wrench will be a factor in how easy or how hard it is to
turn the bolt. A longer lever arm will make turning the bolt easier than a short one.
Have ever walked up to a door that has a horizontal bar across the middle and
were unsure of which side of the door to push on to get it to open?
If you push from close to the hinges (short lever arm) the door is very hard to
open. However if you push from the other end of the bar (long lever arm) the
door opens normally.
Why does having a longer lever arm make opening the door easier?
Torque (t) is how effectively a force causes rotation. The amount of torque applied
is determined by multiplying the force (F) we are pushing with times the length of
the lever arm. The higher the torque, the easier the rotation becomes.
If we are pushing (or pulling) perpendicular (90o) to the axis of rotation, then the
length of the lever arm is just its radius (r). However if we are pushing (or pulling)
at some smaller angle () then the length of the lever are is: r sin.
t = Fr sin
Changing the angle we are pushing (or pulling) on the lever effectively changes
the length of the lever arm. Why would we not always push or pull at 90o?
Since force is in newtons and the radius of the lever arm should be in meters, the
units for torque are newton-meters (N∙m)
Example: A plumber is tightening a pipe. He is using a pipe wrench that is 0.35 m
long. If the plumber pushes with a force of 52.4 N at an angle of 75.5o from
perpendicular, how much torque is he applying to the pipe?

r = 0.35 m
 = 75.5o
F = 52.4 N
Click to see solution
t = Fr sin
t = (52.4 N)(0.35 m) sin(75.5o)
t = 17.7558 N∙m = 18 N∙m
See example on page 202 and practice problems on page 203.
If more than one force is acting on an object, we need to find the net torque (tnet)in
order to determine how the forces will affect the object.
In many problems it will simply be a matter of adding the torques together to find
the net torque.
tnet = t1 + t2 + …..
If there are multiple forces acting on an object and it is not rotating then we know
that the net torque is equal to zero, and we can use that to solve the problem. For
example if only two forces are acting on an object and the object is not rotating
(tnet = 0), then we know that the two torques must be equal but acting in opposite
directions. tnet = t1 + t2 = 0
so t1 = -t2
Note that this means that the one of the two torques must be negative (clockwise
direction). In many problems we can just pick which one we want to be in the
negative (or clockwise) direction while the other will then be in the positive (or
counterclockwise) direction.
Two children are sitting on a seesaw that is 2.5 m long. If child one has a mass of 65
kg and child two has a mass of 51 kg, where should the pivot point of the seesaw be
located to keep them balanced?
Since the seesaw is balanced:
t 1 = F 1 r 1 t 2 = F2 r 2
tnet = t1 + t2 = 0
where F1 = m1g and F2 = m2g
Click to see solution
and r1 + r2 = 2.5 m
Let’s assume that child 1 (creates t1) is in the clockwise (or negative) direction.
F1 = m1g = (65 kg)(9.80 m/s2) = 637 N
F2 = m2g = (51 kg)(9.80 m/s2) = 499.8 N
r1 + r2 = 2.5 m so r1 = (2.5 m – r2)
t1 = -F1r1
t 2 = F 2r2
t1 = (-637 N)r1 = (-637 N)(2.5 m – r2) = – 1592.5 N∙m + (637 N)r2
t2 = (499.8 N)r2
so t1 + t2 = – 1592.5 N∙m + (637 N)r2 + (499.8 N)r2 = 0
t1 + t2 = – 1592.5 N∙m + (1136.8 N)r2 = 0
r2 = (1592.5 N ∙m)/(1136.8 N) = 1.4 m
and
r1 = 2.5 m – 1.4 m = 1.1 m
See example problem on page 204 and practice problems on page 205
Moment of Inertia (I) is a measure of an object’s resistance to rotation.
The Moment of Inertia of a point mass (i.e. we can ignore the mass of the
material that is connecting the object to the axis of rotation) is equal to the
mass of the object times the square of the object’s distance from the axis of
rotation.
I = mr2 units are kg∙m2
}
r
See that the rod connecting the ball to the axis
of rotation is show as being thin (a hint that
we can ignore its mass)
Axis of Rotation
Note that the equation for the moment of inertia depends upon the shape of
the rotating object. Table 8-2 has the equations for the moments of inertia
for a variety of shapes.
The major issue has to due with the idea that when the mass of an object is
distributed over an area, then it is not a simple matter to determine how that
particular distribution of mass impacts the moment of inertia.
A thin rod that is 1.05 m long has a 0.54 kg ball attached at each end. Find the
moment of inertia when the rod is rotated about the exact center and when the rod
is rotated about one end as illustrated in the diagrams below.
Assume the rod’s mass can be ignored-this means these are point mass problems
I = mr2
r = 0.525 m in the first problem Click to see solution
(r = 1.05 m in the second problem)
For the first problem, break it into two halves so that Itotal = I1 + I2
So
I1 = (0.54 kg)(0.525 m)2 = 0.1488 kg∙m2 and I2 = (0.54 kg)(0.525 m)2 = 0.1488 kg∙m2
Itotal = I1 + I2 = 0.1488 kg∙m2 + 0.1488 kg∙m2 = 0.2976 kg∙m2 = 0.30 kg∙m2
For the second problem, I = mr2 = (0.54 kg)(1.05 m)2 = 0.595 kg∙m2 = 0.60 kg∙m2
Note that we are assuming that the ball on the axis of rotation can be ignored (no
radius was given for it, so we could not use the uniform sphere equation).
See Example problem on page 207 and practice problems on page 208
Newton’s second law for rotational motion: the angular acceleration (a) of an object
is equal to the net torque (tnet) on the object divided by the moment of inertia (I)
a = tnet
I
Example: A solid disk has a mass of 23 kg and a diameter of 0.50 m. If you want to
accelerate it from a stop to an angular frequency of 3.5 rev/s over a 12 s time period,
what torque must be applied to the disk?
Needed equations:
m = 23 kg
r = ½ d = 0.25 m
f = 3.5 rev/s
Dt = 12 s
D
I for a solid disk = ½
a=
 = 2pf
Dt
Click to see solution
f = 2pf = (2p)(3.5 rev/s) = 21.99 rad/s
i = 0
D = f - i = 21.99 rad/s
mr2
21.99 rad/s
D
a=
=
= 1.83 rad/s2
12 s
Dt
I=½
mr2
= ½ (23 kg)(0.25
m)2
= 0.719
kg∙m2
tnet = aI = (1.83 rad/s2)(0.719 kg/m2)
= 1.3 kg∙m2/s2 = 1.3 N∙m
See example problem on page 209 and practice problems on page 210
Center of Mass: the center of mass of an object is the point in an object that moves
in the same way that a point particle would move.
Said another way, the center of mass of an object is the point in an object where the
mass appears to be located.
Imagine throwing a hammer. Unlike a baseball which will have its center of mass at
the center of the ball (because a baseball is spherically symmetrical), a hammer’s
center of mass will be located somewhere close to the head of the hammer rather
than in the middle of the handle. This is because the head of the hammer has a
greater mass than the handle. If we watch the hammer carefully as if flies through
the air, we would notice that it is rotating end over end about that single point. If we
threw a ball with equal mass to the hammer, the ball would follow the same path as
that point on the hammer. (See figure 8-11 a on page 211.)
Figures 8-11 b and c on page 211 illustrate a method for finding the center of mass
for an object by suspending the object by a string. The idea is that the force of
gravity will act on the object as if all of the mass were located at the center of mass.
The location of the center of mass in a moving object will have a lot to do with how
stable the motion of the object is.
Imagine putting two objects on a flat turntable. If we turn on the turntable and start
to slowly speed it up, we could see how the location of the center of mass of an
object effects how likely the object is to fall over as it travels in a circle.
Imagine two cylinders that have the same mass, but one has a rather large diameter
and is short while the other has a small diameter and is tall. Which of these would
fall over first as the turntable started to speed up? Why did you pick the one you
did?
Imagine two trucks that have the same mass, but one has a six inch lift kit and large
tires. Which truck would be more likely to roll over when they went around the
same flat curve?
Why do they build racetracks with steeply banked turns?
What could happen if you tried to drive through one of those turns going slow?
All of the issues on the previous slide have to do with the stability of an object in
motion (or at rest in some instances).
In general, where would we want the center of mass to be located in an object to give
it the greatest stability?
Why does a blocker in the NFL try to stay low to the ground when blocking?
Why do tall glasses have wide bases?
Why does a running back try to lower their shoulders before they hit a tackler?
Which box would be easier to tip over, a tall one with a narrow base or a short one
with the wide base (assume they have the same mass)?
An object is said to be in static equilibrium if the net force exerted on the object is
zero (translational equilibrium) and the net torque exerted on the object is zero
(rotational equilibrium).
An object in static equilibrium will have both its velocity and angular velocity either
equal to zero or being constant.
In other words, the object can not be undergoing any kind of acceleration. The idea
than an object is in static equilibrium can be used to help solve problems like the one
below.
A board of uniform density that has a mass of 4.35 kg and is 3.57 m long is
supported by two posts. The leftmost post is 1.15 m from the end. The rightmost
post is 0.480 m from the end. If the board is to remain level and not moving, what
force must each of the posts exert on the board?
In this problem, being in static equilibrium means that the net force acting on the
board must be zero and that the net torque acting on the board must also be zero.
A board of uniform density that has a mass of 4.35 kg and is 3.57 m long is
supported by two posts. The leftmost post is 1.15 m from the end. The rightmost
post is 0.480 m from the end. If the board is to remain level and not moving, what
force must each of the posts exert on the board?
To start this problem realize that there are three forces involved in the problem: The
force of gravity pulling downward on the board from the center of mass and the
contact forces pushing upward from each post.
Since the system is in equilibrium, the net force on the board must be zero.
In addition, the net torque must be zero. Choosing one of the posts as the axis of
rotation will simplify the mathematics so lets assume that the board would rotate
about the point where it touches the leftmost point (choosing this axis of rotation
means that the torque caused by the contact force created by the post touching the
board is zero).
m = 4.35 kg
rR = 3.57 m – 1.15 m – 0.480 m = 1.94 m
rg = (3.57 m/2) – 1.15 m = 0.635 m
rR
FL
FR
rg
Fg
Fg = -mg = - (4.35 kg)(9.80 m/s2) = - 42.63 N
Click to see solution
tnet = tL + tg + tR = 0 and tL = 0 so tg = -rgFg and tR = rRFR and tR = - tg
rRFR = -rgFg therefore FR = (-rgFg)/rR = (-0.635 m)(- 42.63 N)/(1.94 m) =13.95 N
Fnet = FL + Fg + FR = 0 so FL = - Fg - FR = -(- 42.63 N) - (13.95 N) = 28.68 N
FL = 28.7 N and
FR = 14.0 N
See example on pages 214-215 and practice problems on page 215
If you are on an object that is rapidly rotating, any motion you see occurring on that
object may not appear to follow the laws of physics that we have been discussing.
This is because you are part of a rotating frame of reference which is an accelerated
frame of reference and Newton’s Laws do not apply.
There are two famous “forces” that are not really forces that come about because of
rotating frames of reference.
Centrifugal “Force”, as we have seen in an earlier chapter, is the apparent “force”
that pushes an object away from the axis of rotation. But there is no actual force
“pushing or pulling” the object outward. The effect is due to acceleration and inertia
acting in other directions.
Coriolis “Force” is the apparent “force” that seems to push an object off line due to
the rotation of the Earth. In this case, the object moves off line because any object
launched from the Earth’s surface will have a rotational component in it due to the
rotation of the Earth itself.
Coriolis “Force”:
Imagine shooting a super powerful cannon due north from somewhere on the
equator.
The cannon will only provide a force in the northward direction. So if the Earth
were standing still, the projectile would travel due north.
However, the Earth is always rotating, so the projectile also has a component of
velocity that is directed eastward. As the projectile travels, the eastward component
of its velocity remains constant (ignoring wind resistance). However, the Earth that
the projectile is traveling over will have a smaller and smaller eastward component
to its motion (why?). As a result, the eastward component of the projectile will
cause it to drift off target to the east the further it travels north.
What would happen to a projectile fired due south from the north pole?
The Coriolis “Force” has an impact on weather because this same effect causes
winds from the north or south to drift in opposite directions (east and west) which
can cause them to appear to spin as they move past each other.