PHY6426/Fall 2007: CLASSICAL MECHANICS
HOMEWORK ASSIGNMENT #2: SOLUTIONS
due by 9:35 a.m. Mon 09/10
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
1. [50
points]
Following
the
discussion
in
the
class
and
notes
posted
at
http://www.phys.ufl.edu/˜maslov/classmech/pendulum.pdf, find the period of oscillations for a pendulum which starts its motion at the angle φ0 to the vertical and with the initial kinetic energy T0 . Obtain an
asymptotic result for the period in the limit φ0 1 and T0 mgl.
Solution
Energy conservation
m 2 2
` φ̇ − mg` cos φ = T0 − mg` cos φ0
2
r
2g
φ̇ = −
(cos φ − cos φ0 + t0 ),
`
where t0 = T0 /mg`, where the choice of the sign corresponds to φ0 > 0 and subsequent downward motion. The
period is then
s Z
φmax
1
`
dφ √
,
T =4
2g 0
cos φ − cos φ0 + t0
where φmax is the maximum angle determined from the condition
cos φmax = cos φ0 − t0 .
The condition −1 ≤ cos φmax ≤ 1 implies that t0 ≤ cos φ0 + 1. For larger kinetic energies, the motion becomes
unbounded. Equation for the period can be re-written as
s Z
φmax
1
`
dφ √
T =4
,
2g 0
cos φ − cos φmax
upon which the problem reduces to the one considered in the class with the only change φ0 → φmax . Following
the same steps as before:
2 φmax
2 φ
cos φ − cos φmax = 2 sin
− sin
2
2
s Z
` φmax
dφ q
T = 2
g 0
1
sin2 (φmax /2) − sin2 (φ/2)
s Z
` φmax
1
dφ
q
= 2
2 (φ/2)
sin
g 0
sin (φmax /2) 1 −
sin2 (φmax /2)
New variable
sin (φ/2)
, 0 ≤ α ≤ π/2
sin (φmax /2)
1 cos (φ/2) dφ
cos αdα =
2 sin (φmax /2)
cos αdα
dφ
cos αdα
cos αdα
= 2q
= 2
= 2q
sin (φmax /2)
cos (φ/2)
2
1 − sin (φ/2)
1 − sin2 α sin2 (φ
sin α =
.
max /2)
2
Period
s Z
` π/2
T = 4
dα q
g 0
= 4
s
1
2
1 − sin α sin2 (φmax /2)
(1)
`
K (sin (φmax /2)) ,
g
where K (k) is the elliptic integral of the 1st kind.
When φ0 1 and t0 1, φmax is small too. Expanding both cosines in the relation
cos φmax = cos φ0 − t0 ,
we obtain
1−
φ2max
φ2
= 1 − 0 − t0
2
2
φ2max = φ20 + 2t0 .
The square root in Eq.(1) can be expanded for for small sin2 (φmax /2) ≈ φ2max /4, which leads to
s Z
1
` π/2
dα q
T = 4
g 0
1 − sin2 α sin2 (φ
/2)
max
≈
≈
=
=
=
s Z
` π/2
1
2
2
4
dα 1 + sin α sin (φmax /2) . . .
g 0
2
s Z
` π/2
1
4
dα 1 + sin2 αφ2max . . .
g 0
8
s ` π 1π 2
4
+
φ
...
g 2
8 4 max
s `
1
1 + φ2max . . .
2π
g
16
s `
1 2 1
2π
1 + φ0 + t 0 . . . .
g
16
8
The final result
s 1 2 1 T0
`
1 + φ0 +
T = 2π
...
g
16
8 mg`
2. [50 points] A disk rolls without slipping along a horizontal plane. The disk is constrained to remain vertical.
Let be ψ be the angle between the plane of the disk and the x axis of a fixed frame and θ be the angle measuring
spinning of the disk about its center (see Fig. 1). Following the variational formulation, presented in the class,
and neglecting the gravitational force, show that the center of mass of the disk moves in a circle, if ψ̇ (t = 0) 6= 0,
and along a straight line, if ψ̇ (t = 0) = 0.
Solution
A disk rolls without slipping along a horizontal plane. The disk is constrained to remain vertical. Let be ψ be
the angle between the plane of the disk and the x axis of a fixed frame and θ be the angle measuring spinning
of the disk about its center (see Fig. 1). Following the variational formulation, presented in the class, and
neglecting the gravitational force, show that the center of mass of the disk moves in a circle, if ψ̇ (t = 0) 6= 0,
and along a straight line, if ψ̇ (t = 0) = 0.
3
Solution
Lagrangian
L=
1
1
1
1
mẋ2 + mẏ + I0 θ̇2 + I1 ψ̇ 2
2
2
2
2
Components of the angular velocity
ωx = −θ̇ sin ψ
ωy = θ̇ cos ψ
Rolling constraint
~v = ω
~ × ~a = ωy Rx̂ − ωx Rŷ,
where ~a is the vector from the contact point to the center of rotation. In our case, ~a = (0, 0, R) , where R is the
radius of the disk. Constraint functions
f1 = Rθ̇ cos ψ − ẋ
f2 = Rθ̇ sin ψ − ẏ
Lagrange’s equations:
∂f2
∂f1
+ µ2
= −µ1
∂ ẋ
∂ ẋ
mÿ = . . . = −µ2
∂f1
∂f2
I0 θ̈ = µ1
+ µ2
= µ1 R cos ψ + µ2 R sin ψ
∂ θ̇
∂ θ̇
I1 ψ̈ = 0.
mẍ = µ1
From the last equation, it follows that ψ̇ = const or ψ = ψ (0)+ ψ̇ (0) t. From the contraint equations f1 = f2 = 0,
ẋ
θ̇
ẏ
R sin θ =
θ̇
R cos ψ =
Substituting these two equations into the equation for θ̈ gives
I0 θ̇θ̈ = µ1 ẋ + µ2 ẏ
or, subsituting the µ1,2 from the equations for ẍ and ÿ
I0 θ̇θ̈ = −m (ẋẍ + ÿ ẏ) .
Integrating,
2
I0 θ̇ + m (ẋ)2 + (ẏ)2 = C = const
Recalling that ẋ2 = θ̇2 R2 cos2 ψ and ẏ 2 = θ̇2 R2 sin2 ψ, we see that
from which it follows that
2
I0 θ̇ + mR2 θ̇2 = const,
θ̇ = const
4
and
θ = θ (0) + θ̇ (0) t.
This means that
or
ẋ = θ̇R cos ψ (t) = θ̇ (0) R cos ψ(0) + ψ̇ (0) t
ẏ = θ̇R sin ψ (t) = θ̇ (0) R sin ψ(0) + ψ̇ (0) t
θ̇ (0) R
sin ψ(0) + ψ̇ (0) t
ψ̇ (0)
θ̇ (0) R
cos ψ(0) + ψ̇ (0) t
y(t) = y (0) −
ψ̇ (0)
x(t) = x (0) +
2
2
[x (t) − x (0)] + [y (t) − y (0)] = R2
θ̇2 (0)
ψ̇ 2 (0)
.
ψ̇ (0) = 0 requires a special treatment. Coming back to Eqs. for ẋ and ẏ and setting ψ̇ (0) in there gives
ẋ = θ̇R cos ψ (t) = θ̇ (0) R cos (ψ(0))
ẏ = θ̇R sin ψ (t) = θ̇ (0) R sin (ψ(0))
or
x (t) = x (0) + θ̇ (0) R cos (ψ(0)) t
y (t) = y (0) + θ̇ (0) R sin (ψ(0)) t,
which describes a motion along a straight line.
5
z
y
ψ
θ
x
FIG. 1:
z
y
ψ
θ
x
FIG. 2: