v ∆θ
a =
[where we have replaced the tan(∆θ ) in equation 18-4 above with ∆θ ].
c
lim
t
∆ → 0 ∆t
∆θ
∆θ
The constant v
=
can be taken outside the limit yielding a
v lim
. But the lim
is the
c
∆t→0
t
∆
∆t→ 0
t
∆
rate of change of the angle θ, which is, by definition, the angular velocity w. Thus
a =v w
c
v
According to equation 18-3, v = rw . Solving that for w we find that w = r . Substituting this
into our expression for ac yields
v 2
a =
(18-5)
c
r
Please sound the drum roll! This is the result we have been seeking. Note that by substituting
r w for v, we can also write our result as
2
a = r w (18-6)
c
It should be pointed out that, despite the fact that we have been focusing our attention on the case
in which the particle moving around the circle is moving at constant speed, the particle has
centripetal acceleration whether the speed is changing or not. If the speed of the particle is
changing, the centripetal acceleration at any instant is (still) given by 18-5 with the v
being the
speed of the particle at that instant (and in addition to the centripetal acceleration, the particle
also has some along-the-circular-path acceleration known as tangential acceleration). The case
that we have investigated is, however the remarkable case. Even if the speed of the particle is
constant, the particle has some acceleration just because the direction of its velocity is
continually changing. What’s more, the centripetal acceleration is not a constant acceleration
because its direction is continually changing. Visualize it. If you are driving counterclockwise
(as viewed from above) around a circular track, the direction in which you see the center of the
circle is continually changing (and that direction is the direction of the centripetal acceleration).
When you are on the easternmost point of the circle the center is to the west of you. When you
are at the northernmost point of the circle the center is to the south of you. When you are at the
westernmost point of the circle, the center is to the east of you. And when you are at the
southernmost point of the circle, the center is to the north of you.
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Chapter 19 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration
19 Rotational Motion Variables, Tangential Acceleration,
Constant Angular Acceleration
Because so much of the effort that we devote to dealing with angles involves acute angles,
when we go to the opposite extreme, e.g. to angles of thousands of degrees, as we often
do in the case of objects spinning with a constant angular acceleration, one of the most
common mistakes we humans tend to make is simply not to recognize that when someone
asks us; starting from time zero, how many revolutions, or equivalently how many turns
or rotations an object makes; that someone is asking for the value of the angular
displacement ∆θ. To be sure, we typically calculate ∆θ in radians, so we have to convert
the result to revolutions before reporting the final answer, but the number of revolutions
is simply the value of ∆θ.
In the last chapter we found that a particle in uniform circular motion has centripetal acceleration
given by equations 18-5 and 18-6:
v 2
a =
2
a = r w
c
r
c
It is important to note that any particle undergoing circular motion has centripetal acceleration,
not just those in uniform (constant speed) circular motion. If the speed of the particle (the value
v 2
of v
a =
in
) is changing, then the value of the centripetal acceleration is clearly changing.
c
r
One can still calculate it at any instant at which one knows the speed of the particle.
If, besides the acceleration that the particle has just because it is moving in a circle, the speed of
the particle is changing, then the particle also has some acceleration directed along (or in the
exact opposite direction to) the velocity of the particle. Since the velocity is always tangent to
the circle on which the particle is moving, this component of the acceleration is referred to as the
tangential acceleration of the particle. The magnitude of the tangential acceleration of a particle
in circular motion is simply the absolute value of the rate of change of the speed of the particle
dv
a =
. The direction of the tangential acceleration is the same as that of the velocity if the
t
dt
particle is speeding up, and in the direction opposite that of the velocity if the particle is slowing
down.
Recall that, starting with our equation relating the position s of the particle along the circle to the
angular position θ of a particle, s = rθ , we took the derivative with respect to time to get the
relation v = r w . If we take a second derivative with respect to time we get
dv
dw
= r
dt
dt
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Chapter 19 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration
dw
On the left we have the tangential acceleration a of the particle. The
on the right is the
t
dt
time rate of change of the angular velocity of the object. The angular velocity is the spin rate, so
dw
a non-zero value of
means that the imaginary line segment that extends from the center of
dt
dw
the circle to the particle is spinning faster or slower as time goes by. In fact,
is the rate at
dt
which the spin rate is changing. We call it the angular acceleration and use the symbol a (the
dv
dw
Greek letter alpha) to represent it. Thus, the relation
= r
can be expressed as
dt
dt
a = r a (19-1)
t
A Rotating Rigid Body
The characterization of the motion of a rotating rigid body has a lot in common with that of a
particle traveling on a circle. In fact, every particle making up a rotating rigid body is
undergoing circular motion. But different particles making up the rigid body move on circles of
different radii and hence have speeds and accelerations that differ from each other. For instance,
each time the object goes around once, every particle of the object goes all the way around its
circle once, but a particle far from the axis of rotation goes all the way around circle that is
bigger than the one that a particle that is close to the axis of rotation goes around. To do that, the
particle far from the axis of rotation must be moving faster. But in one rotation of the object, the
line from the center of the circle that any particle of the object is on, to the particle, turns through
exactly one rotation. In fact, the angular motion variables that we have been using to
characterize the motion of a line extending from the center of a circle to a particle that is moving
on that circle can be used to characterize the motion of a spinning rigid body as a whole. There
is only one spin rate for the whole object, the angular velocity w, and if that spin rate is
changing, there is only one rate of change of the spin rate, the angular acceleration a. To
specify the angular position of a rotating rigid body, we need to establish a reference line on the
rigid body, extending away from a point on the axis of rotation in a direction perpendicular to the
axis of rotation. This reference line rotates with the object. Its motion is the angular motion of
the object. We also need a reference line segment that is fixed in space, extending from the same
point on the axis, and away from the axis in a direction perpendicular to the axis. This one does
not rotate with the object. Imagining the two lines to have at one time been collinear, the net
angle through which the first line on the rigid body has turned relative to the fixed line is the
angular position θ of the object.
The Constant Angular Acceleration Equations
While physically, there is a huge difference, mathematically, the rotational motion of a rigid
body is identical to motion of a particle that only moves along a straight line. As in the case of
linear motion, we have to define a positive direction. We are free to define the positive direction
whichever way we want for a given problem, but we have to stick with that definition throughout
the problem. Here, we establish a viewpoint some distance away from the rotating rigid body,
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Chapter 19 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration
but on the axis of rotation, and state that, from that viewpoint, counterclockwise is the positive
sense of rotation, or alternatively, that clockwise is the positive sense of rotation. Whichever
way we pick as positive, will be the positive sense of rotation for angular displacement (change
in angular position), angular velocity, angular acceleration, and angular position relative to the
reference line that is fixed in space. Next, we establish a zero for the time variable; we imagine a
stopwatch to have been started at some instant that we define to be time zero. We call values of
angular position and angular velocity, at that instant, the initial values of those quantities.
Given these criteria, we have the following table of corresponding quantities. Note that a
rotational motion quantity is in no way equal to its linear motion counterpart, it simply plays a
role in rotational motion that is mathematically similar to the role played by its counterpart in
linear motion.
Linear Motion Quantity
Corresponding Angular Motion Quantity
x
θ
v
w
a
a
The one variable that the two different kinds of motion do have in common is the stopwatch
reading t.
Recall that, by definition,
dθ
w =
dt
and
dw
a =
dt
While it is certainly possible for a to be a variable, many cases arise in which a is a constant.
Such a case is a special case. The following set of constant angular acceleration equations apply
in the special case of constant angular acceleration: (The derivation of these equations is
mathematically equivalent to the derivation of the constant linear acceleration equations. Rather
than derive them again, we simply present the results.)
1
2
θ = θ + w t + a t
o
o
(19-2)
2
The Constant
w + w
o
θ = θ +
t (19-3)
o
Angular
2
Acceleration
Equations
w = w
(19-4)
o + a t
w 2 = w 2 +
(19-5)
o
2a θ
∆
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Chapter 19 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration
Example 19-1
The rate at which a sprinkler head spins about a vertical axis increases steadily for
the first 2.00 seconds of its operation such that, starting from rest, the sprinkler
completes 15.0 revolutions clockwise (as viewed from above) during that first
2.00 seconds of operation. A nozzle, on the sprinkler head, at a distance of 11.0 cm
from the axis of rotation of the sprinkler head, is initially due west of the axis of
rotation. Find the direction and magnitude of the acceleration of the nozzle at the
instant the sprinkler head completes its second (good to three significant figures)
rotation.
Solution: We’re told that the sprinkler head spin rate increases steadily, meaning that we are
dealing with a constant angular acceleration problem, so, we can use the constant angular
acceleration equations. The fact that there is a non-zero angular acceleration means that the
nozzle will have some tangential acceleration a . Also, the sprinkler head is spinning at the
t
instant in question so the nozzle will have some centripetal acceleration a . We’ll have to find
c
both a and a and add them like vectors to get the total acceleration of the nozzle. Let’s get
t
c
started by finding the angular acceleration a. We start with the first constant angular
acceleration equation (equation 19-2):
0
0
1
2
θ =θ +w t + a t
o
o
2
The initial angular velocity w is given as zero. We have defined the initial angular
o
position to be zero. This means that, at time t = 2.00 s, the angular position θ is
2π rad
15.0 rev = 15.0 rev
= 94.25 rad.
rev
Solving equation 19-2 above for a yields:
2θ
a =
2
t
2 (94 25
. rad)
a =
2
(2.00 s)
rad
a = 47.12
2
s
Substituting this result into equation 19-1:
a = r a
t
gives us
2
a = (.110 m) 47.12 rad/s
t
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Chapter 19 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration
which evaluates to
m
a = 5.18
.
t
2
s
Now we need to find the angular velocity of the sprinkler head at the instant it completes
2.00 revolutions. The angular acceleration a that we found is constant for the first fifteen
revolutions, so the value we found is certainly good for the first two turns. We can use it in the
fourth constant angular acceleration equation (equation 19-5):
0
w2 = w 2
o + 2a
θ
∆
2π rad
where θ
∆ = rev
2
= 2 00
.
rev
= 4.00π rad
rev
w = 2a θ
∆
w = 2 (94 25
. rad/s2) 4 00
. π rad
w = 48.67 rad / s
(at that instant when the sprinkler head completes its 2nd turn)
Now that we have the angular velocity, to get the centripetal acceleration we can use
equation 18-6:
2
a = r w
c
2
a = .110 m (48.67 rad /s)
c
m
a = 260
.
.6
c
2
s
Given that the nozzle is initially at a point due west of the axis of rotation, at the end of
2.00 revolutions it will again be at that same point.
NORTH
at
EAST
a
c
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Chapter 19 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration
Now we just have to add the tangential acceleration and the centripetal acceleration vectorially to
get the total acceleration. This is one of the easier kinds of vector addition problems since the
vectors to be added are at right angles to each other.
North
a
at = 5.18 m/s2
θ
East
ac = 260.6 m/s2
From Pythagorean’s theorem we have
2
2
a =
a + a
c
t
2 2
2 2
a = (260.6 m / s ) + (5 18
. m / s )
a = 261 m/s2
From the definition of the tangent of an angle as the opposite over the adjacent:
at
tanθ =
ac
− a
1
t
θ = tan
ac
2
−
5 18
.
s
/
m
1
θ = tan
2
260 6
.
s
/
m
o
θ = 1 14
.
Thus
a = 261 m/ s2 at 1.14° North of East
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Chapter 19 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration
When the Angular Acceleration is not Constant
The angular position of a rotating body undergoing constant angular acceleration is given, as a
function of time, by our first constant angular acceleration equation, equation 19-2:
1
2
θ = θ + w t + a t
o
o
2
If we take the 2nd derivative of this with respect to time, we get the constant a. (Recall that the
dw
first derivative yields the angular velocity w
a =
and that
.) The expression on the right
dt
side of
1
2
θ = θ + w t + a t contains three terms: a constant, a term with t to the first power, and
o
o
2
a term with t to the 2nd power. If you are given θ in terms of t, and it cannot be rearranged so
that it appears as one of these terms or as a sum of two or all three such terms; then; a is not a
constant and you cannot use the constant angular acceleration equations. Indeed, if you are
being asked to find the angular velocity at a particular instant in time, then you’ll want to take
dθ
the derivative
and evaluate the result at the given stopwatch reading. Alternatively, if you
dt
are being asked to find the angular acceleration at a particular instant in time, then you’ll want to
2
d θ
take the second derivative
and evaluate the result at the given stopwatch reading.
2
dt
Corresponding arguments can be made for the case of w. If you are given w as a function of t
and the expression cannot be made to “look like” the constant angular acceleration equation
w = w
then you are not dealing with a constant angular acceleration situation and you
o + a t
should not use the constant angular acceleration equations.
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23
Chapter 20 Torque & Circular Motion
20 Torque & Circular Motion
The mistake that crops up in the application of Newton’s 2nd Law for Rotational
Motion involves the replacement of the sum of the torques about some particular
axis, ∑τ , with a sum of terms that are not all torques. Oftentimes, the errant
o
sum will include forces with no moment arms (a force times a moment arm is a
torque, but a force by itself is not a torque) and in other cases the errant sum will
include a term consisting of a torque times a moment arm (a torque is already a
torque, multiplying it by a moment arm yields something that is not a torque).
Folks that are in the habit of checking units will catch their mistake as soon as
they plug in values with units and evaluate.
We have studied the motion of spinning objects without any discussion of torque. It is time to
address the link between torque and rotational motion. First, let’s review the link between force
and translational motion. (Translational motion has to do with the motion of a particle through
space. This is the ordinary motion that you’ve worked with quite a bit. Until we started talking
about rotational motion we called translational motion “motion.” Now, to distinguish it from
rotational motion, we call it translational motion.) The real answer to the question of what
causes motion to persist, is nothing—a moving particle with no force on it keeps on moving at
constant velocity. However, whenever the velocity of the particle is changing, there is a force.
The direct link between force and motion is a relation between force and acceleration. The
relation is known as Newton’s 2nd Law of Motion which we have written as equation 14-1:
= 1
a
∑ F
m
in which,
a is the acceleration of the object, how fast and which way its velocity is changing
1
m is the mass, a.k.a. inertia, of the object.
can be viewed as a sluggishness factor, the
m
1
bigger the mass m, the smaller the value of
and hence the smaller the acceleration of the
m
object for a given net force. (“Net” in this context just means “total”.)
∑ F is the vector sum of the forces acting on the object, the net force.
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Chapte