Physics 312: Lecture 1Cartesian and Polar Coordinates

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:

zyx ˆ ,ˆ ,ˆ

zyxr ˆ ˆ ˆ zyx

zyxF ˆ ˆ ˆ zyx FFF

zyxr ˆ ˆ ˆ zyx

raF mm zyxzyx ˆ ˆ ˆ ˆ ˆ ˆ zmymxmFFF zyx

zmF

ymF

xmF

z

y

x

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 masson 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

Nf

0cos

sin

mgNF

xmNmgF

y

x

cossin mgmgxm

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 gx

tgx )cos(sin

2)cos(sin2

1tgx

200 )cos(sin

2

1tgtvxx

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,tvxvx 00 ;

mgym

xm

0

x

y

v0

mg

water

h

2

2

1 ; gthygty

initial x(0)=0, initial vx(0)=v0

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.

m 224m/s 10

m 200m/s 50

2200

g

hvtvx

g

htgth

gthy

2

2

1

02

1

2

2

Notice that we are NOT justplugging in to formulas. Weare deriving the formulas.

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 polarcoordinates, (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 22

xy

yxrry

rx

x

r=|r|

y

φr ˆ ,ˆ

rφ

September 2, 2010

Polar Coordinate Unit Vectors

x

y

x

yx

y

Cartesian unit vectors are constant

rφ

Polar coordinate unit vectors change (direction) with time

rφ

r

φ

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

rr ˆ

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.

φrF ˆˆ FFr

r φ

rr ˆrrF m

dt

drr

rrr

ˆˆ

r

r

September 2, 2010

Derivatives of Polar Unit Vectors:

rφ

r

φ

As the coordinate r changes from time t1 to time t2= t1 + t, the unit vector changes by:

φr ˆˆ

r

r r r

Recall: arc length = rIn this case, = and r = 1ˆ r

φr ˆˆ 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

φr

ˆˆ

dt

d

φrrrv ˆˆˆ rrrdt

d

dt

dr

rrvrvr ;

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.

φ

t rφ

ˆˆ

dt

d

φrrra ˆˆ rrdt

d

dt

d

dt

dφ

φ rφ

φ

φ r

dt

drrr

dt

drrrr

dt

d φφ

rrφra

ˆˆ

ˆˆˆˆ

rφ

r

φ

φ

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.

rφ

ˆˆ

dt

d

dt

drrr

dt

drrrr

dt

d φφ

rrφra

ˆˆ

ˆˆˆˆ

φr

ˆˆ

dt

d

φr

rφφra

ˆ2ˆ

ˆˆˆˆ2

2

rrrr

rrrrr

φrφra ˆˆˆˆ 22 rrrr rvrar /22 ra

rrφ2 r

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.

rrmF

rrmFm r

2

2

aF

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 coordinater = 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 .

rrmF

rrmFm r

2

2

aF

N

mg

mRFmRFr 2

NmgFr cos sinmgF

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 equationfor a harmonic oscillator (see text for more).

mRmgmRNmg sin cos 2

sinR

g

const 00

sin R

g

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 = cosy = sinz = z

φz

(b) Describe the three unit vectors and write the expansion of the position vector r in terms of these unit vectors.

zφρ ˆ,ˆ,ˆ

zρr ˆˆ z