PC235 Winter 2013
Classical Mechanics
Assignment #9 Solutions
#1 (10 points) JRT Prob. 11.2
A massless spring (force constant k1 ) is suspended from the ceiling,
with a mass m1 hanging from its lower end. A second massless spring (force
constant k2 ) is suspended from m1 , and a second mass m2 is suspended from
the second spring’s lower end. Assuming that the masses move only in a
vertical direction and using coordinates y1 and y2 measured from the masses’
equilibrium positions, show that the equations of motion can be written in
the matrix form Mÿ = −Ky, where y is the 2 × 1 column made up of y1
and y2 . Find the matrices M and K.
Solution
Let ŷ1 and ŷ2 be the extensions of the two springs from their unstretched
lengths and y10 and y20 be their values at equilibrium. The displacements
from equilibrium are
y1 = ŷ1 − y10
and y2 = ŷ2 − y20 .
(1)
The net downward forces on the two masses are
F1 = m1 g − k1 ŷ1 + k2 (ŷ2 − ŷ1 ) and F2 = m2 g − k2 (ŷ2 − ŷ1 ),
(2)
and thus the conditions for equilibrium are
m1 g = k1 y10 = k2 (y20 − y10 ) and m2 g = k2 (y20 − y10 ).
(3)
Now, applying Newton’s 2nd law and using eq. (1) to eliminate ŷ1 and ŷ2
from eq. (2), we find that
m1 ÿ1 = F1 = m1 g − k1 (y1 + y10 ) + k2 [y2 − y1 + (y20 − y10 )]
= −k1 y1 + k2 (y2 − y1 )
(4)
(5)
(the equilibrium condition is used to eliminate several terms in the last line.)
Similarly,
m2 ÿ2 = F2 = m2 g − k2 (ŷ2 − ŷ1 ) = −k2 (y2 − y1 ).
(6)
1
The last two results combine to give the matrix equation Mÿ = −Ky, where
k1 + k2 −k2
m1 0
and K =
M=
.
(7)
−k2
k2
0 m2
#2 (15 points) JRT Prob. 11.10
(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 ξ1 = 12 (x1 +
x2 ) and ξ2 = 12 (x1 − x2 ), the equations of motion for ξ1 and ξ2 are
uncoupled.
(c) Write down the general solutions for the normal coordinates and hence
for x1 and x2 (assume that b is small, so that the oscillations are underdamped.)
(d) Find x1 (t) and x2 (t) for the initial conditions x1 (0) = A and x2 (0) =
v1 (0) = v2 (0) = 0, and plot them for 0 ≤ t ≤ 10π using the values
A = k = m = 1, b = 0.1.
Solution
(a) As in section 5.4, we define β = b/2m and ω02 = k/m. Then, the
equations of motion are
ẍ1 = −2β ẋ1 − 2ω02 x1 + ω02 x2
ẍ2 = −2β ẋ2 + ω02 x1 − 2ω02 x2 .
(8)
(b) If you take first the sum and then the difference of these two equations,
you will get the uncoupled equations
ξ¨1 = −2β ξ˙1 − ω02 ξ1
and ξ¨2 = −2β ξ˙2 − 3ω02 ξ2 .
2
(9)
(c) The equation for ξ1 is exactly the equation (5.28) that we found for
a single damped oscillator, and has the solution (5.37), which we can
rewrite as
q
−βt
ξ1 (t) = e (B1 cos ω1 t + C1 sin ω1 t), where ω1 = ω02 − β 2 (10)
ξ2 (t) = e−βt (B2 cos ω2 t + C2 sin ω2 t), where ω2 =
q
3ω02 − β 2 . (11)
The expressions for x1 (t) and x2 (t) follow at once by adding and subtracting these expressions for ξ1 (t) and ξ2 (t).
(d) The given initial conditions imply that ξ1 (0) = ξ2 (0) = A/2, with
both derivatives zero. Therefore, B1 = B2 = A2 , C1 = βA/2ω1 , and
C2 = βA/2ω2 , from which we can write down
β
A −βt
cos ω1 t +
e
sin ω1 t
(12)
ξ1 (t) =
2
ω1
A −βt
β
ξ2 (t) =
cos ω2 t +
e
sin ω2 t .
(13)
2
ω2
1
0.5
0.5
2
1
0
x
x1
The sum and difference of these two functions give us x1 (t) and x2 (t),
which are plotted below.
−0.5
−1
0
−0.5
0
5
10
15
20
25
−1
30
0
5
10
t
15
20
25
30
t
Fig. 1: x1 (t) and x2 (t) for question #2
#3 (15 points) JRT Prob. 11.18
Two equal masses m are constrained to move without friction, one on the
positive x axis and one on the positive y axis. They are attached to two
identical springs (force constant k) whose other ends are attached to the
origin. In addition, the two masses are connected to each other by a third
spring of force constant k ′ . The springs are chosen so that the system is
3
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
L
k’
k
x
k
O
L
Fig. 2: Geometry for Question #3
Solution
Let the equilibrium length of the √
first 2 springs be L and that of the one
that connects the two masses be 2L. 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 = 12 m(ẋ2 + ẏ 2 ) and the total PE
is U = 12 k(x2 + y 2 ) + 12 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
p
p
√
√
z = (L + x)2 + (L + y)2 − 2L ≈
2L2 + 2L(x + y) − 2L (14)
√ p
=
2L
1 + (x + y)/L − 1 (15)
√
1
(16)
≈
2L · (x + y)/L,
2
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 expansion for the square root. Therefore, we can write z 2 = (x + y)2 /2, and the
4
total PE is
1 ′
1
k′
k′
1
2
2
2
2
2
′
k+
x + k+
y + k xy .
U = k(x + y ) + k (x + y) =
2
4
2
2
2
(17)
Writing down Lagrange’s equations for x and y leads to
′
k′
k + k2
m 0
2
and K =
M=
.
(18)
′
k′
0 m
k + k2
2
Next, we set det(K − ω 2 M) = 0, or (mω 2 − k)(mω 2 − k − k ′ ) = 0. Thus, the
normal frequencies are
r
r
k
k + k′
ω1 =
.
(19)
and ω2 =
m
m
For the first normal mode, solving (K − ω12 M)a = 0 gives a1 = −a2 ; 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 ω1 . For the second normal
mode, we have a1 = a2 . 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. 11.26
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 φ1 and
φ2 as generalized coordinates, solve for the normal frequencies of small oscillations, and find and describe the motion in the corresponding normal modes.
Solution
The moment of inertia of the hoop about its edge is I = 2mR2 , so its kinetic
energy is
1
(20)
T1 = I φ̇21 = mR2 φ̇21 .
2
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
5
relative to the pivot point. Thus, the bead’s speed is equal to R φ̇1 + φ̇2 .
The total kinetic energy is therefore
1
(21)
T = mR2 3φ̇21 + 2φ̇1 φ̇2 + φ̇22 .
2
The total potential energy is
U = U1 + U2
= mgR (1 − cos φ1 ) + mgR [(1 − cos φ1 ) + (1 − cos φ2 )]
1
≈
mgR 2φ21 + φ22
2
(22)
(23)
(24)
(we attribute the entire mass of the hoop to a point at its CM). Therefore,
the matrices are
2 0
3 1
2 2
2
,
(25)
and K = mR ω0
M = mR
0 1
1 1
where ω02 = g/R. Setting the determinant of K − ω 2 M to zero, we find the
normal frequencies
1
ω1 = √ ω0
2
and ω2 =
√
2ω0 .
(26)
The first leads to the normal mode where the angles oscillate in phase with
equal amplitudes. The second leads to the normal mode where the angles
oscillate 180 degrees out of phase with the amplitude of φ2 twice that of φ1 .
#5 (10 points) JRT Prob. 11.31
Consider a frictionless rigid horizontal hoop of radius R. Onto this hoop I
thread three beads with masses 2m, m, and m, and, between the beads, three
identical springs, each with force constant k. Solve for the three normal frequencies and find and describe the three normal modes.
Solution
This problem is best solved using Lagrangian methods. The masses are constrained to move along a circular arc of radius R. Therefore, the speed of
6
mass i is Rφ̇i , and the total kinetic energy is
2
2
2 1
2m Rφ̇1 + m Rφ̇2 + m Rφ̇3
T =
2
1
=
mR2 2φ̇21 + φ̇22 + φ̇23 .
2
The three springs contribute to the potential energy. We have
1 2
kR (φ1 − φ2 )2 + (φ2 − φ3 )2 + (φ3 − φ1 )2
U =
2
= kR2 φ21 + φ22 + φ23 − φ1 φ2 − φ2 φ3 − φ3 φ1 .
(27)
(28)
(29)
(30)
The Lagrangian is then
L = T −U
(31)
1
=
mR2 2φ̇21 + φ̇22 + φ̇23 − kR2 φ21 + φ22 + φ23 − φ1 φ2 − φ2 φ3 − φ3 φ(32)
1 .
2
This leads to the Lagrange’s equations
∂L
d ∂L
=
= −kR2 (2φ1 − φ2 − φ3 ) = 2mR2 φ̈1
∂φ1
dt ∂ φ˙1
d ∂L
∂L
=
= −kR2 (−φ1 + 2φ2 − φ3 ) = mR2 φ̈2
∂φ2
dt ∂ φ˙2
∂L
d ∂L
=
= −kR2 (−φ1 − φ2 + 2φ3 ) = mR2 φ̈3 .
∂φ3
dt ∂ φ˙3
(33)
(34)
(35)
We can immediately cancel out the R2 terms from all equations. Then, the
equations become Mφ̈ = −Kφ, where




2k −k −k
2m 0 0
(36)
M =  0 m 0  and K =  −k 2k −k  .
−k −k 2k
0 0 m
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 = ω02 = 1:


2k − 2mω 2
−k
−k

−k
2k − mω 2
−k
(37)
K − ω2M = 
2
−k
−k
2k − mω


2
2(1 − ω )
−1
−1
−1
2 − ω2
−1  .
(38)
= 
−1
−1
2 − ω2
7
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 − 2ω 2 ) (2 − ω 2 )2 − 1 + 1 −(2 − ω 2 ) − 1 − 1 1 + (2 − ω 2 )
= (2 − 2ω 2 ) 3 − 4ω 2 + ω 4 − (2 − ω 2 ) − 1 − 1 − (2 − ω 2 )
= (2 − 2ω 2 ) 3 − 4ω 2 + ω 4 − 6 + 2ω 2
= −2(ω 6 − 5ω 4 + 6ω 2 )
=0
(39)
This produces the characteristic equation ω 6 − 5ω 4 + 6ω 2 = 0. Since there is
no constant term on the left-hand side, one “obvious” normalized frequency
is ω1 = 0. Factoring this solutionpout gives us the p
quadratic equation ω 4 −
5ω 2 + 6 = 0, which gives us ω2 = 2k/m and ω3 = 3k/m. By substituting
ω1 = 0 into the equation (K − ω 2 M)a = 0, we find that a1 = a2 = a3 . 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 ω2 , we find that a1 = −a2 = −a3 ; here, mass 1
oscillates in one direction while the other two oscillate in the other direction
(all amplitudes being equal). For ω3 , a1 = 0 and a2 = −a3 . 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. 10.15
(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 23 M a2 . (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 = 23 M a2 .
8
(b) The initial energy of the cube is pure PE, since there is no motion.
Recall that we calculate the PE of a continuous body by the equivalent
notion that its entire mass is located
√ at its CM. The CM of a cube is
at its center, which is a height a/ 2 above the table when the cube is
at equilibrium on its edge. Just before it hits the table, its CM is a/2
above the table. Thus, the change in PE is
a
M ga √
a
∆U = M g √ −
(40)
=
( 2 − 1).
2
2 2
This energy has all been converted into rotational kinetic energy, T =
1
Iω 2 . Therefore,
2
1 2 1
M ga √
Iω = M a2 ω 2 =
( 2 − 1)
2
3
2
(41)
which is solved to give
ω=
r
3g √
( 2 − 1).
2a
(42)
#7 (10 points) JRT Prob. 10.16
Find the moment of inertia for a uniform cube of mass M and edge
a as in Problem 10.15, and then do the following: The cube is sliding with
velocity v along a flat horizontal frictionless table when it hits a straight very
low step perpendicular to v, and the leading edge comes abruptly to rest.
(a) By considering which quantities are conserved before, during, and after the brief collision, find the cube’s angular velocity just after the
collision.
(b) Find the minimum speed v for which the cube rolls over after hitting
the step.
Solution
(a) The moment of inertia for this cube as it rotates about its edge is
shown in example 10.2 (and required in the previous problem); it is
I = 32 M a2 . During the collision, it is incorrect to assume that kinetic
9
energy is conserved (it might be an inelastic collision). However, angular momentum is always conserved; in particular the component of L
about the edge of the step, which we call Ly (see attached figure). We
find that
X
5π
M av
a
=−
(43)
Ly =
mα rα ×v = M R×v = M √ v sin
4
2
2
(the negative sign simply indicates that L is directed into the page).
Immediately after the collision, the cube is rotating, and we have
2
Ly = Iω0 = M a2 ω0 ,
3
(44)
where the subscript on ω0 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 Ly (that is, conserving angular
3v
momentum), we find that ω0 = 4a
. 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 JRT1015 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,
q we rewrite our velocity as a
√
3g( 2−1)
from that assignment
function of ω0 , and substitute ω0 =
2a
solution. The resulting velocity is
s
√
8ga( 2 − 1)
v=
.
(45)
3
#8 (5 points) JRT Prob. 10.23
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 automatically zero? Prove that Izz = Ixx + Iyy .
Solution
10
Fig. 3: Geometry of question #7
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,
X
X
X
Ixx +Iyy =
mα (yα2 +zα2 )+
mα (zα2 +x2α ) =
mα (x2α +yα2 ) = Izz . (46)
#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, respectively.
(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 x̂, ŷ, and ẑ 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 Ixx . Noting that
11
Fig. 4: Geometry for Question #9
̺ = M/V = M/(abc), we have
Ixx =
=
=
=
=
=
=
Z
a
Z
b
Z
c
dz̺ y 2 + z 2
dy
dx
0
0
0
Z b Z c
Ma
dy
dz y 2 + z 2
abc 0
c
0
Z
1 3
Ma b
2
dy y z + z
abc 0
3
0
Z b
Ma
1
dy cy 2 + c3
abc 0
3
b
M ac 1 3 1 2
y + cy
abc 3
3
0
M ac 1 3 1 2
b + cb
abc 3
3
M 2
b + c2 .
3
(47)
(48)
(49)
(50)
(51)
(52)
(53)
The other two moments of inertia are Iyy = M3 (a2 + c2 ) and Izz =
M
(c2 + a2 ), as can be found using the same equation (or by observing
3
the symmetries inherent to the problem).
12
As for the products of inertia, we have
Z
Z b
Z a Z b Z c
Mc a
dx
dy xy (54)
Ixy = −
dx
dy
dz̺xy = −
abc 0
0
0
0
0
b
Z
Z
Mc a
Mc a
1 2
1 2
= −
(55)
xb
dx xy
dx
=−
abc 0
2
abc 0
2
0
a
1
Mc 1 2 2
Mc 1 2 2
xb
=−
a b = − M ab.
(56)
= −
abc 4
abc 4
4
0
By observing the symmetries inherent to the problem (and by recalling that the inertia matrix must be symmetric), we find that Iyx =
−M ab/4, Ixz = Izx = −M ac/4, and Iyz = Izy = −M bc/4. Therefore,
the inertia matrix can be written


4 (b2 + c2 )
−3ab
−3ac
M
.
−3ab
4 (a2 + c2 )
−3bc
I=
(57)
12
2
2
−3ac
−3bc
4 (a + b )
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). Therefore,
M ωab M ω (a2 + c2 ) M ωbc
.
(58)
L = Iω = −
,
,−
4
3
4
(c) Since rotation about the y−axis does not produce an angular momentum L parallel to ω, it is clear that x̂, ŷ, and ẑ 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. 10.27
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
13
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
Z
Z R Z 2π
ρ dρ dφ
σπR4
2
2
Izz = σ(x + y )dA = σ
ρ2
=
,
(59)
sin α
2 sin α
0
0
since the φ integral evaluates to 2π and the ρ integral evaluates to R4 /4.
2
The area of the cone
can check this by performing
R is A = πR / sin α (you
2
the integral A =
dA). Therefore, σπR / sin α = M , the total mass. All
together, we have Izz = M R2 /2.
The other two moments, Ixx and Iyy , must be identical by symmetry. For
the first of these,
Z
Ixx = σ(y 2 + z 2 )dA.
(60)
The first term here is the same as the second term in Izz . Since the two terms
in Izz are equal (again, by symmetry), we conclude that the first term in Ixx
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 Ixx is h2 /R2 times Izz . All together,
we find that
1
1
h2
Ixx = Iyy =
(61)
+ 2 Izz = M (R2 + 2h2 ).
2 R
4
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,


(R2 + 2h2 )
0
0
M
0 (R2 + 2h2 )
0 .
I=
(62)
4
0
0 2R2
#11 (10 points) JRT Prob. 10.36
A rigid body consists of three equal masses (m) fastened at the positions (a, 0, 0), (0, a, 2a), and (0, 2a, a).
(a) Find the inertia tensor I.
14
Fig. 5: Geometry of question #10
(b) Find the principal moments and a set of orthogonal principal axes.
Solution
(a) The three masses are equal (m1 = m2 = m3 = m) and their positions
are r1 = a(1, 0, 0), r2 = a(0, 1, 2), r3 = a(0, 2, 1). Therefore,
X
Ixx =
mα (yα2 + zα2 ) = ma2 (0 + 5 + 5) = 10ma2
(63)
X
Iyy =
mα (x2α + zα2 ) = ma2 (1 + 4 + 1) = 6ma2
(64)
X
Izz =
mα (x2α + yα2 ) = ma2 (1 + 1 + 4) = 6ma2
(65)
X
Ixy = −
mα xα yα = −ma2 (0 + 0 + 0) = 0
(66)
X
Ixz = −
mα xα zα = −ma2 (0 + 0 + 0) = 0
(67)
X
Iyz = −
mα yα zα = −ma2 (0 + 2 + 2) = −4ma2 .
(68)
That is,

5
0
0
3 −2  .
I = 2ma2  0
0 −2
3

15
(69)
(b) The characteristic equation is
det(I − λ1) = (10ma2 − λ)2 (2ma2 − λ) = 0.
(70)
Therefore, the principal moments are λ1 = λ2 = 10ma2 and λ3 = 2ma2 .
If we set λ = 10ma2 , the equation (I−λ1)ω = 0 yields three equations:
0 = 0, ω2 + ω3 = 0, and ω2 + ω3 = 0. This tells us that ω2 = −ω3 , and
that the two normalized eigenvectors corresponding to λ = 10ma2 are
e1 = (1, 0, 0),
1
and e2 = √ (0, 1, −1)
2
(71)
(note that any two perpendicular directions in the plane defined by e1
and e2 are also suitable principal axes.)
Setting λ = 2ma2 , the equation (I − λ1)ω = 0 yields three equations,
ω1 = 0, ω2 − ω3 = 0, and −ω2 + ω3 = 0. When normalized, this defines
the principal axis
1
e3 = √ (0, 1, 1).
(72)
2
16