```Generalized Coordinates
Rectangular coordinates are the most straightforward coordinates, but not necessarily the easiest.
&quot;It would be very desirable and convenient to have a general method for setting up equations of
motion directly in terms of any convenient set of generalized coordinates. Furthermore, it is
desirable to have uniform methods of writing down, and perhaps solving, the equations of motion
in terms of any coordinate system.&quot;
1. The number of generalized coordinates must equal the number of rectangular coordinates.
[This is not always three. If we have two objects we are following, each has three coordinates so
the total number of coordinates is six!]
2. Constraints are conditions which restrict the possible set of values of the coordinates.
[Example: in 2-D circular motion, the radius is constrained (by something) to be constant, with
only the angle, θ, being allowed to vary.]
----------Examples:
Polar coordinates are an example of a 2-D system of coordinates that are not rectangular:
r = r(x,y) = (x&sup2;+y&sup2;)
 q1 = q1(x,y)
and x = r cos(θ)  x = x(q1,q2)
θ = θ(x,y) = arctan(y/x)  q2 = q2(x,y)
and y = r sin(θ)  y = y(q1,q2).
Another example involving time: rotating polar coordinates
r = r(x,y,t) = (x&sup2;+y&sup2;)
 q1 = q1(x,y,t) and x = x(r,θ,t) = r cos(θ+t)
θ = θ(x,y,t) = arctan(y/x) - t  q2 = q2(x,y,t) and y = y(r,θ,t) = r sin(θ+t)
{note: sign difference due to fact that if we let β = θ+t, then θ = β-t}
----------3. We also need generalized velocities for energy considerations (for use in conservation of
energy) and for momentum considerations (for use in Newtons Second Law: ΣF = dp/dt).
For one particle in rectangular form: KE = &frac12;mv&sup2; = &frac12;m(x’&sup2;+y’&sup2;+z’2)
where x’ = dx/dt.
For more than one particle in rectangular form: KE = T = Σi[&frac12;mi{xi’&sup2;+yi’&sup2;+zi’&sup2;}].
Since x1 = x1(q1, q2, ... , qn, t) we have
dx1 = (x1/q1)dq1 + (x1/q2)dq2 + ... (x1/qn)dqn + (x1/t)dt
so x1’ = dx1/dt
= [1/dt]*[(x1/q1)dq1 + (x1/q2)dq2 + ... (x1/qn)dqn + (x1/t)dt)
= Σi(x1/dqi)qi’ + x1/t
------------Example: Rotating Polar Coordinates
x = r cos(t+θ) 
x’ = (x/r)r’ + (x/θ)θ’ + (x/t)
= r’cos(t+θ) + θ’(-r sin(t+θ)) + (-r sin(t+θ))
x component of r’
and of rθ’ and of r .
------------4. The KE involves not just x’, but x’&sup2;. What does this look like?
x’&sup2; = [Σk{(x/qk)qk’ + x/t}] * [Σ{(x/q)q’ + x/t}]
= [ΣkΣ{(x/qk)(x/q)qk’q’}] + [Σk{(x/qk)(x/t)qk’}]
+ [Σ{(x/q)(x/t)q’}] + (x/t)&sup2; .
Since the k and the  are what are called dummy indices (because they are summed over
completely) and do not matter whether they are called k or , the second and third terms are
exactly the same and can be combined. Thus the expression becomes:
x’&sup2; = [ΣkΣ{(x/qk)(x/q)qk’q’}] + 2[Σk{(x/qk)(x/t)qk’}] + (x/t)&sup2;
If we have more than one particle, all the “x” symbols above are replaced by “xi”.
We can now write the expression for KE (sometimes referred to as T):
T = Σi{&frac12;mi(xi’&sup2; + yi’&sup2; + zi’&sup2;)} = ΣkΣl{&frac12;Akqk’q’} + ΣkBkqk’ + To
where
Ak = Σi{mi[(xi/qk)(xi/q) + (yi/qk)(yi/q) + (zi/qk)(zi/q)]}
Bk = Σi{mi[(xi/qk)(xi/t) + (yi/qk)(yi/t) + (zi/qk)(zi/t)]}
To = Σi{&frac12;mi[(xi/t)&sup2; + (yi/t)&sup2; + (zi/t)&sup2;]} .
NOTE: Ak = Ak . If Ak = δk (where δk = 1 if k=, and δk = 0 otherwise; it is called the Dirac
delta function), then the coordinates are said to be orthogonal.
Suppose we have a non-orthogonal system. v = u’u + w’w
angle between u and w is α which is NOT 90. Then:
where the
v&sup2; = v &middot; v = (u’u + w’w) &middot; (u’u + w’w) = u’&sup2; + w’&sup2; + 2u’w’ cos(α) .
Notice that this term has an Ak that is not zero if k≠ !
-------------EXAMPLE:
For the case of one particle, use the above to show what the kinetic energy looks like in the
rotating polar coordinates. In particular,
a) find A11, A12, A21 and A22; show that this system is orthogonal (A12 = A21 = 0);
b) find B1 and B2;
c) find To;
d) write down the final Kinetic Energy; identify each term in the expression.
For rotating polar coordinates (2-D):
q1 = r = r(x,y,t) = (x&sup2;+y&sup2;)
and
x = x(r,θ,t) = r cos(θ+t)
q2 = θ = θ(x,y,t) = arctan(y/x) - t and
y = y(r,θ,t) = r sin(θ+t)
Kinetic Energy:
T = Σi{&frac12;mi(x’&sup2; + y’&sup2; + z’&sup2;)} = ΣkΣl{&frac12;Akqk’q’} + ΣkBkqk’ + To
where
Ak = Σi{mi[(xi/qk)(xi/q) + (yi/qk)(yi/q) + (zi/qk)(zi/q)]}
Bk = Σi{mi[(xi/qk)(xi/t) + (yi/qk)(yi/t) + (zi/qk)(zi/t)]}
To = Σi{&frac12;mi[(xi/t)&sup2; + (yi/t)&sup2; + (zi/t)&sup2;]} .
We have only 1 particle, so there is no need to take the sum over i, and all the mi’s will just be m,
all of the xi’s will be x, and all of the yi’s will be y.
We only have 2-D, so there are no z terms.
For rotating polar coordinates,
(x/q1) = (x/r) = ([r cos(θ+t)]/r) = cos(θ+t)
(x/q2) = (x/θ) = ([r cos(θ+t)]/θ) = -r sin(θ+t)
(x/t)
= (x/t) = ([r cos(θ+t)]/t) = -rω sin(θ+t)
(y/q1) = (y/r) = ([r sin(θ+t)]/r) = sin(θ+t)
(y/q2) = (y/θ) = ([r sin(θ+t)]/θ) = r cos(θ+t)
(y/t)
= (y/t) = ([r sin(θ+t)]/t) = rω cos(θ+t)
a) Therefore, we have:
A11 = m[(x/q1)(x/q1) + (y/q1)(y/q1) = m[cos2(θ+t) + sin2(θ+t)] = m.
A12 = m[(x/q1)(x/q2) + (y/q1)(y/q2)] = A21
= m[{cos(θ+t)}*{-r sin(θ+t)} + {sin(θ+t)}*{ r cos(θ+t)}] = 0.
A22 = m[(x/q2)(x/q2) + (y/q2)(y/q2) = m[r2sin2(θ+t) + r2cos2(θ+t)] = mr2.
b)
B1 = m[(x/q1)(x/t) + (y/q1)(y/t)]
= m[{cos(θ+t)}*{ -rω sin(θ+t)} + { sin(θ+t)}*{ rω cos(θ+t )} ] = 0.
B2 = m[(x/q2)(x/t) + (y/q2)(y/t)]
= m[ {-r sin(θ+t)}*{ -rω sin(θ+t)} + { r cos(θ+t)}*{ rω cos(θ+t)}]
= mr2ω .
c)
To = Σi{&frac12;mi[(xi/t)&sup2; + (yi/t)&sup2; + (zi/t)&sup2;]}
= &frac12;m[ {-rω sin(θ+t)}&sup2; + { rω cos(θ+t)}&sup2; ] = &frac12;mr2ω2 .
d)
T = Σi{&frac12;mi(x’&sup2; + y’&sup2; + z’&sup2;)} = ΣkΣl{&frac12;Akqk’q’} + ΣkBkqk’ + To
= &frac12; [A11q1’q1’ + A12q1’q2’ + A21q2’q1’ + A22q2’q2’] + B1q1’ + B2q2’ + To
= &frac12; mr’2 + 0 + 0 + &frac12; mr2θ’2 + 0 + mr2ωθ’ + &frac12; mr2ω2
= &frac12; mr’2 + &frac12; mr2(θ’+ω)2
where r’ is the radial speed, and r(θ’+ω) is the tangential speed.
-----------
WRITTEN HOMEWORK PROBLEMS #5a and #6a:
(You should be able to do part a of problem 5 and part a of problem 6 now.)
Problem #5: (Problem 9-1 in Symon)
Consider the coordinates u and w that are defined in terms of the polar coordinates r and θ:
u = ln(r/a) – θ cot(ζ)
and
w = ln(r/a) + θ tan(ζ)
where a and ζ are constants.
Sketch the curves of constant u and w: this is already done for you - see notes on next
section and/or the excel spreadsheet “Spiral Coordinates” on the course web page..
a) Find the kinetic energy for a particle of mass m in terms of u, w, du/dt, and dw/dt .
b) Find expressions for Qu and Qw in terms of the polar force components Fr and Fθ .
c) Find pu and pw .
** d) extra credit: find the forces Qu and Qw required to make the particle move with
constant speed, ds/dt, along a spiral of constant u=uo .
HINT: You can try to find the inverse transformations to get
x(u,w) and y(u,w), and then proceed as indicated above. Or it
may be easier to get the kinetic energy in polar form (r,θ); find
the inverse transformations: r(u,w) and θ(u,w); then find
r’(u,w,u’,w’) and θ’(u,w,u’,w’); then write the kinetic energy in
terms of u,w,u’,w’.
Problem #6: (Problem 9-3 in Symon)
For plane (2-D) parabolic coordinates f and h defined such that: x = f – h and y = 2(fh)1/2
(see the excel spreadsheet “Parabolic Coordinates” accessed from the course web page to see an
image);
a) find the expression for the kinetic energy in terms of f, h, f’, and h’;
b) find the expression for the momenta: pf and ph ;
c) write out the Lagrange equations in these coordinates if we assume the particle is not acted on
by any force.
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