A horizontal disk of rotational inertia
2
4 25
. kg ⋅ m with respect to its axis of symmetry is
spinning counterclockwise about its axis of symmetry, as viewed from above, at
15.5 revolutions per second on a frictionless massless bearing. A second disk, of rotational
inertia
2
1.80 kg ⋅ m with respect to its axis of symmetry, spinning clockwise as viewed from
above about the same axis (which is also its axis of symmetry) at 14.2 revolutions per second,
is dropped on top of the first disk. The two disks stick together and rotate as one about their
common axis of symmetry at what new angular velocity (in units of radians per second)?
I = 1.80 kg⋅m2
w
2
2
I
= 4.25 kg⋅m2
1
w
1
w′
Some preliminary work (expressing the given angular velocities in units of rad/s):
rev 2π rad
rad
rev 2π rad
rad
w = 15 5
.
= 97.39
w =14 2
.
= 89 22
.
1
2
s rev
s
s rev
s
Now we apply the principle of conservation of angular momentum for the special case in which
there is no transfer of angular momentum to or from the system from outside the system.
Referring to the diagram:
L
= L′
We define counterclockwise, as viewed from
above, to be the “+” sense of rotation.
I w − I w = (I + I )w′
1
1
2
2
1
2
I w − I w
1
1
2
2
w′ =
I + I
1
2
(4 25
. kg ⋅ m2 ) 97.39 rad /s − 1
( .80 kg ⋅ m2 ) 89 22
.
rad / s
w′ =
4.25 kg ⋅ m2 + 1.80 kg ⋅ m2
rad
w′ = 41.9
(Counterclockwise as viewed from above.)
s
29
Chapter 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics
6 One-Dimensional Motion (Motion Along a Line): Definitions
and Mathematics
A mistake that is often made in linear motion problems involving acceleration, is using
the velocity at the end of a time interval as if it was valid for the entire time interval. The
mistake crops up in constant acceleration problems when folks try to use the definition of
∆x
average velocity v =
in the solution. Unless you are asked specifically about
t
∆
average velocity, you will never need to use this equation to solve a physics problem.
Avoid using this equation—it will only get you into trouble. For constant acceleration
problems, use the set of constant acceleration equations provided you.
Here we consider the motion of a particle along a straight line. The particle can speed up and
slow down and it can move forward or backward but it does not leave the line. While the
discussion is about a particle (a fictitious object which at any instant in time is at a point in space
but has no extent in space—no width, height, length, or diameter) it also applies to a rigid body
that moves along a straight line path without rotating, because in such a case, every particle of
the body undergoes one and the same motion. This means that we can pick one particle on the
body and when we have determined the motion of that particle, we have determined the motion
of the entire rigid body.
So how do we characterize the motion of a particle? Let’s start by defining some variables:
t How much time t has elapsed since some initial time. The initial time is often referred to
as “the start of observations” and even more often assigned the value 0. We will refer to
the amount of time t that has elapsed since time zero as the stopwatch reading. A time
interval ∆t (to be read “delta t”) can then be referred to as the difference between two
stopwatch readings.
x
Where the object is along the straight line. To specify the position of an object on a line,
one has to define a reference position (the start line) and a forward direction. Having
defined a forward direction, the backward direction is understood to be the opposite
direction. It is conventional to use the symbol x to represent the position of a particle.
The values that x can have, have units of length. The SI unit of length is the meter. (SI
stands for “Systeme International,” the international system of units.) The symbol for the
meter is m. The physical quantity x can be positive or negative where it is understood
that a particle which is said to be minus five meters forward of the start line (more
concisely stated as x = −5 m) is actually five meters behind the start line.
30
Chapter 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics
v How fast and which way the particle is going—the velocity1 of the object. Because we
are considering an object that is moving only along a line, the “which way” part is either
forward or backward. Since there are only two choices, we can use an algebraic sign
(“+” or “−”) to characterize the direction of the velocity. By convention, a positive value
of velocity is used for an object that is moving forward, and a negative value is used for
an object that is moving backward. Velocity has both magnitude and direction. The
magnitude of a physical quantity that has direction is how big that quantity is, regardless
of its direction. So the magnitude of the velocity of an object is how fast that object is
going, regardless of which way it is going. Consider an object that has a velocity of
5 m/s. The magnitude of the velocity of that object is 5 m/s. Now consider an object that
has a velocity of −5 m/s. (It is going backward at 5 m/s.) The magnitude of its velocity
is also 5 m/s. Another name for the magnitude of the velocity is the speed. In both of the
cases just considered, the speed of the object is 5 m/s despite the fact that in one case the
velocity was −5 m/s. To understand the “how fast” part, just imagine that the object
whose motion is under study has a built-in speedometer. The magnitude of the velocity,
a.k.a. the speed of the object, is simply the speedometer reading.
a Next we have the question of how fast and which way the velocity of the object is
changing. We call this the acceleration of the object. Instrumentally, the acceleration of
a car is indicated by how fast and which way the tip of the speedometer needle is
moving. In a car, it is determined by how far down the gas pedal is pressed or, in the
case of car that is slowing down, how hard the driver is pressing on the brake pedal. In
the case of an object that is moving along a straight line, if the object has some
acceleration, then the speed of the object is changing.
1 Looking ahead: The velocity of an object in the one-dimensional world of this chapter is, in the three-dimensional
world in which we live, the x component of the velocity of the object. For the case of an object whose velocity has
only an x component, to get the velocity of the object through three dimensional space, you just have to multiply the
x component of the velocity by the unit vector i. In vector notation, saying that an object has a velocity of 5m/si
means the object is moving with a speed of 5m/s in the +x direction and saying that an object has a velocity of
−5m/si means the object is moving with a speed of 5m/s in the −x direction.
31
Chapter 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics
Okay, we’ve got the quantities used to characterize motion. Soon, we’re going to develop some
useful relations between those variables. While we’re doing that, I want you to keep these four
things in mind:
1. We’re talking about an object moving along a line.
2. Being in motion means having your position change with time.
3. You already have an intuitive understanding of what instantaneous velocity is because
you have ridden in a car. You know the difference between going 65 mph and
15 mph and you know very well that you neither have to go 65 miles nor travel for an
hour to be going 65 mph. In fact, it is entirely possible for you to have a speed of
65 mph for just an instant (no time interval at all)—it’s how fast you are going (what
your speedometer reading is) at that instant. To be sure, the speedometer needle may
be just “swinging through” that reading, perhaps because you are in the process of
speeding up to 75 mph from some speed below 65 mph, but the 65 mph speed still
has meaning and still applies to that instant when the speedometer reading is 65 mph.
Take this speed concept with which you are so familiar, tack on some directional
information, which for motion on a line just means, specify “forward” or “backward;
and you have what is known as the instantaneous velocity of the object whose motion
is under consideration.
A lot of people say that the speed of an object is how far that object travels in a
certain amount of time. No! That’s a distance. Speed is a rate. Speed is never how
far, it is how fast. So if you want to relate it to a distance you might say something
like, “Speed is what you multiply by a certain amount of time to determine how far
an object would go in that amount of time if the speed stayed the same for that entire
amount of time.” For instance, for a car with a speed of 25 mph, you could say that
25 mph is what you multiply by an hour to determine how far that car would go in an
hour if it maintained a constant speed of 25 mph for the entire hour. But why explain
it in terms of position? It is a rate. It is how fast the position of the object is
changing. If you are standing on a street corner and a car passes you going 35 mph, I
bet that if I asked you to estimate the speed of the car that you would get it right
within 5 mph one way or the other. But if we were looking over a landscape on a day
with unlimited visibility and I asked you to judge the distance to a mountain that was
35 miles away just by looking at it, I think the odds would be very much against you
getting it right to within 5 miles. In a case like that, you have a better feel for “how
fast” than you do for “how far.” So why define speed in terms of distance when you
can just say that the speed of an object is how fast it is going?
4. You already have an intuitive understanding of what acceleration is. You have been
in a car when it was speeding up. You know what it feels like to speed up gradually
(small acceleration) and you know what it feels like to speed up rapidly (big, “pedal-
to-the-metal,” acceleration).
32
Chapter 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics
All right, here comes the analysis. We have a start line (x=0) and a positive direction (meaning
the other way is the negative direction).
x
0
Consider a moving particle that is at position x when the clock reads t and at position x when
1
1
2
the clock reads t .
2
x
x
x
0
1
2
The displacement of the particle is, by definition, the change in position ∆x = x − x of the
2
1
particle. The average velocity v is, by definition,
∆x
v =
(6-1)
∆t
where ∆t = t − t is the change in clock reading. Now the average velocity is not something
2
1
that one would expect you to have an intuitive understanding for, as you do in the case of
instantaneous velocity. The average velocity is not something that you can read off the
speedometer, and frankly, it’s typically not as interesting as the actual (instantaneous) velocity,
but it is easy to calculate and we can assign a meaning to it (albeit a hypothetical meaning). It is
the constant velocity at which the particle would have to travel if it was to undergo the same
displacement ∆x = x − x in the same time ∆t = t − t at constant velocity. The importance of
2
1
2
1
the average velocity in this discussion lies in the fact that it facilitates the calculation of the
instantaneous velocity.
Calculating the instantaneous velocity in the case of a constant velocity is easy. Looking at what
we mean by average velocity, it is obvious that if the velocity isn’t changing, the instantaneous
velocity is the average velocity. So, in the case of a constant velocity, to calculate the
instantaneous velocity, all we have to do is calculate the average velocity, using any
displacement with its corresponding time interval, that we want. Suppose we have
position vs. time data on, for instance, a car traveling a straight path at 24 m/s.
33
Chapter 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics
Here’s some idealized fictitious data for just such a case.
Data Reading
Time [seconds]
Position [meters]
Number
0
0
0
1
0.100
2.30
2
1.00
23.0
3
10.0
230
4
100.0
2300
Remember, the speedometer of the car is always reading 24 m/s. (It should be clear that the car
was already moving as it crossed the start line at time zero—think of time zero as the instant a
stopwatch was started and the times in the table as stopwatch readings.) The position is the
distance forward of the start line.
Note that for this special case of constant velocity, you get the same average velocity, the known
value of constant speed, no matter what time interval you choose. For instance, if you choose the
time interval from 1.00 seconds to 10.0 seconds:
∆x
v =
(Average velocity.)
t
∆
x − x
3
2
v =
t − t
3
2
m
230
− 23. m
0
v =
10. s
0 − 1.
s
00
m
v = 23 0
.
s
and if you choose the time interval 0.100 seconds to 100.0 seconds:
∆x
v =
t
∆
x − x
4
1
v =
t − t
4
1
m
2300
− 2.
m
30
v =
100. s
0 − 0.
s
100
m
v = 23 0
.
s
34
Chapter 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics
The points that need emphasizing here are that, if the velocity is constant then the calculation of
the average speed yields the instantaneous speed (the speedometer reading, the speed we have an
intuitive feel for), and when the velocity is constant, it doesn’t matter what time interval you use
to calculate the average velocity; in particular, a small time interval works just as well as a big
time interval.
So how do we calculate the instantaneous velocity of an object at some instant when the
instantaneous velocity is continually changing? Let’s consider a case in which the velocity is
continually increasing. Here we show some idealized fictitious data (consistent with the way an
object really moves) for just such a case.
Velocity (This is what
Time since object was
Position (distance
we are trying to
Data Reading Number
at start line.
ahead of start line)
calculate. Here are the
[s]
[m]
correct answers.) [m/s]
0
0
0
10
1
1
14
18
2
1.01
14.1804
18.08
3
1.1
15.84
18.8
4
2
36
26
5
5
150
50
What I want to do with this fictitious data is to calculate an average velocity during a time
interval that begins with t = 1 s and compare the result with the actual velocity at time t = 1 s.
The plan is to do this repeatedly, with each time interval used being smaller than the previous
one.
Average velocity from t = 1 s to t = 5 s:
∆x
v =
t
∆
x − x
5
1
v =
t − t
5
1
m
150
−
m
14
v =
s
5 − s
1
m
v = 34
s
Note that this value is quite a bit larger than the correct value of the instantaneous velocity at
t = 1 s (namely 18 m/s). It does fall between the instantaneous velocity of 18 m/s at t = 1 s and
the instantaneous velocity of 50 m/s at t = 5 seconds. That makes sense since, during the time
interval, the velocity takes on various values which for 1 s < t < 5 s are all greater than 18 m/s but
less than 50 m/s.
35
Chapter 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics
For the next two time intervals in decreasing time interval order (calculations not shown):
Average velocity from t = 1 to t = 2 s: 22 m/s
Average velocity from t = 1 to t = 1.1 s: 18.4 m/s
And for the last time interval, we do show the calculation:
∆x
v =
t
∆
x − x
2
1
v =
t − t
2
1
14 1804
.
m − 14 m
v =
1.
s
01 − s
1
m
v = 18.04
s
Here I copy all the results so that you can see the trend:
Average velocity from t = 1 to t = 5 s: 34 m/s
Average velocity from t = 1 to t = 2 s: 22 m/s
Average velocity from t = 1 to t = 1.1 s: 18.4 m/s
Average velocity from t = 1 to t = 1.01 s: 18.04 m/s
Every answer is bigger than the instantaneous velocity at t = 1s (namely 18 m/s). Why?
Because the distance traveled in the time interval under consideration is greater than it would
have been if the object moved with a constant velocity of 18 m/s. Why? Because the object is
speeding up, so, for most of the time interval the object is moving faster than 18 m/s, so, the
average value during the time interval must be greater than 18 m/s. But notice that as the time
interval (that starts at t = 1s) gets smaller and smaller, the average velocity over the time interval
gets closer and closer to the actual instantaneous velocity at t = 1s. By induction, we conclude
that if we were to use even smaller time intervals, as the time interval we chose to use was made
smaller and smaller, the average velocity over that tiny time interval would get closer and closer
to the instantaneous velocity, so that when the time interval got to be so small as to be virtually
indistinguishable from zero, the value of the average velocity would get to be indistinguishable
from the value of the instantaneous velocity. We write that:
∆x
v = lim
∆ →0 ∆
t
t
(Note the absence of the bar over the v. This v
is the instantaneous velocity.) This expression
for v
is, by definition, the derivative of x with respect to t. The derivative of x with respect to t is
d x
written as
which means that
dt
36
Chapter 6 One-Dimensional Motion (Motion Along a Line): Definitions and Mathematics
d x
v =
(6-2)
dt
d x
Note that, as mentioned,
is the derivative of x with respect to t. It is not some variable d
dt
times x all divided by d times t. It is to be read “dee ex by dee tee” or, better yet, “the derivative
of x with respect to t.” Conceptually what it means is, starting at that value of time t at which
you wish to find the velocity, let t change by a very small amount. Find the (also very small)
amount by which x changes as a result of the change in t and divide the tiny change in x by the
<