Equations of Motion in Classical Mechanics: A Comprehensive Guide, Lecture notes of Analytical mechanics

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A Guided Tour of

Analytical Mechanics

with animations in MAPLE

Rouben Rostamian

Department of Mathematics and Statistics

UMBC

[email protected]

December 2, 2018

ii

• 1 An introduction through examples Preface vii
  • 1.1 The simple pendulum à la Newton
  • 1.2 The simple pendulum à la Euler
  • 1.3 The simple pendulum à la Lagrange
  • 1.4 The double pendulum
  • Exercises
• 2 Work and potential energy
  • Exercises
• 3 A single particle in a conservative force field
  • 3.1 The principle of conservation of energy
  • 3.2 The scalar case
  • 3.3 Stability
  • 3.4 The phase portrait of a simple pendulum
  • Exercises
• 4 The Kapitsa pendulum
  • 4.1 The inverted pendulum
  • 4.2 Averaging out the fast oscillations
  • 4.3 Stability analysis
  • Exercises
• 5 Lagrangian mechanics
  • 5.1 Newtonian mechanics
  • 5.2 Holonomic constraints
  • 5.3 Generalized coordinates
  • 5.4 Virtual displacements, virtual work, and generalized force
  • 5.5 External versus reaction forces
  • 5.6 The equations of motion for a holonomic system
  • Exercises
• 6 Constraint reactions
  • Exercises
• 7 Angular velocity
  • Exercises
• 8 The moment of inertia tensor iv Contents
  • 8.1 A brief introduction to tensor algebra
    • 8.1.1 Tensor algebra
    • 8.1.2 Connection with R^3 and 3 × 3 matrices
    • 8.1.3 Symmetric tensors
  • 8.2 The moment of inertia tensor
  • 8.3 Translation of the origin
  • 8.4 The principal moments of inertia
  • Exercises
• 9 Rigid body dynamics through the Gibbs-Appell formulation
  • 9.1 A formula for the Gibbs function
  • 9.2 Example: Euler’s equations of motion
  • 9.3 Rotation about a fixed point
• 10 The Gibbs-Appell formulation of dynamics
  • 10.1 Gibbs-Appell according to Lurie [ 12 ]
    • 10.1.1 Acceleration in generalized coordinates
      • namics 10.1.2 Ideal constraints and the fundamental equation of dy-
    • 10.1.3 Virtual work and generalized force
    • 10.1.4 Constraints
    • 10.1.5 Virtual displacements
    • 10.1.6 Back to the fundamental equation: Part
    • 10.1.7 Back to the fundamental equation: Part
    • 10.1.8 The Gibbs-Appell equations of motion
    • 10.1.9 Quasi-velocities
    • 10.1.10 Appell’s equations of motion in terms of quasi-velocities
  • 10.2 Gibbs-Appell according to Gantmacher [ 8 ]
    • 10.2.1 Pseudocoordinates
    • 10.2.2 Work and generalized forces
    • 10.2.3 Newton’s equations in pseudocoordinates
    • 10.2.4 The energy of the acceleration
  • 10.3 A modification noted by Desloge
  • 10.4 The simple pendulum via Gibbs-Appell
  • 10.5 The Caplygin sleigh .ˇ
  • 10.6 The Caplygin sleigh revisited .ˇ
  • 10.7 The problem from page 63 of Gantmacher
• 11 Rigid body dynamics
  • 11.1 Three frames of reference
  • 11.2 The energy of acceleration for a rigid body
  • 11.3 The rolling coin
  • 11.4 The three frames
  • 11.5 The angular velocity
  • 11.6 The no-slip constraint
  • 11.7 The acceleration of the coin’s center
  • 11.8 The rotational acceleration
  • Exercises
• 12 Quaternions Contents v
  • 12.1 The quaternion algebra
  • 12.2 The geometry of the quaternions
    • 12.2.1 The reflection operator
    • 12.2.2 The rotation operator
  • 12.3 Angular velocity
  • 12.4 A differential equation for the quaternion rotation
  • 12.5 Unbalanced ball rolling on a horizontal plane
    • 12.5.1 The no-slip condition
    • 12.5.2 The Gibbs function and the equations of motion
  • Exercises

Preface

Unless otherwise specified, by “solving a problem” I mean performing all the steps laid out below:

1. Select configuration parameters.
2. Define the position vectors r 1 , r 2 ,... of the point masses in terms of the generalized coordinates q 1 , q 2 ,....
3. Compute the velocities of the point masses:

vi = r˙i =

j

∂ ri ∂ qj

q^ ˙j , i = 1, 2,....

4. Compute the kinetic energy T = (^12)

i mi ‖vi ‖

(^2) , the potential energy V , and the Lagrangian L = T − V.

5. Form the equations of motion (a system of second order differential equations (DEs)) in the unknowns q 1 (t ), q 2 (t ),.. .:

d d t

∂ L

∂ q˙j

∂ L

∂ qj

, j = 1, 2,....

If done by hand, this step would be the most labor-intensive part of the calculations. The calculations can get unbearably complex and can easily lead to formulas that fill more than one page. Fortunately we can relegate the tedious computations to MAPLE.^1

6. Solve the system of DEs. Except for a few special cases, such system are generally not solvable in terms of elementary function. One solves them numerically with the help of specialized software such as MAPLE (or MATHEMATICA). The software replaces the continues time variable t by a closely spaced “time ticks” t 0 , t 1 , t 2 ,... which span the time interval of interest, say [0, T ], and then it ap- plies some rather sophisticated numerical algorithms to evaluate the unknowns q 1 (t ), q 2 (t ),... at those time ticks. The result may be presented as:

(a) a table of numbers; but that’s not very illuminating, so it’s rarely done that way; (^1) Nowadays MAPLE and MATHEMATICA are the two dominant Computer Algebra Systems. If you are famil- iar with MATHEMATICA, you should be able to translate the MAPLE commands in this book into the equivalent MATHEMATICA commands.

vii

viii Preface

(b) as a set of plots of qj versus t. This is the most common way. Both MAPLE (and MATHEMATICA) can do this easily; or (c) as a computer animation, which is the most “user friendly” choice but which takes some work—and a certain amount of know-how—to produce. I will show you how to do this in MAPLE.

2 Chapter 1. An introduction through examples

ϕ

m g

x

y

ϕ

w = m g j

− τ er

er

e ϕ i

j

Figure 1.1: On the left is a depiction of the physical shape of the pendulum. On the right we see the mathematical machinery devised to analyze the pendulum’s motion. The unit vectors i and j are attached to the fixed Cartesian coordi- nates system and are stationary; the unit vectors er and e ϕ move with the pendulum. The weight of the bob is w = m g j.

bob’s initial velocity is small. Then the pendulum will oscillate back and forth about the stable configuration, similar to what we see in a grandfather clock. If the initial velocity is slightly larger, the pendulum will undergo wider oscillations. If, however, the initial velocity is sufficiently large, the pendulum will not oscillate at all. It will swing about pivot, reach the unstable equilibrium position at the top and go past it, fall down from the other side, and return to its initial position, having made a complete 360 degree rotation about the pivot. At this point the pendulum finds itself in the same condition that it had at the initial time, therefore it will repeat what it did the first time around. In the absence of energy dissipating factors, the rotations about the pivot will continue indefinitely. To make a mathematical model of the pendulum, we introduce the Cartesian coordi- nates xy with the origin at the pendulum’s pivot, and the y axis pointing down. We also introduce the stationary unit vectors i and j along the x and y axes, and the moving unit vectors er along the pendulum’s rod and e ϕ which is perpendicular to it, as shown in Fig- ure 1.1. It it evident that the vectors er and e ϕ may be expressed as linear combinations of the vectors i and j:

er = i sin ϕ + j cos ϕ , e ϕ = −i cos ϕ + j sin ϕ.

Furthermore, let us observe that their time derivatives are related through

e˙r = i ϕ ˙ cos ϕ − j ϕ ˙ sin ϕ = − ϕ ˙e ϕ , e˙ ϕ = i ϕ ˙ sin ϕ + j ϕ ˙ cos ϕ = ϕ ˙er. (1.1)

The bob’s position vector r(t ) relative to the origin is r = ℓ er , where ℓ is the length of the rod, and therefore the bob’s velocity v = r˙ and acceleration a = v˙ may be computed easily with the help of (1.1):

v = r˙ = ( ℓ er )·^ = − ℓ ϕ ˙e ϕ , a = v˙ = (− ℓ ϕ ˙e ϕ )·^ = − ℓ ϕ ¨e ϕ − ℓ ϕ ˙ e˙ ϕ = − ℓ ϕ ¨e ϕ − ℓ ϕ ˙^2 er.

We see that the bob’s acceleration has a component along e ϕ and another along er. Newton’s law of motion asserts that ma = F , where F is the resultant of all forces acting on the bob. Referring to Figure 1.1 we see that the forces acting on the bob consist

1.2. The simple pendulum à la Euler 3

of weight w and the tension − τ er along the rod,^3 where τ generally varies with time and is unknown. It follows that

m(− ℓ ϕ ¨e ϕ − ℓ ϕ ˙^2 er ) = w − τ er.

The weight, however, is w = m g j, where m is the mass of the bob and g is the acceler- ation due to gravity. We replace w with its decomposition w = m g j = (m g cos ϕ )er + (m g sin ϕ )e ϕ in the equation of motion, and collect the coefficients of er and and e ϕ , and arrive at (^)  m ℓ ϕ ¨ + m g sin ϕ

e ϕ +

m ℓ ϕ ˙^2 + m g cos ϕ − τ

er = 0. Since er and e ϕ are orthogonal, hence linearly independent, each of the expressions in the square brackets is zero. We conclude that

m ℓ ϕ ¨ + m g sin ϕ = 0, m ℓ ϕ ˙^2 + m g cos ϕ − τ = 0. (1.2)

The first equation is a second order differential equation in the unknown ϕ. It has a unique solution for any initial condition

ϕ ( 0 ), ˙ ϕ ( 0 ) , although the solution is not expressible in terms of elementary functions. In practice, one solves the equation through a numerical approximation algorithm on a computer. Once the solution ϕ (t ) is obtained, it may be substituted in the second equation to evaluate the tension τ (t ) in the rod, should it be of interest.

1.2 The simple pendulum à la Euler

In the previous section we assumed, without explanation, that the force within the pen- dulum’s rod points along the rod; see Figure 1.1 where that force is shown as the vector − τ er. That assumption seems to be so “obvious” that many textbooks on mechanics and its applications present it without as much as a comment. A close scrutiny, however, shows that the assumption is far from obvious, and in fact, it is not a logical consequence of any of Newton’s laws of motion. Antman [ 2 ] presents a critical analysis of this issue and concludes that the proper approach is through an application of Euler’s law of motion, which states that the rate of change of the pendulum’s angular momentum equals the resultant torque applied to it.

1.3 The simple pendulum à la Lagrange

In this section we rederive the differential equation of motion of the simple pendulum through Lagrange’s analytical approach. We no longer need the vectors er and e ϕ. In- stead, we write the bob’s position vector r directly in terms of its i and j components:

r = ( ℓ sin ϕ )i + ( ℓ cos ϕ )j,

and then differentiate to find the velocity

v = r˙ = ( ℓ ϕ ˙ cos ϕ )i − ( ℓ ϕ ˙ sin ϕ )j.

It follows that that ‖v‖^2 = ℓ^2 ϕ ˙^2. To proceed further, we introduce a few definitions and assertions whose motivations and explanations will emerge only in subsequent chapters. (^3) The assertion that the force exerted on the bob by the rod lies along the rod requires justification. See the next section for elaboration.

1.4. The double pendulum 5

m 1 g

ϕ 1 ℓ 2

m 2 g

ϕ 2

x

y

r 1

m 1 g j

ϕ 1

m 2 g j

ϕ 2

i

j (^) r 2

Figure 1.2: On the left is a depiction of the physical shape of the double pendulum. On the right we see the pendulum’s mathematical model given by the position vectors r 1 and r 2 of the two bobs.

To make a mathematical model of a double pendulum, we follow the ideas sketched in the previous section. Specifically, we introduce the xy Cartesian coordinates and the sta- tionary unit vectors i and j as shown in Figure 1.2, and then express the position vectors r 1 and r 2 of the two bobs in terms of their components relative to i and j: r 1 = ( ℓ 1 sin ϕ 1 )i + ( ℓ 1 cos ϕ 1 )j, r 2 = r 1 + ( ℓ 2 sin ϕ 2 )i + ( ℓ 2 cos ϕ 2 )j. (1.6) Then we find the velocities of the bobs through differentiation: v 1 = ( ℓ 1 ϕ ˙ 1 cos ϕ 1 )i − ( ℓ 1 ϕ ˙ 1 sin ϕ 1 )j, v 2 = v 1 + ( ℓ 2 ϕ ˙ 2 cos ϕ 2 )i − ( ℓ 2 ϕ ˙ 2 sin ϕ 2 )j.

We see that ‖v 1 ‖^2 = ℓ^21 ϕ ˙^21. Computing ‖v 2 ‖^2 takes only a little bit more work. We observe that v 2 = v 1 + v˜, where ˜v = ( ℓ 2 ϕ ˙ 2 cos ϕ 2 )i − ( ℓ 2 ϕ ˙ 2 sin ϕ 2 )j. Therefore ‖v 2 ‖^2 = ‖v 1 ‖^2 + ‖v˜‖^2 + 2 v 1 · v˜ = ℓ^21 ϕ ˙^21 + ℓ^22 ϕ ˙^22 + 2

( ℓ 1 ϕ ˙ 1 cos ϕ 1 )i − ( ℓ 1 ϕ ˙ 1 sin ϕ 1 )j

( ℓ 2 ϕ ˙ 2 cos ϕ 2 )i − ( ℓ 2 ϕ ˙ 2 sin ϕ 2 )j

= ℓ^21 ϕ ˙^21 + ℓ^22 ϕ ˙^22 + 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 (cos ϕ 1 cos ϕ 2 + sin ϕ 1 sin ϕ 2 ). = ℓ^21 ϕ ˙^21 + ℓ^22 ϕ ˙^22 + 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 cos( ϕ 2 − ϕ 1 ). We conclude that the double pendulum’s kinetic energy is

T =

m 1 ℓ^21 ϕ ˙^21 +

m 2

ℓ^21 ϕ ˙^21 + ℓ^22 ϕ ˙^22 + 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 cos( ϕ 2 − ϕ 1 )

(m 1 + m 2 ) ℓ^21 ϕ ˙ 12 +

m 2 ℓ^22 ϕ ˙ 22 + m 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 cos( ϕ 2 − ϕ 1 ).

As to the potential energy, let us recall that a mass’s potential energy in a constant gravitational field is the product of its weight and its elevation above a certain reference point. In the case of a double pendulum, it is easiest to set the reference point at the origin of the coordinates; see Figure 1.2. Then the j components of the vectors r 1 and r 2 provide the elevations of the bobs below the reference point, therefore their elevations above the reference point will require a sign reversal. Referring to (1.6) we see that V = −m 1 g cos ϕ 1 − m 2 g

ℓ 1 cos ϕ 1 + ℓ 2 cos ϕ 2

= −(m 1 + m 2 )g cos ϕ 1 − m 2 g ℓ 2 cos ϕ 2. Thus, the double pendulum’s Lagrangian, L = T − V , takes the form

L( ϕ 1 , ϕ 2 , ˙ ϕ 1 , ˙ ϕ 2 ) =

(m 1 + m 2 ) ℓ^21 ϕ ˙^21 +

m 2 ℓ^22 ϕ ˙ 22 + m 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 cos( ϕ 2 − ϕ 1 )

• (m 1 + m 2 )g cos ϕ 1 + m 2 g ℓ 2 cos ϕ 2.

6 Chapter 1. An introduction through examples

In the previous section’s simple pendulum, the Lagrangian L( ϕ , ˙ ϕ ) was a function two variables. In the present case, the Lagrangian L( ϕ 1 , ϕ 2 , ˙ ϕ 1 , ˙ ϕ 2 ) is a function of four variables. In general, if a mechanical system’s geometric configuration is specified through n variables q 1 ,... , qn , then its Lagrangian is a function of 2n variables q 1 ,... , qn , ˙q 1 ,... , ˙qn. The equivalent of the single equation of motion (1.4) now is a system of n equations, called the mechanical system’s Euler–Lagrange equations:

d d t

∂ L

∂ q˙i

∂ L

∂ qi

, i = 1,... , n.

The variable q 1 ,... , qn are called the system’s generalized coordinates. Applied to the case of double pendulum, the Euler–Lagrange equations lead to

d d t

∂ L

∂ ϕ ˙ 1

∂ L

∂ ϕ 1

d d t

∂ L

∂ ϕ ˙ 2

∂ L

∂ ϕ 2

To evaluate these explicitly, we begin by computing

∂ L ∂ ϕ ˙ 1

= (m 1 + m 2 ) ℓ^21 ϕ ˙ 1 + m 2 ℓ 1 ℓ 2 ϕ ˙ 2 cos( ϕ 2 − ϕ 1 ),

∂ L ∂ ϕ ˙ 2

= m 2 ℓ^22 ϕ ˙ 2 + m 2 ℓ 1 ℓ 2 ϕ ˙ 1 cos( ϕ 2 − ϕ 1 ),

∂ L ∂ ϕ 1

= m 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 sin( ϕ 2 − ϕ 1 ) − (m 1 + m 2 )g sin ϕ 1 ,

∂ L ∂ ϕ 2

= −m 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 sin( ϕ 2 − ϕ 1 ) − m 2 g ℓ 2 sin ϕ 2.

We conclude that the differential equations of motion are  (m 1 + m 2 ) ℓ^21 ϕ ˙ 1 + m 2 ℓ 1 ℓ 2 ϕ ˙ 2 cos( ϕ 2 − ϕ 1 )

= m 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 sin( ϕ 2 − ϕ 1 ) − (m 1 + m 2 )g sin ϕ 1 ,  m 2 ℓ^22 ϕ ˙ 2 + m 2 ℓ 1 ℓ 2 ϕ ˙ 1 cos( ϕ 2 − ϕ 1 )

= −m 2 ℓ 1 ℓ 2 ϕ ˙ 1 ϕ ˙ 2 sin( ϕ 2 − ϕ 1 ) − m 2 g ℓ 2 sin ϕ 2.

Exercises

1.1. Pendulum with a mobile pivot. Figure 1.3 shows a pendulum whose pivot is allowed to move horizontally without friction. The pivot has mass m 1 while the bob has mass m 2. Find the equations of motion of the pendulum. 1.2. A spherical pendulum. The motion of the simple pendulum of length ℓ intro- duced in this chapter was confined to a single vertical plane, and therefore the pendulum’s bob moved along a circular arc of radius ℓ. If off-plane motions are permitted, then the bob will move on a sphere of radius ℓ centered at the pivot. In that setting the pendulum is called a spherical pendulum; see Figure 1.4. Derive the equations of motion of the spherical pendulum. 1.3. Bead on a spinning hoop. A circular wire hoop of radius R spins about a vertical diameter at a constant angular velocity Ω. A bead of mass m can slide without fric- tion along the hoop. The hoop’s radius that connects to the bead makes an angle

8 Chapter 1. An introduction through examples

R

m g

ϕ

Figure 1.5: Bead on a rotating hoop (Exercise 1.3).

ϕ

Ω

O

A

m 1

B

m 1

m 2 S

Figure 1.6: A simplified Watt governor (Exercise 1.4).

x

y

z

ρ (t )

P

m 1

Q m 2

ϕ (t )

Figure 1.7: The point P slides on the table. The point Q moves vertically (Exercise 1.5).

Chapter 2

Work and potential energy

Work is the product of force and its displacement. To be precise, the infinitesimal work dW performed in displacing a force F by an infinitesimal distance d r is dW = F · d r. If the point of the application of the force moves along a path C in space, then the work performed along the path is the line integral

W =

C

F · d r. (2.1)

If you are repositioning a massive desk in an office, for example, then work measures the amount of effort exerted by you in performing the task. Expanding upon the moving of the desk scenario, suppose that you intend to move the desk from a point A to a point B. It should be obvious that the amount of work performed will vary, depending on the path along which you move the desk between A and B. Chances are that the shortest (straight line) path will require lesser effort than a long path that winds around the office. There are many interesting and important situations where, unlike the moving of the desk example, the work performed in goings from a point A to a point B is independent of the path taken between A and B. The most elementary example is the raising or lowering of a weight. To see how it works, set up a Cartesian coordinate system in space so that the x and y axes are horizontal, and the z axis points up. Let ra = (xa , ya , za ) and rb = (xb , yb , zb ) be the position vectors^5 of the starting and ending points A and B, and let r = 〈x, y, z〉 be the position vector of a generic point along a path C (A, B) with endpoints A and B. Suppose that we move an object of mass m along that path. The force of the object’s weight is F = 〈0, 0, −m g 〉, where g is the acceleration of gravity. The work performed along the path is

W =

C (A,B)

F · d r

C (A,B)

〈0, 0, −m g 〉 · 〈d x, d y, d z〉 =

C (A,B)

−m g d z = −m g (zb − za ).

We see that the work in moving the weight from A to B is expressed in terms the z co- ordinates of the endpoints, thus it is the same on all conceivable paths that go from A to B. (^5) A position vector of a point P (x, y, ) is the vector r = 〈x, y, z〉 that extends from the origin to the point P.

9

11

A

P Q

r r^ +

he

e

C

(A

,^ r

Figure 2.1: If V (P ) is evaluated by integration along the path C (A, r), then V (Q) may be evaluated by integrating along that same path, and then continuing along the straight line segment P Q of length h in the direction e.

• The assumption that the domain D is connected is needed to ensure that a path may be established between any pair of points in D. That’s what enabled us to sketch the curve C (A, r) that connects the points A and P in Figure 2.1.
• The assumption that the domain D is open means that a ball of positive radius may be placed around any point P ∈ D so that the ball lies entirely within D. It’s that property which ensures that moving away from P by a small distance h, as we did in Figure 2.1, we land safely on a point Q which lies within D.

Remark 2.2. Adding a constant to the potential function V does not affect the equality F (r) = ∇V (r). Thus, a vector field’s potential, if it has one, is defined modulo an additive constant.

Example 2.2. Earlier in the this section we observed that the force field corresponding to an object’s weight is F = 〈0, 0, −m g 〉. We see that F (r) = −∇V (r) where V (x, y, z) = m g z. We will use m g z as the potential of a weight throughout these notes. Note the effect of the minus sign in (2.2); in its absence the potential of a weight would have been −m g z.

Example 2.3. In the previous example we assumed that the acceleration of gravity g is a constant. That’s a good assumption if the changes in height during the motion are small relative to the radius of the Earth. In general, the gravitational force that a point mass M exerts on a point mass m drops as the inverse square of the distance. Specifically, Newton’s law of gravitation says

F (r) = −

GM m ‖r‖^2

r ‖r‖

where r is m’s position vector relative to M , and G is the universal gravitational constant. The inverse square law is manifested through the ‖r‖^2 term that appears in the denomi- nator inside the parentheses. The factor r / ‖r‖ is a unit vector that points from M to m. It is possible to show (see Exercise ??) that F is derived from a potential.

Theorem 2.4. Suppose the force field F is derived from a potential V. Then the work per-

12 Chapter 2. Work and potential energy

formed in moving the force along any path from a point ra to rb is given by

W = V (ra ) − V (rb ). (2.4)

Proof. We have F (r) = −∇V (r) therefore

F · d r = −∇V (r) · d r = −

D ∂ V

∂ x 1

∂ V

∂ xn

E

· 〈d x 1 ,... , d xn 〉

 ∂ V

∂ x 1

d x 1 + · · · +

∂ V

∂ xn

d xn

= −dV ,

therefore W =

C

F · d r =

∫ (^) rb

ra

−dV = V (ra ) − V (rb ).

Exercises

2.1. Verify that the gravitational potential field F (r) in (2.3) is derived from the poten- tial V (r) =

GM m ‖r‖