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Chapter 2Particles and Cylindrical Polar Coordinates

TOPICS

Here, we discuss the cylindrical polar coordinate system and how it is used in par-ticle mechanics. This coordinate system and its associated basis vectors {er,e ,Ez}are vital to understand and practice.

It is a mistake to waste your time memorizing formulae here. Instead, focus onunderstanding the material. You will repeat it countless times which will naturallydevelop the ability to derive the results from scratch.

2.1 The Cylindrical Polar Coordinate System

Consider the pendulum system shown in Figure 2.1. Here, a particle of mass m isattached by an inextensible string of length L to a fixed point O. Assuming that thestring remains taut, then the distance from O to the particle remains constant: ||r||=

Ex

Ey

Og

mg

Tension in string

t

Fig. 2.1 An example of the motion of a pendulum. The behavior of the tension T in the stringduring this motion is also shown.

O.M. OReilly, Engineering Dynamics: A Primer, DOI 10.1007/978-1-4419-6360-4 2, 17c Springer Science+Business Media, LLC 2010

18 2 Particles and Cylindrical Polar Coordinates

Ex

Ey

ere

Fig. 2.2 The unit vectors er and e .

L. This system is a prototypical example of a situation where a polar coordinatesystem can be effectively used.

To define a cylindrical polar coordinate system {r, ,z}, we start with a Cartesiancoordinate system {x,y,z} for the three-dimensional space E3. Using these coordi-nates, we define r, , and z as

r =

x2 + y2, = tan1(y

x

), z = z.

The coordinate r 0. Apart from the points {x,y,z} = {0,0,z} , given r, , and z,we can uniquely determine x, y, and z:

x = r cos( ), y = r sin( ), z = z.

Here, is taken to be positive in the counterclockwise direction.If we now consider the position vector r of a point in this space, we have, as

always,r = xEx + yEy + zEz.

We can write this position vector using cylindrical polar coordinates by substitutingfor x and y in terms of r and :

r = r cos( )Ex + r sin( )Ey + zEz.

Before we use this representation to establish expressions for the velocity and ac-celeration vectors, it is convenient to introduce the unit vectors er and e :

ereEz

=

cos( ) sin( ) 0sin( ) cos( ) 0

0 0 1

ExEyEz

.

Two of these vectors are shown in Figure 2.2.

2.2 Velocity and Acceleration Vectors 19

Note that {er,e ,Ez} are orthonormal and form a right-handed basis1 for E3. Youshould also be able to see that

ExEyEz

=

cos( ) sin( ) 0sin( ) cos( ) 0

0 0 1

ereEz

.

Because er = er( ) and e = e ( ), these vectors change as changes:

derd = sin( )Ex + cos( )Ey = e ,ded = cos( )Ex sin( )Ey = er.

It is crucial to note that is measured positive in the counterclockwise direction.Returning to the position vector r, it follows that

r = xEx + yEy + zEz = r cos( )Ex + r sin( )Ey =rer

+zEz = rer + zEz.

Furthermore, because {er,e ,Ez} is a basis, we then have, for any vector b, that

b = brer + b e + bzEz = bxEx + byEy + bzEz.

It should be clear that br = b er, b = b e , and bz = b Ez.

2.2 Velocity and Acceleration Vectors

Consider a particle moving in space: r = r(t). We recall that

r = rer + zEz = xEx + yEy + zEz.

As the particle is in motion, its coordinates are functions of time: x = x(t), y = y(t),r = r(t), = (t), and z = z(t). To calculate the (absolute) velocity vector v of theparticle, we differentiate r(t):

v =drdt =

drdt er + r

derdt +

dzdt Ez.

Now, using the chain rule, er = der/d = e . Also,

ded = er,

derd = e .

1 Details of these results are discussed in Section A.4 of Appendix A.


It follows that

v =drdt er + r

ddt e +

dzdt Ez

=dxdt Ex +

dydt Ey +

dzdt Ez.

To calculate the (absolute) acceleration vector a, we differentiate v with respectto time:

a =dvdt =

ddt

(drdt er

)+

ddt

(r

ddt e

)+

d2zdt2 Ez.

Using the chain rule to determine the time derivatives of the vectors er and e ,and after collecting terms in the expressions for a, the final form of the results isobtained:

a =

(d2rdt2 r

(ddt

)2)

er +(

rd2dt2 + 2

drdt

ddt

)e +

d2zdt2 Ez

=d2xdt2 Ex +

d2ydt2 Ey +

d2zdt2 Ez.

We have also included the representations for the velocity and acceleration vectorsin Cartesian coordinates to emphasize the fact that the values of these vectors do notdepend on the coordinate system used.

2.2.1 Common Errors

In my experience, the most common error with using cylindrical polar coordinatesis to write r = rer + e + zEz. This is not true. Another mistake is to differentiateer and e incorrectly with respect to time. Last, but not least, many people presumethat all of the results presented here apply when is taken to be positive in theclockwise direction. Alas, this is not the case.

2.3 Kinetics of a Particle

Consider a particle of mass m. Let F denote the resultant external force acting onthe particle, and let G = mv be the linear momentum of the particle. Eulers first law(which is also known as Newtons second law or the balance of linear momentum)postulates that

F = dGdt = ma.

2.4 The Planar Pendulum 21

With respect to a Cartesian basis F = ma is equivalent to three scalar equations:

Fx = max = mx, Fy = may = my, Fz = maz = mz,

where F = FxEx + FyEy + FzEz and a = axEx + ayEy + azEz.With respect to a cylindrical polar coordinate system, the single vector equation

F = ma is equivalent to three scalar equations:

(F = ma) er : Fr = m(

d2rdt2 r

(ddt

)2)

,

(F = ma) e : F = m(

rd2dt2 + 2

drdt

ddt

),

(F = ma) Ez : Fz = md2z

dt2 .

Finally, we recall for emphasis the relations

er = cos( )Ex + sin( )Ey,e = sin( )Ex + cos( )Ey,Ex = cos( )er sin( )e ,Ey = sin( )er + cos( )e .

You will use these relations countless times in an undergraduate engineering dy-namics course.

2.4 The Planar Pendulum

The planar pendulum is a classical problem in mechanics. As shown in Figure 2.3,a particle of mass m is suspended from a fixed point O either by an inextensiblemassless string or rigid massless rod of length L. The particle is free to move on aplane (z = 0), and during its motion a vertical gravitational force mgEy acts on theparticle.

We ask the following questions: what are the equations governing the motionof the particle and what is the tension in the string or rod? The answers to thesequestions are used to construct the motion of the particle and the plot of tension asa function of time that were shown in Figure 2.1 at the beginning of this chapter.


Ter

mgEyNEz

Ex

Ey

g O

Particle of mass m

Massless rod or string of length L

Fig. 2.3 The planar pendulum and the free-body diagram of the particle of mass m.

2.4.1 Kinematics

We begin by establishing some kinematical results. We note that r = Ler. Differ-entiating with respect to t, and noting that L is constant, gives us the velocity v.Similarly, we obtain a from v:

v = Lderdt = L

ddt e ,

a = Ld2dt2 e + L

ddt

dedt = L

d2dt2 e L

(ddt

)2er.

Alternatively, one can get these results by substituting r = L and z = 0 in the generalexpressions recorded in Section 2.2. I do not recommend this approach inasmuch asit emphasizes memorization.

2.4.2 Forces

Next, as shown in Figure 2.3, we draw a free-body diagram. There is a tension forceTer and a normal force NEz acting on a particle. The role of the tension force isto ensure that the distance of the particle from the origin is L and the normal forceensures that there is no motion in the direction of Ez. These two forces are knownas constraint forces. They are indeterminate (we need to use F = ma to determinethem). One should also note that the gravitational force has the representations

mgEy = mgsin( )er mgcos( )e .

2.5 Summary 23

2.4.3 Balance Law

The third step is to write down the balance of linear momentum (F = ma):

Ter + NEz mgEy = mL e mL 2er.

We obtain three scalar equations from this vector equation:

mL = mgcos( ), T = mL 2 mgsin( ), N = 0.

2.4.4 Analysis

The first of these equations is a second-order differential equation for (t):

mL = mgcos( ).

Given the initial conditions (t0) and (t0), one can solve this equation and deter-mine the motion of the particle. Next, the second equation gives the tension T in thestring or rod once (t) is known:

Ter = (mL 2 mgsin( ))er.

A representative example of the behavior of T during a motion of the pendulum isshown in Figure 2.1. This figure was constructed by first numerically solving theordinary differential equations for (t) and then computing the corresponding T (t).

For a string, it is normally assumed that T > 0, and for some motions of the stringit is possible that this assumption is violated. In this case, the particle behaves as ifit were free to move on the plane and r = L. Regardless, the normal force NEz iszero in this problem.

2.5 Summary

In this chapter, the cylindrical polar coordinate system {r, ,z} was introduced. Toassist with certain expressions, the vectors er = cos( )Ex + sin( )Ey and e =sin( )Ex + cos( )Ey were introduced. It was also shown that the position vec-tor of a particle has the representations

r = rer + zEz =

x2 + y2er + zEz= r cos( )Ex + r sin( )Ey + zEz= xEx + yEy + zEz.


By differentiating r with respect to time, the velocity and acceleration vectors wereobtained. These vectors have the representations

v =drdt = rer + r

e + zEz= xEx + yEy + zEz,

a =dvdt =

(r r 2)er +

(r + 2r

)e + zEz

= xEx + yEy + zEz.

To establish these results, the chain rule and the important identities er = e ande = er were used.

Using a cylindrical polar coordinate system, F = ma can be written as three scalarequations:

Fr = m(r r 2) ,

F = m(r + 2r

),

Fz = mz.

These equations were illustrated using the example of the planar pendulum.

2.6 Exercises

The following short exercises are intended to assist you in reviewing the presentchapter.

2.1. Using Figure 2.2, verify that er = cos( )Ex +sin( )Ey and e =sin( )Ex +cos( )Ey. Then, by considering cases where er lies in the second, third, andfourth quadrants, verify that these definitions are valid for all values of .

2.2. Starting from the definitions er = cos( )Ex +sin( )Ey and e =sin( )Ex +cos( )Ey, show that er = e and e = er. In addition, verify thatEx = cos( )er sin( )e and Ey = sin( )er + cos( )e .

2.3. Calculate the velocity vectors of particles whose position vectors are 10er and5er + tEz, where = t. Why do all of these particles move with constantspeed ||v|| yet have a nonzero acceleration?

2.4. The position vector of a particle of mass m that is placed at the end of arotating telescoping rod is r = 6ter, where = 10t + 5 (radians). Calculatethe velocity and acceleration vectors of the particle, and determine the force Fneeded to sustain the motion of the particle. What is the force that the particleexerts on the telescoping rod?

2.5. In solving a problem, one person uses cylindrical polar coordinates whereasanother uses Cartesian coordinates. To check that their answers are identical,they need to examine the relationship between the Cartesian and cylindrical

2.6 Exercises 25

polar components of a certain vector, say b = brer +b e . To this end, showthat

bx = b Ex = br cos( )b sin( ), by = b Ey = br sin( )+ b cos( ).

2.6. Consider the projectile problem discussed in Section 1.6 of Chapter 1. Usinga cylindrical polar coordinate system, show that the equations governing themotion of the particle are

mrmr 2 = mgsin( ), mr + 2mr = mgcos( ), mz = 0.

Notice that, in contrast to using Cartesian coordinates to determine the gov-erning equations, solving these differential equations is nontrivial.

2.7. Consider a spherical bead of mass m and radius R that is placed inside a longcylindrical tube. The inner radius of the tube is R, and the tube is pivoted sothat it rotates in a horizontal plane. Furthermore, the contact between the tubeand the bead is smooth. Here, the bead is modeled as a particle of mass m.Now suppose that the tube is whirled at a constant angular speed (radiansper second). The whirling motion of the tube is such that the velocity vectorof the bead is v = rer + re . Show that the equation governing the motionof the bead is

r 2r = 0,and the force exerted by the tube on the particle is mgEz + 2mre .

2.8. Consider the case where the bead is initially at rest relative to the whirlingtube at a location r0 = L. Using the solution to the differential equation r 2r = 0 recorded in Section A.5.3 of Appendix A, show that, unless L = 0,the bead discussed in the previous exercise will eventually exit the whirlingtube.

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2 Particles and Cylindrical Polar Coordinates2.1 The Cylindrical Polar Coordinate System2.2 Velocity and Acceleration Vectors2.2.1 Common Errors

2.3 Kinetics of a Particle2.4 The Planar Pendulum2.4.1 Kinematics2.4.2 Forces2.4.3 Balance Law2.4.4 Analysis

2.5 Summary2.6 Exercises

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