2. September 2, 2010
Cartesian Coordinates
You know the Cartesian coordinate system as the coordinate
system, so that forces can be written in terms of their components,
and the position vector is
The acceleration is found by differentiating the position vector twice
So the vector equation of motion becomes
Whenever you see a vector equation like this, you should consider it as
three simultaneous equations, one for each component:
z
y
x ˆ
,
ˆ
,
ˆ
z
y
x
r ˆ
ˆ
ˆ z
y
x
z
y
x
F ˆ
ˆ
ˆ z
y
x F
F
F
z
y
x
r ˆ
ˆ
ˆ z
y
x
r
a
F
m
m
z
y
x
z
y
x ˆ
ˆ
ˆ
ˆ
ˆ
ˆ z
m
y
m
x
m
F
F
F z
y
x
z
m
F
y
m
F
x
m
F
z
y
x
3. 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 , with
coefficient of kinetic friction , subject to gravity.
As always, the first step is to choose a coordinate
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
on the plane produces a corresponding normal force
perpendicular to the plane.
Because there is friction in the problem, the motion (obviously down
the plane) produces an opposing force f with magnitude N up the
plane.
The x and y components of the equation of motion are then:
which leads to the equation
x
y
mg
N
f
0
cos
sin
mg
N
F
x
m
N
mg
F
y
x
cos
sin mg
mg
x
m
4. September 2, 2010
Example 1.1, cont’d
Eliminating the common term m, we have this equation:
As usual, to get the position we integrate this equation twice to get
(remember that the block starts from rest, so the initial velocity is
zero and the constant of integration is therefore zero). Finally
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
)
cos
(sin
g
x
t
g
x )
cos
(sin
2
)
cos
(sin
2
1
t
g
x
2
0
0 )
cos
(sin
2
1
t
g
t
v
x
x
5. 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):
Choose a coordinate system (x horizontal, y positive upward)
Write down Newton’s second law for x and y
Integrate both twice,
t
v
x
v
x 0
0;
mg
y
m
x
m
0
x
y
v0
mg
water
h
2
2
1
; gt
h
y
gt
y
initial x(0)=0, initial vx(0)=v0
initial y(0)=h, initial vy(0)=0
6. 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.
m
224
m/s
10
m
200
m/s
50
2
2
0
0
g
h
v
t
v
x
g
h
t
gt
h
gt
h
y
2
2
1
0
2
1
2
2
Notice that we are NOT just
plugging in to formulas. We
are deriving the formulas.
7. September 2, 2010
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
We now wish to introduce unit vectors for these new polar
coordinates, (r, ), which we will write . Remember,
these have to be 1 unit long, and point in the direction of r and .
1.7 Two-Dimensional Polar
Coordinates
)
/
arctan(
sin
cos 2
2
x
y
y
x
r
r
y
r
x
x
r=|r|
y
φ
r ˆ
,
ˆ
r̂
φ̂
8. September 2, 2010
Polar Coordinate Unit Vectors
x̂
ŷ
x̂
ŷ
x̂
ŷ
Cartesian unit vectors
are constant
r̂
φ̂
Polar coordinate unit vectors
change (direction) with time
r̂
φ̂
r̂
φ̂
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 length:
There is a major difference between the behavior of the cartesian unit
vectors and the corresponding ones for polar coordinates.
r
r
r
ˆ
9. September 2, 2010
Polar Coordinate Unit Vectors
Since and 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:
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 F might be the force of air
resistance as the stone flies through the air.
The position vector is then particularly simple:
But to solve Newton’s 2nd
law, , we need the second derivative of r.
Let’s just take the derivative using the product rule:
where we keep the second term because now the unit vector is not
constant.
To see what the derivative of the unit vector is, let’s look at how
changes.
φ
r
F ˆ
ˆ
F
Fr
r̂ φ̂
r
r ˆ
r
r
F
m
dt
d
r
r
r
r
r
ˆ
ˆ
r̂
r̂
10. September 2, 2010
Derivatives of Polar Unit Vectors:
r̂
φ̂
r̂
φ̂
As the coordinate r changes from time t1 to time t2= t1 + t, the unit
vector changes by:
φ
r ˆ
ˆ
r̂
r̂ r̂ r̂
Recall: arc length = r
In this case, = and r = 1
ˆ
r
φ
r ˆ
ˆ t
We can rewrite , hence or, after taking the limit as t
approaches zero,
Thus, our first derivative is , so the components of v
are
t
φ
r
ˆ
ˆ
dt
d
φ
r
r
r
v ˆ
ˆ
ˆ
r
r
r
dt
d
dt
dr̂
r
r
v
r
vr
;
11. September 2, 2010
Derivatives of Polar Unit Vectors:
Now we need to take another derivative to get
This is going to involve a time derivative of the unit vector, for which we use
much the same procedure as before.
Since the unit vector is perpendicular to the unit vector, we
have the same geometry as before, except rotated 90 degrees.
The change is now in the direction, and its length is again
, so finally we have: .
All that remains is to do a careful, term-by-term derivative to get
And then plug in our new-found expressions for the derivatives of the unit
vectors.
φ̂
t
r
φ
ˆ
ˆ
dt
d
φ
r
r
r
a ˆ
ˆ
r
r
dt
d
dt
d
dt
dφ̂
φ̂ r̂
φ̂
φ̂
φ̂
r̂
dt
d
r
r
r
dt
d
r
r
r
r
dt
d φ
φ
r
r
φ
r
a
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
r̂
φ̂
r̂
φ̂
φ̂
12. September 2, 2010
Acceleration in Polar Coordinates
Starting from the expression on the previous slide:
and our expressions for the unit vector derivatives:
we have finally:
Admittedly this looks a little complicated, so let’s look at some special
cases:
r = constant (i.e. stone on a string):
Here, is the centripetal acceleration and is any angular
acceleration I might impose in swinging the stone.
When r is not constant, all terms are necessary. The acceleration term
involving is probably not surprising, but the term is much harder to
understand. This is the Coriolis Force, which will be introduced in Chapter 9.
r
φ
ˆ
ˆ
dt
d
dt
d
r
r
r
dt
d
r
r
r
r
dt
d φ
φ
r
r
φ
r
a
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
φ
r
ˆ
ˆ
dt
d
φ
r
r
φ
φ
r
a
ˆ
2
ˆ
ˆ
ˆ
ˆ
ˆ
2
2
r
r
r
r
r
r
r
r
r
φ
r
φ
r
a ˆ
ˆ
ˆ
ˆ 2
2
r
r
r
r
r
v
r
ar /
2
2
r
a
r̂
r
φ̂
2
r
13. 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.
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.
r
r
m
F
r
r
m
F
m r
2
2
a
F
14. September 2, 2010
Example 1.2: An Oscillating
Skateboard
We start with Newton’s 2nd
law in polar coordinates.
Statement of Problem:
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:
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 varies.
Newton’s law takes on the simple form:
These state how the skateboard accelerates under forces Fr and F. The
radial forces are and the azimuthal force is just .
r
r
m
F
r
r
m
F
m r
2
2
a
F
N
mg
mR
F
mR
Fr
2
N
mg
Fr
cos
sin
mg
F
15. September 2, 2010
Example 1.2: An Oscillating
Skateboard, cont’d
Solution, cont’d:
Equating these two, we have
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:
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 = 0, for example, then .
In particular, if we place it there at rest, so , then the skateboard will
remain there, i.e. = 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 = 0 is a stable equilibrium.
We can make the problem easier to solve if we consider only small deviations
from zero, so that . This yields the equation
for a harmonic oscillator (see text for more).
mR
mg
mR
N
mg
sin
cos 2
sin
R
g
const
0
0
sin
R
g
16. September 2, 2010
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 = (, , z). (a) Make a sketch to
illustrate the three cylindrical coordinates. Give expressions for , , z in
terms of x, y, z.
r
z
x = cos
y = sin
z = z
̂
φ̂
ẑ
(b) Describe the three unit vectors and write
the expansion of the position vector r in terms of
these unit vectors.
z
φ
ρ ˆ
,
ˆ
,
ˆ
z
ρ
r ˆ
ˆ z
