11 Home exercise sheet

11
Home exercise sheet
Exercise 11.1:
A unsuspecting solid spherical planet of mass M0 rotates with angular velocity ω0 . Suddenly,
a giant asteroid of mass αM0 smashes into and sticks to the planet at a location which is at
polar angle θ relative to the initial rotational axis. The new mass distribution is no longer
spherically symmetric, and the rotational axis will precess. Recall Eulers equation
~
dL
~ =N
~ ext
+ω
~ ×L
dt
(1)
for rotations in a body-fixed frame.
1. What is the new inertia tensor Iαβ along principle center-of-mass frame axes? Dont
forget that the CM is no longer at the center of the sphere! Recall I = 25 M R2 for a
solid sphere.
2. What is the period of precession of the rotational axis in terms of the original length
of the day ω2π0 ?
Solution:
1. Let’s choose body-fixed axes with ẑ pointing from the center of the planed to the smoldering asteroid. The CM lies a distance
d=
αM0 · R + M0 · 0
α
R
=
(1 + α)M0
1+α
(2)
from the center of the sphere. Thus, relative to the center of the sphere, we have




1 0 0
1 0 0
2
I = M0 R2 0 1 0 + αM0 R2 0 1 0
(3)
5
0 0 1
0 0 0
Now we shift to a frame with the CM at the origin, using the parallel axis theorem,
CM
2
~
~
Iαβ (d) = Iαβ + M d δαβ − dα dβ
(4)
1
Thus, with d~ = dẑ,

CM
Iαβ





1 0 0
1 0 0
1 0 0
2
=
M0 R2 0 1 0 + αM0 R2 0 1 0 − (1 + α)M0 d2 0 1 0 (5)
5
0 0 1
0 0 0
0 0 0

2
α
+ 1+α
0
0
5
2
2
α
0
+ 1+α 0 
(6)
= M0 R
5
2
0
0
5
In the absence of external torgues, Euler’s equations along principal axes read
dω1
= (I2 − I3 ) ω2 ω3
dt
dω2
I2
= (I3 − I1 ) ω3 ω1
dt
dω3
= (I1 − I2 ) ω1 ω2
I3
dt
I1
(7)
(8)
(9)
Since I1 = I2 , ω3 (t) = ω3 (0) = ω0 cos θ is a constant. We then obtain ω˙1 = Ωω2 , and
ω˙2 = −Ωω1 , with
Ω−
5α
I2 − I3
ω3 =
ω3
I1
7α + 2
(10)
2. The period of precession τ in units of the pre-cataclysmic day is
τ
ω
7α + 2
= =
T
Ω
5α cos θ
Exercise 11.2: Rotating Disk
2
(11)
A disk of mass m and radius a is mounted in the middle of a massless rod of length l. The
plane of the disk is tilted by an angle θ away from the normal to the rod. Initially the disk
is at rest tilted up, as shown in the figure. At t = 0 it begins to rotate around the axis of the
rod with an angular velocity ω in the direction shown. The rod is supported at the two ends
and forces F1 (t) and F2 (t) are such that it rotates without wobble with a constant angular
velocity ω.
1. What are the directions of the principal axes of inertia of the disk?
2. Calculate the principal moments of inertia of the disk.
3. Determine the torques needed to keep the disk spinning without wobble.
4. Determine the components F1x (t), F1y (t), F2x (t), F2y (t) in the fixed frame of reference
that are needed to keep the disk spinning without wobble.
Solution:
1. One axis is perpendicular to the plane of the disk, call it ê3 , the other 2 are orthogonal
to each other and lie in the plane of the disk. Take ê1 to be horizontal out of the page
and ê2 up in the plane of the page before the rotation starts.
2.
I3
I1
a
ma2
2πr2 rdr =
2
0
Z 2π Z a
m
ma2
= I2 = 2
(r sin θ)2 rdrdθ =
πa 0
4
0
m
=
πa2
Z
3
(12)
(13)
3. We first write the components of the angular velocity ω in the body frame of reference
ω3 = ω cos θ
ω2 = −ω sin θ
ω1 = 0
(14)
(15)
(16)
We now use Euler equations to find the torgues
N1
N2
N3
ma2
= I1 ω˙1 − ω2 ω3 (I2 − I3 ) = −ω sin θ cos θ
4
= I2 ω˙2 − ω3 ω1 (I3 − I1 ) = 0
= I3 ω˙3 − ω1 ω2 (I1 − I2 ) = 0
2
(17)
(18)
(19)
4. The torques are in the rotating frame and need to be projected back to the lab frame.
With the disk oriented as shown: (F2x − F1x ) l = N1 ; (F1y − F2y ) l = N2 ;
After rotating by 90 degrees (F1y − F2y ) l = N1 ; (F1x − F2x ) l = N2 ;
ma2 ω 2
sin θ cos θ cos ωt
4
ma2 ω 2
(F1y − F2y ) l = N2 cos ωt + N1 sin ωt = −
sin θ cos θ sin ωt
4
(F1x − F2x ) l = −N1 cos ωt + N2 sin ωt =
(20)
(21)
mg ma2 ω 2
mg ma2 ω 2
+
sin θ cos cos ωt, F2x =
−
sin θ cos θ cos ωt (22)
2
8l
2
4
ma2 ω 2
ma2 ω 2
= −
sin θ cos cos ωt, F2y =
sin θ cos θ cos ωt.
(23)
8l
8l
F1x =
F1y
Exercise 11.3: Rotating Rectangle
Consider a rectangulare cube of density ρ and dimensions a × b × c, as depicted in the figure.
4
1. Compute the inertia tensor Iαβ along body-fixed principle axes, with the origin at the
center of mass.
2. Shifting the origin to the center of either of the b × c faces, and keeping the axes
parallel,compute the new inertia tensor.
3. A mass less torsional fiber is (masslessly) welded along the diagonal of either b × c face.
The potential energy in this fiber is given by U (θ) = 12 Y θ2 , where Y is a constant and
θ is the angle of rotation of the fiber. Neglecting gravity, find an expression for the
oscillation frequency of the system.
Solution:
1. We first compute Izz
CM
Izz
Z
a
2
=ρ
Z
b
2
dx
− a2
Z
c
2
dy
− 2b
− 2c
1
dz x2 + y 2 = M a2 + b2
12
(24)
where M = ρabc. Corresponding expressions hold for the other moments of inertia.
Thus,
 2

b + c2
0
0
1
a2 + c 2
0 
(25)
I CM = M  0
12
2
2
0
0
a +b
2. We shift the origin by a distance d~ = − 21 ax̂ and use the parallel axis theorem,
Iαβ = Iαβ (0) + M d~2 δαβ − dα dβ
(26)
resulting in


b2 + c 2
0
0
4a2 + c2
0 
I= 0
2
0
0
4a + b2
(27)
3. Let θ be the twisting angle of the fiber. The kinetic energy in the fiber is
1
Iαβ ωα ωβ
2
1
=
nα Iαβ nβ θ̇2
2
T =
5
(28)
(29)
where
n̂ = √
bŷ
cẑ
+√
b2 + c 2
b2 + c 2
(30)
We then find
1
1
b2 c 2
2
Iaxis ≡ nα Iαβ nβ = M a + M 2
3
6 b + c2
q
Y
The frequency of oscillations is then Ω = Iaxis
, or
s
Ω=
b2 + c 2
6Y
· 2 2
M 2a (b + c2 ) + b2 c2
6
(31)
(32)