Download Notes for Class Meeting 3: Consequences of Newton`s Laws

Notes for Class Meeting 3: Consequences of Newton’s Laws
Circular Motion
As mentioned the previous notes, we will need only two simple examples of motion
to discuss the topics we wish to cover in this course. In the previous note, we discussed
linear motion with constant acceleration. Here we will discuss circular motion with
constant speed. In the previous notes, we defined acceleration as the change in velocity
with time,

 Δv
a=
.
(2.3)
Δt
We also noted that velocity is a vector quantity. It has both a magnitude and a direction.
In circular motion with constant speed, the magnitude of the velocity does not change,
but the direction does. Thus there is acceleration according to our definition, and by
Newton’s second law, a force is needed to coerce the object to follow a circular path
rather than the straight path that it would follow in the absence of a force. The magnitude
of the acceleration is given by the formula
v2
a= ,
(3.1)
r
where v is the speed and r is the radius of the circle. The direction of the acceleration is
toward the center of the circle (hence the name “centripetal acceleration”). The
corresponding centripetal force is given by multiplying the acceleration by the mass, thus
mv 2
F=
,
(3.2)
r
with, of course, the force pointing toward the center of the circle.
It is important for you to understand this concept, but it is not crucial for you to
understand its derivation. However, the derivation is relatively simple (it only uses some
high school geometry and does not need calculus, as stated by March), so I have attached
it as an appendix for those of you who are interested in seeing it.
Dimensional Analysis
As an aside, this is a good place to discuss dimensional analysis. When a physicist
does not know how to calculate some quantity exactly, he or she can still get an estimate
by what is known as dimensional analysis. The technique is to list all of the variables
that can contribute to a result and then try to find the often unique combination of them
that can give the right dimensions for the quantity in question. This will then give the
correct formula up to a numerical constant, which is often, but not always, of order one.
Let’s try this for the centripetal force. From the definition of force (i.e., Newton’s 2nd
law) the units of force are mass (m) times acceleration ( d / t 2 ) which equals md / t 2 ,
where m stands for mass, d stands for distance, and t stands for time. The only possible
variables that could contribute to the force are the mass (m), the speed (d/t), and the
radius of the circle (d). If you try to combine these variables, you will find that there is
only one combination that can give the units of force, and that is the combination given
by the right side of Eq. 3.2. There could be a numerical coefficient and you have to do
the full derivation to determine what it is. In this case, it is simply one.
Galilean Relativity
A statement of Galilean relativity is “If Newton’s laws are valid in one frame, they
are valid in every frame moving with constant velocity with respect to that frame.” It is

easy to show this. To calculate a velocity in the second frame, add v0 to every velocity in
  

the first frame. For a change in velocity, let Δv = v f − vi , where v f is the velocity at the

end of the change and vi is the velocity at the start of the change. With this, the
demonstration that Newton’s laws have Galilean relativity follows.
• Newton’s 1st law: This is obviously true.
• Newton’s 2nd law:
 
 
 
 


Δ(v f − vi )
Δ[ v f + v0 − (vi + v0 )]
Δ(v f − vi )
Δv

F = ma = m
=m
→m
=m
Δt
Δt
Δt
Δt
• Newton’s 3rd law:


Δ(m1v1 + m2 v2 + …) = 0 ⇒




Δ[m1 (v1 f − v1i ) + m2 (v2 f − v2i ) + …) = 0 →

  

 

Δ[m1 (v1 f + v0 − v1i − v0 ) + m2 (v2 f + v0 − v2i − v0 ) + …) = 0 ⇒


Δ(m1v1 + m2 v2 + …) = 0
(3.2)
(3.3)
This means that there is no experiment you can do to determine which frame is at rest
and which frame is moving. Such frames are called inertial frames. This is not true of an
inertial frame and an accelerating frame, because in the latter case one feels a force.
Galilean relativity will be one of the two postulates from which Einstein will derive
the special theory of relativity, except that he will replace “Newton’s laws” with the more
general “the laws of physics.”
Time Symmetry
Newton’s laws and (with one minor exception that we will mention at the end of the
course) all of the known laws of physics are symmetric with respect to the direction of
time. If we reverse the direction of time, all velocities change sign:




Δd
Δd
Δd
 Δd

(3.4)
v=
=
→ time reversal →
=−
= −v
Δt t f − ti
−t f + ti
Δt
Accelerations, on the other hand, do not change sign, since time enters them twice. It is
thus easy to see that Newton’s laws are valid for either direction of time. If one could
reverse time at some instant, all motions would proceed backwards and the past would be
reconstructed.
This raises the question of what determines the direction of time, or to put it more
graphically, why we remember the past, but not the future. We will discuss this issue in a
couple of weeks.
Appendix: Derivation of Eq. 3.1
vf
Δv
r
ΔL
θ
r
vi
vf
θ
vi
Since both the spatial triangle and velocity triangles are isosceles triangles with the
same included angle, they are similar triangles. Therefore,
ΔL Δv
.
(3.5)
=
r
v
Dividing both sides of Eq. 3.5 by Δt ,
1 ΔL 1 Δv
1
1
v2
=
⇒
v= a ⇒ a= .
(3.6)
r Δt v Δt
r
v
r
As Δt → 0 , θ → 0 and Δv becomes perpendicular to vi and v f since the sum of the
angles of a triangle equal 180o. Since vi and v f are perpendicular to the radii, Δv , and
thus a points to the center of the circle.