HW#3—Solutions
—Phys410—Fall 2011
Prof. Ted Jacobson
Room 4115, (301)4056020
www.physics.umd.edu/grt/taj/410a/
[email protected]
7.20 (
particle on a helix
)
In cylindrical coordinates we have in general
T
=
1
2
m
( ˙
ρ
2
+
ρ
2
˙
φ
2
+ ˙
z
2
). Here
ρ
=
R
is constrained, and
φ
=
z/λ
, so
T
=
1
2
m
(1 +
R
2
/λ
2
) ˙
z
2
. Also the potential energy is
U
=
mgz
relative to an arbitrary
z
= 0 point. So
L
==
1
2
m
(1 +
R
2
/λ
2
) ˙
z
2

mgz
.
Lagrange’s equation gives ¨
z
=

g/
(1 +
R
2
/λ
2
). When
R
→
0 this becomes freefall,

g
, which makes sense because the helix becomes a straight vertical line.
7.21 (
bead on a spinning rod
) Let’s add two interesting parts (b): Suppose the rod has
length
L
from the pivot to one end, and the bead is released with zero radial velocity
at some initial radius
r
0
.
In the limit that
r
0
goes to zero, what velocity does the
bead have when it reaches the end of the rod? Give the components of the velocity
along the rod and perpendicular to the rod. Compare your result to what you would
expect from dimensional analysis. (c) (i) Explain why the Hamiltonian is a conserved
quantity, and show that it is equal to
E

Ω
J
, where
E
and
J
are the kinetic energy
and angular momentum of the bead, and Ω is the angular velocity of the rod (called
lower case omega in the book). (ii) Use this conservation law to evaluate the kinetic
energy when the bead leaves the rod, and show that it agrees with what you get using
your results from part (b).
Solution
:
(a)
L
=
1
2
m
( ˙
r
2
+
r
2
ω
2
).
The angular part of the kinetic energy looks
like an upside down harmonic oscillator. The Lagrange equation is ¨
r
=
ω
2
r
, whose
solutions are
r
(
t
) =
A
cosh(
ωt
) +
B
sinh(
ωt
). Then ˙
r
(
t
) =
ωA
sinh(
ωt
) +
ωB
cosh(
ωt
),
so ˙
r
(0) =
B
, so if the particle is released with no radial motion from
r
0
the solution
is
r
(
t
) =
r
0
cosh(
ωt
). For
ωt
1 this becomes
r
(
t
)
’
(
r
0
/
2)
e
ωt
. The only force in
this situation is the force of constraint which acts at any instant perpendicular to
the rod, in the
φ
direction. Thus although ¨
r
is nonzero, the radial component of the
acceleration vector is actually zero.
How can that be?
It’s because it also has an
equal but opposite, inward pointing radial centripetal component to its acceleration,
rω
2
. In the rotating frame of reference (we’ll get to that later), this means there is an
apparent “centrifugal force”
mrω
2
, pointing outward.
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 Fall '11
 Jacobson
 Physics, mechanics, Energy, Kinetic Energy, Potential Energy, Cos, Lagrangian mechanics

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