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PC235 Winter 2013
Classical Mechanics
Assignment #9 Solutions
#1 (10 points) JRT Prob. 11.
A massless spring (force constant k 1
) is suspended from the ceiling,
with a mass m 1
hanging from its lower end. A second massless spring (force
constant k 2
) is suspended from m 1
, and a second mass m 2
is suspended from
the second spring’s lower end. Assuming that the masses move only in a
vertical direction and using coordinates y 1 and y 2 measured from the masses’
equilibrium positions, show that the equations of motion can be written in
the matrix form M¨y = −Ky, where y is the 2 × 1 column made up of y 1
and y 2
. Find the matrices M and K.
Solution
Let ˆy 1
and ˆy 2
be the extensions of the two springs from their unstretched
lengths and y 10
and y 20
be their values at equilibrium. The displacements
from equilibrium are
y 1 = ˆy 1 − y 10 and y 2 = ˆy 2 − y 20. (1)
The net downward forces on the two masses are
F 1 = m 1 g − k 1 yˆ 1 + k 2 (ˆy 2 − yˆ 1 ) and F 2 = m 2 g − k 2 (ˆy 2 − yˆ 1 ), (2)
and thus the conditions for equilibrium are
m 1
g = k 1
y 10
= k 2
(y 20
− y 10
) and m 2
g = k 2
(y 20
− y 10
Now, applying Newton’s 2nd law and using eq. (1) to eliminate ˆy 1
and ˆy 2
from eq. (2), we find that
m 1
y¨ 1
= F
1
= m 1
g − k 1
(y 1
) + k 2
[y 2
− y 1
− y 10
)] (4)
= −k 1
y 1
(y 2
− y 1
(the equilibrium condition is used to eliminate several terms in the last line.)
Similarly,
m 2
y¨ 2
= F
2
= m 2
g − k 2
(ˆy 2
− yˆ 1
) = −k 2
(y 2
− y 1
The last two results combine to give the matrix equation My¨ = −Ky, where
M =
m 1 0
0 m 2
and K =
k 1 + k 2 −k 2
−k 2
k 2
#2 (15 points) JRT Prob. 11.
(a) Write down the equations of motion corresponding to eq. (11.2) for
the case of two equal-mass carts with three identical springs, but with
each cart subjected to a linear resistive force −bv (same coefficient b
for both carts).
(b) Show that if you change variables to the normal coordinates ξ 1
1
2
(x 1
x 2 ) and ξ 2 =
1
2
(x 1 − x 2 ), the equations of motion for ξ 1 and ξ 2 are
uncoupled.
(c) Write down the general solutions for the normal coordinates and hence
for x 1
and x 2
(assume that b is small, so that the oscillations are un-
derdamped.)
(d) Find x 1
(t) and x 2
(t) for the initial conditions x 1
(0) = A and x 2
v 1
(0) = v 2
(0) = 0, and plot them for 0 ≤ t ≤ 10 π using the values
A = k = m = 1, b = 0.1.
Solution
(a) As in section 5.4, we define β = b/ 2 m and ω
2
0
= k/m. Then, the
equations of motion are
x ¨ 1 = − 2 β x˙ 1 − 2 ω
2
0
x 1 + ω
2
0
x 2
x ¨ 2
= − 2 β x˙ 2
2
0
x 1
− 2 ω
2
0
x 2
(b) If you take first the sum and then the difference of these two equations,
you will get the uncoupled equations
ξ 1
= − 2 β
ξ 1
− ω
2
0
ξ 1
and
ξ 2
= − 2 β
ξ 2
− 3 ω
2
0
ξ 2
in equilibrium with all three springs relaxed (length equal to unstretched
length). What are the normal frequencies? Find and describe the normal
modes. Consider only small displacements from equilibrium.
y
x
L
L
O
k
k
k’
Fig. 2: Geometry for Question #
Solution
Let the equilibrium length of the first 2 springs be L and that of the one
that connects the two masses be
2 L. Let x and y be the displacements
of the two masses from their equilibrium positions; these will be our two
generalized coordinates. The total KE is T =
1
2
m( ˙x
2
2 ) and the total PE
is U =
1
2
k(x
2
2 ) +
1
2
k
′ z
2 , where z is the extension of the diagonal spring
(which is a function of x and y; that will take a bit of work to figure out).
Since we are interested only in small oscillations, we can write
z =
p
(L + x)
2
2 −
2 L ≈
p
2 L
2
2 L (14)
2 L
p
1 + (x + y)/L − 1
2 L ·
(x + y)/L, (16)
where for the last expression on the first line we dropped terms higher than
1st order in x and y and for the final expression we used the binomial expan-
sion for the square root. Therefore, we can write z
2
= (x + y)
2
/2, and the
total PE is
U =
k(x
2
2
) +
k
′
(x + y)
2
=
k +
k
′
x
2
k +
k
′
y
2
′
xy
Writing down Lagrange’s equations for x and y leads to
M =
m 0
0 m
and K =
k +
k
′
2
k
′
2
k
′
2
k +
k
′
2
Next, we set det(K − ω
2
M) = 0, or (mω
2
− k)(mω
2
− k − k
′
) = 0. Thus, the
normal frequencies are
ω 1
r
k
m
and ω 2
r
k + k
′
m
For the first normal mode, solving (K − ω
2
1
M)a = 0 gives a 1
= −a 2
; the
masses oscillate with equal amplitudes but out of step. In this mode, the
diagonal spring’s length remains constant (at least in the small-oscillation
approximation), which is why k
′ is irrelevant to ω 1
. For the second normal
mode, we have a 1 = a 2. Here, the masses move with equal amplitudes and
both in step (both x and y increase together and decrease together). In this
mode, the diagonal spring does stretch and compress; this extra contribution
to the PE increases the frequency of oscillations.
#4 (10 points) JRT Prob. 11.
A bead of mass m is threaded on a frictionless circular wire hoop of radius R
and mass m. The hoop is suspended at the point A and is free to swing in its
own vertical plane as shown in Fig. 11.20 of the text. Using the angles φ 1
and
φ 2
as generalized coordinates, solve for the normal frequencies of small oscil-
lations, and find and describe the motion in the corresponding normal modes.
Solution
The moment of inertia of the hoop about its edge is I = 2mR
2 , so its kinetic
energy is
T
1
I
φ
2
1
= mR
2 ˙ φ
2
1
The velocity of the bead is (for small oscillations) the sum of the velocity of
the bead relative to the hoop’s center and the velocity of the hoop’s center
mass i is R
φ i
, and the total kinetic energy is
T =
2 m
R
φ 1
2
R
φ 2
2
R
φ 3
2
mR
2
φ
2
1
φ
2
2
φ
2
3
The three springs contribute to the potential energy. We have
U =
kR
2
(φ 1
− φ 2
2
− φ 3
2
− φ 1
2
= kR
2
φ
2
1
2
2
2
3
− φ 1
φ 2
− φ 2
φ 3
− φ 3
φ 1
The Lagrangian is then
L = T − U (31)
mR
2
φ
2
1
φ
2
2
φ
2
3
− kR
2
φ
2
1
2
2
2
3
− φ 1
φ 2
− φ 2
φ 3
− φ 3
φ 1
This leads to the Lagrange’s equations
∂L
∂φ 1
d
dt
∂L
φ 1
= −kR
2
(2φ 1
− φ 2
− φ 3
) = 2mR
2 ¨ φ 1
∂L
∂φ 2
d
dt
∂L
φ 2
= −kR
2
(−φ 1 + 2φ 2 − φ 3 ) = mR
2 ¨ φ 2 (34)
∂L
∂φ 3
d
dt
∂L
φ 3
= −kR
2
(−φ 1
− φ 2
) = mR
2 ¨ φ 3
We can immediately cancel out the R
2 terms from all equations. Then, the
equations become M
φ = −Kφ, where
M =
2 m 0 0
0 m 0
0 0 m
and K =
2 k −k −k
−k 2 k −k
−k −k 2 k
We then need to find all ω such that det(K−ω
2
M) = 0. We start by dividing
this matrix by m and making the substitution k/m = ω
2
0
K − ω
2
M =
2 k − 2 mω
2
−k −k
−k 2 k − mω
2 −k
−k −k 2 k − mω
2
2(1 − ω
2
) − 1 − 1
− 1 2 − ω
2 − 1
− 1 − 1 2 − ω
2
The characteristic equation is found by setting the determinant of this matrix
to zero. The algebra proceeds as follows:
det(K − ω
2
M) = (2 − 2 ω
2
)
(2 − ω
2
)
2
− 1
−(2 − ω
2
) − 1
1 + (2 − ω
2
)
= (2 − 2 ω
2
)
3 − 4 ω
2
4
− (2 − ω
2
) − 1 − 1 − (2 − ω
2
)
= (2 − 2 ω
2
)
3 − 4 ω
2
4
− 6 + 2ω
2
= −2(ω
6
− 5 ω
4
2
)
This produces the characteristic equation ω
6 − 5 ω
4
2 = 0. Since there is
no constant term on the left-hand side, one “obvious” normalized frequency
is ω 1
= 0. Factoring this solution out gives us the quadratic equation ω
4
−
5 ω
2
- 6 = 0, which gives us ω 2
p
2 k/m and ω 3
p
3 k/m. By substituting
ω 1 = 0 into the equation (K − ω
2
M)a = 0 , we find that a 1 = a 2 = a 3. That
is, the masses move with constant speed around the hoop at their equilibrium
separation (and hence none of the springs are ever stretched or compressed -
thus the zero frequency). For ω 2
, we find that a 1
= −a 2
= −a 3
; here, mass 1
oscillates in one direction while the other two oscillate in the other direction
(all amplitudes being equal). For ω 3 , a 1 = 0 and a 2 = −a 3. Here, the heavier
mass is stationary while the other two oscillate with equal amplitudes and
completely out of phase. These solutions are similar to those found for the
linear triatomic molecule in the class notes.
#6 (5 points) JRT Prob. 10.
(a) Write down the integral for the moment of inertia of a uniform cube of
side a and mass M , rotating about an edge, and show that it is equal
to
2
3
M a
2
. (1 point)
(b) If I balance this cube on an edge in unstable equilibrium on a rough
table, it will eventually topple and rotate until it hits the table. By
considering the energy of the cube, find its angular velocity just before
it hits the table. (4 points)
Solution
(a) This is example 10.2 from the text. For a uniform solid cube rotating
around its edge, I =
2
3
M a
2
.
energy is conserved (it might be an inelastic collision). However, angu-
lar momentum is always conserved; in particular the component of L
about the edge of the step, which we call L y
(see attached figure). We
find that
L
y
X
m α
r α
×v = M R×v = M
a
v sin
5 π
M av
(the negative sign simply indicates that L is directed into the page).
Immediately after the collision, the cube is rotating, and we have
L
y
= Iω 0
M a
2
ω 0
where the subscript on ω 0 indicates that this is the angular velocity
immediately after the collision; ω will decrease over time as the cube
tips. Equating these two expressions for L y
(that is, conserving angular
momentum), we find that ω 0
3 v
4 a
. This tells us the angular frequency
of the initial rotation as a function of the initial velocity of the cube.
(b) To continue this problem, realize that we are running problem JRT10-
15 in reverse. There, we balanced the cube on its edge (with ω = 0
initially), and calculated ω at the point where it fell on its side. Here,
we wish to start with an (unknown) angular velocity that results in
ω = 0 at the tipping point. Therefore, we rewrite our velocity as a
function of ω 0
, and substitute ω 0
q
3 g(
√
2 −1)
2 a
from that assignment
solution. The resulting velocity is
v =
s
8 ga(
#8 (5 points) JRT Prob. 10.
Consider a rigid plane body or “lamina,” such as a flat piece of sheet
metal, rotating about a point O in the body. If we choose axes so that the
lamina lies in the xy plane, which elements of the inertia tensor are auto-
matically zero? Prove that I zz
= I
xx
+ I
yy
Solution
Fig. 3: Geometry of question #
Since the whole body lies in the plane z = 0, the four products of inertia
involving z are all zero: Ixz = Iyz = Izx = Izy = 0. For the same reason,
I
xx
+I
yy
X
m α
(y
2
α
+z
2
α
X
m α
(z
2
α
+x
2
α
X
m α
(x
2
α
+y
2
α
) = I
zz
#9 (10 points)
A rectangular “brick” of mass M is positioned with one corner at the origin
and with side lengths a, b, and c in the x−, y−, and z−directions, respec-
tively.
(a) Calculate the inertia tensor I with respect to the origin. (10 points)
(b) Suppose that the brick is rotating with angular velocity ω, about the
y−axis. Find the resulting angular momentum L. (3 points)
(c) Discuss whether or not ˆx, ˆy, and ˆz constitute a set of principal axes
for the brick. (2 points)
Solution
(a) Note that this is essentially Example 10.2 from the text. We just need
to change the limits of integration to account for the fact that we hae
a rectangular brick and not a cube. We’ll start with I xx
. Noting that
As for the products of inertia, we have
I
xy
Z
a
0
dx
Z
b
0
dy
Z
c
0
dz̺ xy = −
M c
abc
Z
a
0
dx
Z
b
0
dy xy (54)
M c
abc
Z
a
0
dx
xy
2
b
0
M c
abc
Z
a
0
dx
xb
2
M c
abc
x
2
b
2
a
0
M c
abc
a
2
b
2
M ab. (56)
By observing the symmetries inherent to the problem (and by recall-
ing that the inertia matrix must be symmetric), we find that I yx
−M ab/4, Ixz = Izx = −M ac/4, and Iyz = Izy = −M bc/4. Therefore,
the inertia matrix can be written
I =
M
4 (b
2
2 ) − 3 ab − 3 ac
− 3 ab 4 (a
2
2
) − 3 bc
− 3 ac − 3 bc 4 (a
2
2 )
It is easy to verify that if the brick is changed to a cube - that is, if
a = b = c, I takes the same form as in example 10.2 of the text.
(b) This particular angular momentum can be written ω = (0, ω, 0). There-
fore,
L = Iω =
M ωab
M ω (a
2
2 )
M ωbc
(c) Since rotation about the y−axis does not produce an angular momen-
tum L parallel to ω, it is clear that ˆx, ˆy, and ˆz do not constitute a set
of principal axes for the brick. You can also claim that this is the case
because of the non-zero off-diagonal elements of I.
#10 (10 points) JRT Prob. 10.
Find the inertia tensor for a uniform, thin hollow cone, such as an
ice-cream cone, of mass M , height h, and base radius R, spinning about its
pointed end.
Solution
Since this is a thin cone, it has an areal mass density σ (mass/area). All
points on the cone can be described by the two cylindrical coordinates ρ and
φ (you can use z and φ instead, but it’s very slightly less convenient). Refer
to the figure below, and imagine dividing the surface into strips as shown,
and then dividing the strips into small increments of angle dφ. Then, the
element of area is dA = (dρ/ sin α)(ρ dφ), where α is the half-angle of the
cone. The moment about the z axis is therefore
I
zz
Z
σ(x
2
2
)dA = σ
Z
R
0
Z
2 π
0
ρ
2
ρ dρ dφ
sin α
σπR
4
2 sin α
since the φ integral evaluates to 2π and the ρ integral evaluates to R
4 /4.
The area of the cone is A = πR
2
/ sin α (you can check this by performing
the integral A =
R
dA). Therefore, σπR
2 / sin α = M , the total mass. All
together, we have I zz
= M R
2 / 2.
The other two moments, Ixx and Iyy, must be identical by symmetry. For
the first of these,
Ixx =
Z
σ(y
2
2
)dA. (60)
The first term here is the same as the second term in I zz
. Since the two terms
in I zz
are equal (again, by symmetry), we conclude that the first term in I xx
is equal to Izz /2. In the second term of Ixx we can replace z by ρh/R, from
which we see that the second term in I xx
is h
2 /R
2 times I zz
. All together,
we find that
I
xx
= I
yy
h
2
R
2
I
zz
M (R
2
2
). (61)
Finally, all of the off-diagonal terms in I are zero by rotational symmetry
about the z axis (as was the case with the solid cone). Therefore,
I =
M
(R
2
2 ) 0 0
0 (R
2
2 ) 0
0 0 2 R
2
#11 (10 points) JRT Prob. 10.
A rigid body consists of three equal masses (m) fastened at the posi-
tions (a, 0 , 0), (0, a, 2 a), and (0, 2 a, a).
(a) Find the inertia tensor I.
(b) The characteristic equation is
det(I − λ 1 ) = (10ma
2
− λ)
2
(2ma
2
− λ) = 0. (70)
Therefore, the principal moments are λ 1
= λ 2
= 10ma
2 and λ 3
= 2ma
2 .
If we set λ = 10ma
2 , the equation (I−λ 1 )ω = 0 yields three equations:
0 = 0, ω 2
= 0, and ω 2
= 0. This tells us that ω 2
= −ω 3
, and
that the two normalized eigenvectors corresponding to λ = 10ma
2 are
e 1
= (1, 0 , 0), and e 2
(note that any two perpendicular directions in the plane defined by e 1
and e 2
are also suitable principal axes.)
Setting λ = 2ma
2 , the equation (I − λ 1 )ω = 0 yields three equations,
ω 1 = 0, ω 2 − ω 3 = 0, and −ω 2 + ω 3 = 0. When normalized, this defines
the principal axis
e 3
