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Physics Courseware Support: Mechanics

  

Explore. While learning, having success is a secondary goal: using your curiosity as a compass is what matters. Things can be done in many different ways, take full permission to make mistakes. Computer algebra can transform the algebraic computation part of physics into interesting discoveries and fun.

  

 

  

The following material assumes knowledge of how to use Maple. If you feel that is not your case, for a compact introduction on reproducing in Maple the computations you do with paper and pencil, see sections 1 to 5 of the Mini-Course: Computer Algebra for Physicists. Also, the presentation assumes an understanding of the subjects and the style is not that of a textbook. Instead, it focuses on conveniently using computer algebra to support the practice and learning process. The selection of topics follows references [1] and [2] at the end. Maple 2023.0 includes Part I. Part II is forthcoming.

 

Part I

1. 

Position, velocity and acceleration in Cartesian, cylindrical and spherical coordinates

a. 

The position rtas a function of time

b. 

The velocity vt

c. 

The acceleration at

d. 

Deriving these formulas

e. 

Velocity and acceleration in the case of 2-dimensional motion on the x, y plane

1. 

The equations of motion

a. 

A single particle

i. 

The equations of motion - vectorial form

ii. 

The case of constant acceleration

iii. 

Motion under gravitational force close to the Earth's surface

iv. 

Motion under gravitational force not close to the Earth's surface

A. 

Circular motion

B. 

Escape velocity

i. 

Different acceleration in different regions

ii. 

The equations of motion using tensor notation

A. 

Cartesian coordinates

B. 

Curvilinear coordinates

a. 

Many-particle systems

i. 

Center of mass

ii. 

The equations of motion

iii. 

Static: reactions of planes and tensions on cables

a. 

Lagrange equations

i. 

Motion of a pendulum

1. 

Conservation laws

a. 

Work

b. 

Conservation of the total energy of a closed system or a system in a constant external field

c. 

Conservation of the total momentum of a closed system

d. 

Conservation of angular momentum

e. 

Cyclic coordinates

1. 

Integration of the equations of motion

a. 

Motion in one dimension

b. 

Reduced mass

i. 

The two-body problem

ii. 

A many-body problem

a. 

Motion in a central field

b. 

Kepler's problem

1. 

Small Oscillations

a. 

Free oscillations in one dimension

b. 

Forced oscillations

c. 

Oscillations of systems with many degrees of freedom

1. 

Rigid-body motion

a. 

Angular velocity

b. 

Inertia tensor

c. 

Angular momentum of a rigid body

d. 

The equations of motion of a rigid body

1. 

Non-inertial coordinate systems

a. 

Coriolis force and centripetal force

 

Part II (forthcoming)

1. 

The Hamiltonian and equations of motion; Poisson brackets

2. 

Canonical transformations

3. 

The Hamilton-Jacobi equation

 

Position, velocity and acceleration in Cartesian, cylindrical and spherical coordinates

 

Load the Physics:-Vectors package

 

withPhysics:-Vectors

&x,`+`,`.`,Assume,ChangeBasis,ChangeCoordinates,CompactDisplay,Component,Curl,DirectionalDiff,Divergence,Gradient,Identify,Laplacian,,Norm,ParametrizeCurve,ParametrizeSurface,ParametrizeVolume,Setup,Simplify,`^`,diff,int

(1)

 

Depending on the geometry of a problem, it can be convenient to work with either Cartesian or curvilinear coordinates. In an arbitrary reference system, the position in Cartesian coordinates and the basis of unitary vectorsi,j,kis given by

r_ = x _i + y _j + z _k 

r=xi+yj+zk

(2)

 

Problem

Rewrite the position vector r in cylindrical and spherical coordinates

 

Starting from the position in the Cartesian system, now as functions of the time to allow for differentiation, first note that the Cartesian unit vectors i,j,k do not depend on time, they are constant vectors. So rt is entered as

 

restart;withPhysics:-Vectors :

 

r_t = xt _i + yt _j + zt _k

rt=xti+ytj+ztk

(20)

Before proceeding further, use a compact display to more clearly visualize the following expressions. When in doubt about the contents behind a given display, input show as shown below.

CompactDisplayx,y,z,ρ,r,θ,φ,_ρ,_r,_θ,_φt

φtwill now be displayed asφ

(21)

 

For the velocity and acceleration, note the dot notation for derivatives with respect to t

v_t = diffrhs,t

v_t=diffxt,t_i+diffyt,t_j+diffzt,t_k

(22)

 

show

vt=x.ti+y.tj+z.tk

(23)

a_t = diffrhs,t

a_t=diffdiffxt,t,t_i+diffdiffyt,t,t_j+diffdiffzt,t,t_k

(24)

 

The position rtas a function of time

 

Problem

Given the position vector as a function of the time t, rewrite it in cylindrical and spherical coordinates while making the curvilinear unit vectors' time dependency explicit.

 

The velocity vt

 

Problem

Rewrite the velocity vt=r.t in cylindrical and spherical coordinates while making the curvilinear unit vectors' time dependency explicit .

 

The acceleration at

 

Problem

Rewrite the acceleration at=r..tin cylindrical and spherical components while making the curvilinear unit vectors' time dependency explicit.

 

Deriving these formulas

 

All these results for the position r, velocity v and acceleration a are based on the differentiation rules for cylindrical and spherical unit vectors. It is thus instructive to also be able to derive any of these formulas; for that, we need the differentiation rule for the unit vectors. For example, for the spherical ones

 

restart;withPhysics:-Vectors :CompactDisplayx,y,z,ρ,r,θ,φ,_ρ,_r,_θ,_φt, quiet

 

map%diff = diff, _r, _θ,_φt , t

%diff_rt,t=diffθt,t_θt+diffφt,tsinθt_φt,%diff_θt,t=diffθt,t_rt+diffφt,tcosθt_φt,%diff_φt,t=diffφt,t_ρt

(38)

The above result contains, in the last equation, the cylindrical radial unit vector ρt; rewrite it in the spherical basis

_ρt = ChangeBasis_ρt,spherical

_ρt=sinθt_rt+cosθt_θt

(39)

So the differentiation rules for spherical unit vectors, with the result expressed in the spherical system, are

subs,

%diff_rt,t=diffθt,t_θt+diffφt,tsinθt_φt,%diff_θt,t=diffθt,t_rt+diffφt,tcosθt_φt,%diff_φt,t=diffφt,tsinθt_rt+cosθt_θt

(40)

 

Problem

With this information at hand, derive, in steps, the expressions for the velocity and acceleration in cylindrical and spherical coordinates

 

Summary

• 

You can express rt,vt and atin any of the Cartesian, cylindrical or spherical systems via three different methods: 1) using the ChangeBasis command 2) differentiating 3) deriving the formulas by differentiating in steps, starting from the differentiation rules for the curvilinear unit vectors.

Velocity and acceleration in the case of 2-dimensional motion on the x, y plane


Problem

Derive formulas for velocity and acceleration in the case of  2-dimensional motion on the x,y plane, starting from the general 3-dimensional formulas above, e.g. (44) and (51) in spherical coordinates. Specialize the resulting formulas for the case of circular motion.

 

The equations of motion

 

A single particle

 

restart; withPhysics:-Vectors :CompactDisplayr_, p_, F_, L_, N_t

Ntwill now be displayed asN

(62)

The equation of motion of a single particle is Newton's 2nd law

F_t= mdiffr_t,t,t  

F_t=mdiffdiffr_t,t,t

(63)

where r..t=at is the acceleration and mr.t=pt is the linear momentum, so in terms of  p

F_t= diffp_t,t   

F_t=diffp_t,t

(64)

We define the angular momentum L of a particle, and the torque N acting upon it, as

L_t = r_t×p_t

L_t=&xr_t,p_t

(65)

N_t = r_t×F_t 

N_t=&xr_t,F_t

(66)

Differentiating the definition of L

diff,t

diffL_t,t=&xdiffr_t,t,p_t+&xr_t,diffp_t,t

(67)

Since r.=v is parallel to p=mv, the first term in the above cancels, and in the second term, from (64), p.=F

eval,diffr_t,t=0, diffp_t,t=F_t

diffL_t,t=&xr_t,F_t

(68)

from which

subsrhs=lhs,

N_t=diffL_t,t

(69)

 

• 

As discussed below, in the case of a closed system, F=0 and these two equations result in

p.=0,     L.=0

that is, the linear and angular momentum are conserved quantities. Note that L.=0 does not require that F=0, only that r×F=0.

 

The equations of motion - vectorial form

 

Problem

Assuming that the acceleration is known as a function of t, compute:

a) The trajectory rtstarting from at=r..t
b) A solution for each of the three Cartesian components

c) A solution for generic initial conditions

 

The case of constant acceleration

 

Problem

Starting from the vectorial equation (72) for rt, derive the formula for constant acceleration

 

Motion under gravitational force close to the Earth's surface

 

Problem

Derive a formula for motion under gravitational force close to the Earth's surface

 

Motion under gravitational force not close to the Earth's surface

 

The problem of two particles of masses m__1 and m__2 gravitationally attracted to each other, discarding relativistic effects, is formulated by Newton's law of gravity: the particles attract each other - so both move - with a force along the line that joins the particles and whose magnitude is proportional to 1r2, where r represents the distance between the particles (this problem is treated in general form in the more advanced sections).

 

Problem

As a specific case, consider the problem of a particle of mass mM, where M is earth's mass, moving not close to the surface (if compared with the radius of earth).