Analytical Mechanics Spring 2003, Test 3
Problem 1:
A hoop of mass m and radius R rolls without slipping
down a wedge of mass M and angle α, that can slide
frictionlessly on a table.
(a) What are the Lagrangian equations of motion?
(b) What are the generalized momenta?
(c) What are the Hamiltonian equations of motion?
(a) A picture really helps here. I refer you to the book for
the picture. Let's write down the kinetic energy in Cartesian coordinates,
and then convert them later to generalized coordinates that are more
convenient. The reason for this is that our kinetic energy coordinates should
be orthogonal, or we've counted the same motion twice. I know Cartesian are
orthogonal, so I'm safe.
T = Ttranslation + Trotation
T = 1/2 m [(vx-VX)2 + vy2] + 1/2 I ω2 + 1/2 M VX2
where vx is the speed of hoop relative to the wedge, and VX is the
speed of the wedge relative to the fixed tabletop.
Using the coordinate system of ξ-S, where S is the distance along the wedge,
and ξ is the table-top inertial coordinate system, we have:
y = -S sinα
S = Rθ
vx = S' cosα
Then we can convert our kinetic energy into:
T = 1/2 m [ (S' cosα - ξ')2 + (S' sinα)2] + 1/2 mR2(S'/R)2 + 1/2 M ξ'2
The potential energy, starting from the top of the wedge;
U = mgy = -mgS sinα
Finally, L = T - U, and our two Euler equations are:
(i) dL/dS - d/dt( dL/dS') = 0
(ii) dL/dξ - d/dt( dL/dξ') = 0
Which look like:
(i) 2mS" + mξ" cosα - mg sinα = 0
(ii)(m + M)ξ" + mS" cosα = 0
(b) The generalized momenta are found by taking dL/dq' = p, So
(i) pS = dL/dS' = mS' cosα2 - m cosαξ + mS' sin2α + mS'
(ii) pξ = dL/dξ' = (m+M)ξ'
(c) The Hamiltonian equations of motion are found by, q' = dH/dp, and -p'= dH/dq, so
H = T + U = pξ/2(m + M) + a mess in S
(ia) q'S = a mess
(ib) -p'S = -mg sinα (which means gravity is the only external S-force)
(iia) q'ξ = ξ' (which is exactly what we expect)
(iib) -p'ξ = 0 (which must be true because all ξ-forces are internal)
Problem 2:
Is it cheaper to dispose of nuclear waste by dumping it into the sun
or by sending it out of the solar system? How can Jupiter help?
Formulae:
E = -k/2a = 1/2 mv21 - k / r1
a = (r1 + r2) / 2
Thus the velocity at r1 is: vr1=
{(2k/mr1)[r2/(r1+r2)]}1/2
k = Gm1m2
G = 6.67x10-11
ms= 333,480 x 5.98 x 1024kg
res = 1.495 x 1011m
rs= 6.96 x 108
(a) The trick is to recognize that the cost in dumping the waste is
in the Δv needed to either decelerate it from earth orbit (and crash
into the sun) or accelerate from earth orbit to past the orbit of Pluto.
What we want is this ratio (which neatly cancels all the calculations of
earth orbit speed, vE=29.8km/s).
Δvsun = vE{[2r2/(r2+r1)]1/2 - 1}
Δvpluto = vE{[2r3/(r3+r1)]1/2 - 1}
Taking the ratio eliminates the need to calculate vE. Let r1 = 1AU,
r2 = rs = 1/214 AU, and r3=&infty;
Then the ratio becomes: [sqrt(2/214) - 1] / [sqrt(2) - 1] = (.903/.414) = 2.18
So going to the sun is a little more than twice the Δv than heading out of the
solar system. Since Δv is proportional to fuel, and fuel is around $10,000/lb in
LEO, we're talking at least twice as expensive.
(b) Jupiter can do two things: it can speed up a spacecraft by the "slingshot effect",
so that we only need fuel for making it to Jupiter orbit (not past Pluto), or, we can
use Jupiter to redirect our spacecraft (shed angular momentum) and crash right into
the sun. Either way, it makes getting rid of our nuclear fuel the same cost whichever
place we send it, namely, the cost of getting to Jupiter. Calculating that quantity as
a ratio of escape velocity:
Δv = [sqrt(10/6)-1]/.414 = .70
showing that Jupiter is only 70% of the Δv needed for escape, and therefore is a
better candidate for disposal. (For that matter, bury it in Jupiter, it can use the heat.)
Problem 3:
Newton's equation in a non-inertial reference frame looks like:
a = F/m - 2ω x v - R" - α x r - ω x (ω x r)
(a)Identify each term and estimate its relevance for a ball dropped from the Sears tower.
(b)If the latitude of Chicago is 42o what is the deflection for a ball dropped from the Sears tower, 400m high?
(a) Term 1 is the external acceleration, which in the case of our ball, is
just the acceleration due to gravity, g
Term 2 is the Coriolis force, which though small, is the main cause of deflection
of our dropped ball.
Term 3 is the acceleration of our reference frame origin, the center of the earth. It is accelerating around the sun, and the sun around the galaxy, but as we calculated for one
of the homework problems, its 0.06% of gravity and not worth considering.
Term 4 is the angular acceleration of the reference frame, the center of the earth. It changes by milliseconds per year due to water redistribution from oceans to glaciers, so again, its a 1:billion change not worth considering.
Term 5 (in the case that r is perpendicular to ω) is just the centrifugal force, which as we calculated in our homework, is 0.3% of g, and actually gets folded into the
local g vector we use for our plumb line. So we wouldn't even call this a deflection,
since we use a plumb line to tell us which way is "down".
(b) So our equation for the Sears tower becomes: a = g - 2ω x v
Now we follow the homework pretty closely.
g = 0 ex + 0 ey -g ez
ω = -ωsinλex + 0 ey +
ωcosλez
v = 0 ex + 0 ey -gt ez
Carrying out the cross product is the trickiest part of this problem, which can be
done mnemonically by writing the matrix down and taking the determinant (as illustrated
in the text), or more compactly using tensor notation and the Levi-Civita tensor, but
we just write down the result from the homework:
ω x v = -ωg t cosλey
Then putting the pieces together, we have the acceleration as:
ax = x" = 0
ay = y" = 2ωg t cosλ
az = z" = -g
Solution to the z-equation, and using the initial condition that z(0) = h, we have:
z(t) = h - 1/2 gt2
solving for t when the ball hits the ground gives: τ = sqrt(2h/g)
Solving for y(t) = 1/3ωg t3 cosλ
And plugging in for τ gives: y(τ) = 1/3 ω sqrt(8h3/g) cosλ
Finally, inserting ω=2π/(24 x 3600), and h=400m, and g=9.8m/s2, and
λ=42o, we get: 13cm.