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Physics 312: Lecture 1
Cartesian and Polar Coordinates
Cartesian Coordinates

You know the Cartesian coordinate system as the xˆ , yˆ , zˆ coordinate system,
so that forces can be written in terms of their components,
and the position vector is
F  Fx xˆ  Fy yˆ  Fz zˆ
r  x xˆ  y yˆ  z zˆ

The acceleration is found by differentiating the position vector twice
r  x xˆ  y yˆ  z zˆ
So the vector equation of motion F  ma  mr becomes
Fx xˆ  Fy yˆ  Fz zˆ  mx xˆ  my yˆ  mz zˆ
 Whenever you see a vector equation like this, you should consider it as
three simultaneous equations, one for each component:  F  mx
x

 Fy  my
 F  mz
 z

September 2, 2010
Example 1.1





Here is a well-known example to remind you how to solve force problems—
a block m sliding from rest down an incline at angle q, with coefficient of
kinetic friction m, subject to gravity.
As always, the first step is to choose a coordinate
y
N
f
system. Let’s choose x down the incline, and
y perpendicular, with x = 0 at t = 0.
As you should recall, the downward weight of the mass
q
x
on the plane produces a corresponding normal force
mg
perpendicular to the plane.
Because there is friction in the problem, the motion (obviously down the
plane) produces an opposing force f with magnitude mN up the plane.
The x and y components of the equation of motion are then:
Fx  mg sin q  mN  mx
Fy  N  mg cos q  0
which leads to the equation
mx  mg sin q  mmg cos q
September 2, 2010
Example 1.1, cont’d
Eliminating the common term m, we have this equation:
x  g (sin q  m cos q )
 As usual, to get the position we integrate this equation twice to get
x  g (sin q  m cos q )t
(remember that the block starts from rest, so the initial velocity is zero
and the constant of integration is therefore zero). Finally
1
x  g (sin q  m cos q )t 2
2
Again, the constant of integration is zero because we chose our axes with
x = 0 at t = 0.
 Notice that in general we have two unknown “constants of integration” for
this 2nd-order differential equation, and they must be specified by the
“initial conditions.” This is a completely general situation. In this case,
the constants are zero because the initial conditions are x = 0 and v = 0 at t
= 0. The general solution (for x = x0 and v = v0 at t = 0) is

x  x0  v0t 
1
g (sin q  m cos q )t 2
2
September 2, 2010
Example: Problem 1.36

Statement of the Problem, part (a):


A plane, which is flying horizontally at a constant speed v0, and at a height h
above the sea, must drop a bundle of supplies to a castaway on a small raft.
(a) Write down Newton’s second law for the bundle as it falls from the plane,
assuming you can neglect air resistance. Solve your equation to give the
bundle’s position in flight as a function of time t.
Solution to part (a):


mx  0
my   mg

v0
Choose a coordinate system (x horizontal, y positive upward)
Write down Newton’s second law for x and y
x
Integrate both twice,
x  v0 ;
y   gt ;
x  v0t
y  h
h
y
mg
water
initial x(0)=0, initial vx(0)=v0
1 2
gt
2
initial y(0)=h, initial vy(0)=0
September 2, 2010
Example: Problem 1.36, cont’d

Statement of the Problem, part (b):


How far before the raft (measured horizontally) must the pilot drop the bundle if it
is to hit the raft? What is the distance if v0 = 50 m/s, h = 100 m, and g  10 m/s2?
Solution to part (b):

This may take a little thought, but the raft is at position y = 0, so one solution is to
find out when the bundle reaches y = 0 and see how far the bundle moves in x
during that time. That is the distance before the raft when the bundle should be
dropped.
y  h
h
x  v0t  v0
1 2
gt  0
2
1 2
2h
gt  t 
2
g
2h
200 m
 50 m/s
 224 m
g
10 m/s 2
Notice that we are NOT just
plugging in to formulas. We
are deriving the formulas.
September 2, 2010
1.7 Two-Dimensional Polar
Coordinates
Although Newton’s 2nd law takes a simple form in Cartesian coordinates,
there are many circumstances where the symmetry of the problem lends
itself to other coordinates.
 To illustrate this, let’s take a look at 2-d polar coordinates. You should
already have some familiarity with these, but we will see that certain
complexities arise that require a bit of care.
 First of all you can go back and forth between Cartesian and polar
coordinates by the familiar
φ̂

x  r cos f  
 r  x2  y2

y  r sin f  
f  arctan( y / x)

r̂
r=|r|
f
We now wish to introduce unit vectors for these new polar
x
ˆ . Remember,
coordinates, (r, f), which we will write rˆ , φ
these have to be 1 unit long, and point in the direction of r and f.
September 2, 2010
y
Polar Coordinate Unit Vectors


One way to construct a unit vector is to take any vector r and divide by its
length |r|. Clearly, such a unit vector is in the direction of r but has unit
r
length:
rˆ 
r
There is a major difference between the behavior of the cartesian unit
vectors and the corresponding ones for polar coordinates.
ŷ
φ̂
ŷ
x̂
ŷ
r̂
φ̂
φ̂ r̂
x̂
x̂
Cartesian unit vectors
are constant
r̂
Polar coordinate unit vectors
change (direction) with time
September 2, 2010
Polar Coordinate Unit Vectors




Since r̂ and φ̂ are perpendicular vectors in our two-dimensional space, any
vector can be split into components in terms of them. For instance, the
force F can be written: F  Fr rˆ  Ff φˆ
Imagine twirling a stone at the end of a string. Then the force Fr on the
stone is just the tension in the string, and Ff might be the force of air
resistance as the stone flies through the air.
The position vector is then particularly simple: r  rrˆ
But to solve Newton’s 2nd law, F  mr, we need the second derivative of r.
Let’s just take the derivative using the product rule:
drˆ
r  rrˆ  r
dt
where we keep the second term because now the unit vector r̂ is not
constant.
 To see what the derivative of the unit vector is, let’s look at how r̂ changes.
September 2, 2010
dr̂
Derivatives of Polar Unit Vectors:
dt

As the coordinate r changes from time t1 to time t2= t1 + Dt, the unit vector r̂
changes by:
φ̂
r̂ Df Dr̂
φ̂ r̂
r̂
Recall: arc length = rq
In this case, q = Df and r = rˆ  1
Drˆ  Df φˆ
We can rewrite Df  f Dt , hence Drˆ  f Dt φˆ or, after taking the limit as Dt
approaches zero,
drˆ 
 f φˆ
dt
d
rrˆ  rrˆ  rf φˆ , so the components of v
 Thus, our first derivative is v  r 
dt
are v  r; v  rf  r

r
f
September 2, 2010
dφ̂
Derivatives of Polar Unit Vectors:
dt
Now we need to take another derivative to get
d
d
a  r  r 
rrˆ  rf φˆ
dt
dt
 This is going to involve a time derivative of the φ̂ unit vector, for which we use
much the same procedure as before.
Dφ̂
 Since the φ̂ unit vector is perpendicular to the r̂ unit vector, we
Df φ̂
have the same geometry as before, except rotated 90 degrees.
The change Dφ̂ is now in the  r̂ direction, and its length is again
φ̂
Df  f Dt, so finally we have: dφˆ  f rˆ .
dt
φ̂ r̂
 All that remains is to do a careful, term-by-term derivative to get
r̂
d
drˆ  
dφˆ 

a
rrˆ  rf φˆ   rrˆ  r    rf  rf φˆ  rf 
dt
dt  
dt 

 And then plug in our new-found expressions for the derivatives of the unit vectors.







September 2, 2010
Acceleration in Polar Coordinates

Starting from the expression on the previous slide:
d
drˆ  
dφˆ 

a
rrˆ  rf φˆ   rrˆ  r    rf  rf φˆ  rf 
dt
dt  
dt 

ˆ
and our expressions for the unit vector derivatives: dr  f φˆ
dt





 
 


dφˆ
 f rˆ
dt
we have finally: a  rrˆ  rfφˆ  rf  rf φˆ  rf2 rˆ
 r  rf2 rˆ  rf  2rf φˆ


Admittedly this looks a little complicated, so let’s look at some special cases:
 r = constant (i.e. stone on a string):
a  rf2 rˆ  rfφˆ  r 2rˆ  r φˆ
Here, ar  r 2  v 2 / r is the centripetal acceleration and af  r is any

angular acceleration I might impose in swinging the stone.
When r is not constant, all terms are necessary. The acceleration term involving rr̂
is probably not surprising, but the term 2rf φ̂ is much harder to understand. This
is the Coriolis Force, which will be introduced in Chapter 9.
September 2, 2010
Newton’s 2nd Law in Polar
Coordinates

Now that we have the acceleration, we are ready to write down Newton’s 2nd
law in polar coordinates.
 Fr  m r  rf 2
F  ma  
 Ff  m rf  2rf




These expressions are complicated and hard to remember. Fortunately, after
we introduce the Lagrangian approach to solving problems in Chapter 7,
these expressions will automatically appear without having to remember
them. Before then, you can refer to these equations when we need them.
 You may think you would rather avoid these nasty expressions and just do
problems in rectangular coordinates, so we should do a problem that
illustrates the power of polar coordinates.

September 2, 2010
Example 1.2: An Oscillating
Skateboard
 F  mr  rf 
2
We start with Newton’s 2nd law in polar coordinates. F  ma
 Statement of Problem:





 Ff  m rf  2rf
A “half-pipe” at a skateboard park consists of a concrete trough with a
semicircular cross section of radius R = 5 m, as shown in the figure. I hold a
frictionless skateboard on the side of the trough pointing down toward the bottom
and release it. Discuss the subsequent motion using Newton’s second law. In
particular, if I release the board just a short way from the bottom, how long will it
take to come back to the point of release?
Solution:


r
Because the skateboard is constrained to move in a circular
motion, it is easiest to work in polar coordinates. The coordinate
r = R, and the problem becomes one-dimensional—only f varies.
Newton’s law takes on the simple form:
Fr  mRf2
f N
Ff  mRf
mg
These state how the skateboard accelerates under forces Fr and Ff. The radial
forces are Fr  mg cos f  N and the azimuthal force is just Ff  mg sin f .
September 2, 2010


Example 1.2: An Oscillating
Skateboard,
cont’d
Solution, cont’d:

Equating these two, we have
mg cos f  N  mRf2

 mgsin f  mRf
The first can be solved to give the normal force as a function of time, which may
be interesting in some problems, but is not needed in this problem. Therefore,
we can focus on the second equation, which on rearrangement becomes: 
g
f 


R
This is a 2nd-order differential equation whose solution turns out to be rather
complex, but by looking at its behavior we can learn quite a bit about the motion.
If we place the skateboard at f = 0, for example, then f  0  f  const . In
particular, if we place it there at rest, so f  0 , then the skateboard will remain
there, i.e. f = 0 is a point of equilibrium. By virtue of the minus-sign, if we place
the skateboard to the right of the bottom, then it accelerates to the left, and if we
place the skateboard to the left, then it accelerates to the right. This tells us that
f = 0 is a stable equilibrium.
We can make the problem easier to solve if we consider only small deviations
from zero, so that sin f  f . This yields the equation f   g f
R
for a harmonic oscillator (see text for more).
September 2, 2010
sin f
Example: Problem 1.47

Statement of Problem:

Let the position of a point P in three dimensions be given by the vector r = (x, y, z)
in rectangular (Cartesian) coordinates. The same position can be specified by
cylindrical polar coordinates, r = (r, f, z). (a) Make a sketch to illustrate the three
cylindrical coordinates. Give expressions for r, f, z in terms of x, y, z.
r ẑ φ̂
r̂
r
f
z
x = r cosf
y = r sinf
z=z

ˆ , φˆ , zˆ and write
(b) Describe the three unit vectors ρ
the expansion of the position vector r in terms of
these unit vectors.
r  r ρˆ  z zˆ
September 2, 2010