Math 142-2, Homework 2
Solutions
Problem 10.4
Consider Coulomb friction, equation 10.3.
(a) What is the sign of γ?
Consider a mass moving under friction alone, and assume x′ > 0. Then, mx′′ = Ff = −γ or alternatively
mv = −γ. Since energy should be lost (and v > 0), the velocity should be decreasing. Thus, γ > 0.
′
(b) What is the dimension of γ?
γ has units of force, or [kg ms−2 ].
(c) Assume that dx/dt = 0, then Ff could be any value such that |Ff | ≤ γ. What values of x are then
equilibrium positions of the spring-mass system?
The the spring force is −kx. If |kx| ≤ γ, then Ff = kx, and the mass remains at rest. Equilibrium
positions are |x| ≤ γk .
Problem 10.5
Assume the same form of friction as in exercise 10.4. If initially x = x0 > 0 and the velocity is v0 > 0,
then at what time does the mass of a spring-mass system first stop moving to the right? Will the mass
continue to move after that time?
The differential equation is
mx′′ + kx = −γ
which has the general solution
γ
x = A cos ωt + B sin ωt −
k
ω=
r
From x(0) = x0 and x′ (0) = v0 ,
γ
v0
γ
x = x0 +
cos ωt +
sin ωt −
k
ω
k
1
k
m
Differentiate to get velocity and solve for v = 0.
γ
sin ωt + v0 cos ωt
v = −ω x0 +
k
γ
sin ωt = v0 cos ωt
ω x0 +
k
kv0
tan ωt =
ω(kx0 + γ)
1
kv0
−1
t = tan
ω
ω(kx0 + γ)
This is the time at which the mass first stops moving to the right. To determine whether it will move again,
we need to determine whether kx ≤ γ.
Q = tan ωt
kv0
=
ω(kx0 + γ)
1
cos ωt = p
1 + Q2
Q
sin ωt = p
1 + Q2
kv0
tan ωt =
ω(kx0 + γ)
v0
γ
γ
cos ωt +
sin ωt −
x = x0 +
k
ω
k
γ
Q
1
v0
γ
p
p
= x0 +
+
−
2
2
k
ω
k
1+Q
1+Q
γ
v0 Q
1
γ
−
=p
x0 + +
2
k
ω
k
1+Q
1
kv0 Q
− 2γ
kx − γ = p
kx0 + γ +
2
ω
1+Q
kx0 + γ
=p
(1 + Q2 ) − 2γ
1 + Q2
p
= (kx0 + γ) 1 + Q2 − 2γ
To move, we must have kx − γ > 0 or
0 < (kx0 + γ)
2γ < (kx0 + γ)
2
p
p
1 + Q2 − 2γ
1 + Q2
2
4γ < (kx0 + γ) (1 + Q2 )
4γ 2 < (kx0 + γ)2 + Q2 (kx0 + γ)2
4γ 2 ω 2 < (kx0 + γ)2 ω 2 + k 2 v02
4γ 2 ω 2 < (kx0 + γ)2 ω 2 + k 2 v02
0 < (kx0 − γ)(kx0 + 3γ)ω 2 + k 2 v02
This is clearly true if kx0 > γ, so that the mass will move again if this is true. This condition means the
spring is initially stretch out more than friction can resist. It will be even more stretched by the time it
stops, so moving again is inevitable.
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Otherwise, kx0 ≤ γ, and we have a condition on the initial velocity.
ωp
(γ − kx0 )(kx0 + 3γ)
v0 >
k
Problem 10.9
A particle not connected to a spring, moving in a straight line, is subject to a retardation force of
magnitude β(dx/dt)n , with β > 0.
(a) Show that if 0 < n < 1, the particle will come to rest in a finite time. How far will the particle
travel, and when will it stop?
mv ′ = −βv n
β
v −n v ′ = −
m
β
v 1−n
= − t + C1
1−n
m
v 1−n
β
v 1−n
=− t+ 0
1−n
m
1−n
β(1 − n)
1−n
t + v01−n
v
=−
m
1
1−n
β(1 − n)
1−n
t + v0
v= −
m
The particle comes to rest when
0=
β(1 − n)
−
t + v01−n
m
β(1 − n)
t = v01−n
m
mv01−n
t=
β(1 − n)
1
1−n
We get the position by integrating the velocity
1
1−n
+1
− β(1−n)
t + v01−n
m
+ C2
x= β(1−n)
1
+
1
−
1−n
m
2−n
1−n
m
β(1 − n)
1−n
x=−
−
t + v0
+ C2
β(2 − n)
m
m
v 2−n + C2
x0 = −
β(2 − n) 0
2−n !
β(1 − n) 1−n
m
2−n
1−n
v
− v0 −
t
+ x0
x=
β(2 − n) 0
m
3
At the time when the particle comes to rest,


2−n
1−n 1−n
m
v 2−n − v01−n − β(1 − n) mv0
 + x0
x=
β(2 − n) 0
m
β(1 − n)
=
mv02−n
+ x0
β(2 − n)
(b) What happens if n = 1?
mv ′ = −βv 1
β
v −1 v ′ = −
m
β
ln v = − t + C1
m
β
v = v 0 e− m t
β
m
x=
1 − v 0 e− m t + x 0
β
m
x→
+ x0
β
The velocity decays exponentially, never coming to rest. The position exponentially approaches a finite
value.
(c) What happens if 1 < n < 2?
The solution obtained for the first case is valid.
1
1−n
β(1 − n)
1−n
v= −
t + v0
m
2−n !
1−n
m
β(1
−
n)
+ x0
x=
v 2−n − v01−n −
t
β(2 − n) 0
m
x→
mv02−n
+ x0
β(2 − n)
The velocity decays to zero in the limit t → ∞, and the position approaches a finite value.
(d) What happens if n > 2?
The solution obtained for the first case is valid. The velocity decays to zero in the limit t → ∞, but the
position increases without bound.
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Problem 12.6
Show that the local maximum or minimum for the displacement of an underdamped oscillation does not
occur halfway between the times at which the mass passes through its equilibrium position. However,
show that the time period between successive local maxima (or minima) is constant. What is this
period?
The critical points are
ct
x = Ae− 2m sin(ωt + φ0 )
0 = x′
ct
= Ae− 2m cos(ωt + φ0 ) −
ct
c
Ae− 2m sin(ωt + φ0 )
2m
ct
ct
c
Ae− 2m sin(ωt + φ0 ) = Ae− 2m cos(ωt + φ0 )
2m
2m
tan(ωt + φ0 ) =
c
1
2m
−1
t=
tan
− φ0 + nπ
ω
c
The critical points alternate between max and min, so the period between consecutive min and max are
π
π
T = 2π
ω . The half-way point occurs at angle ωt + φ0 + 2 . Noting sin(z + 2 ) = cos z,
ct
π
x = Ae− 2m sin ωt + φ0 +
2
ct
6= 0
= Ae− 2m cos(ωt + φ0 )
since tan(ωt + φ0 ) is finite.
Problem 12.7
Consider a spring-mass system with linear friction (but without gravity). Suppose that there is an
additional force, B cos ω0 t, a periodic forcing function. Assume that B and ω0 are known.
(a) If the coefficient of friction is zero (i.e., c = 0), then determine
the general solution of the differential
p
equation. Show that p
the solution is oscillatory if ω0 6= k/m. Show that the solution algebraically
grows in time if ω0 = k/m
p (this is called resonance and occurs if the forcing frequency ω0 is the same
as the natural frequency k/m).
The solution can be determined by the method of undetermined coefficients.
mx′′ + kx = Bcosω0 t
B
cos ω0 t
x = C1 sin(ωt) + C2 cos(ωt) +
m(ω 2 − ω02 )
r
k
ω=
m
The solution is oscillatory, since it is a combination of sines and cosines. If on the other hand ω = ω0 , the
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solution is different
mx′′ + kx = Bcosωt
x = C1 sin(ωt) + C2 cos(ωt) +
B
t sin ωt
2mω
This is oscillatory but grows in time due to the time scale.
(b) If c2 < 4mk, show that the general solution consists of the sum of oscillatory terms (of frequency ω0 )
and terms which exponentially decay in time. Thus for sufficiently large times the motion is accurately
approximated by an oscillation of constant amplitude. What is that amplitude? Note it is independent
of initial conditions.
mx′′ + cx′ + kx = Bcosωt
x = C1 e−rt sin(ωt) + C2 e−rt cos(ωt) + P cos ω0 t + Q sin ω0 t
√
c2 − 4km
ω=
2m
c
r=
2m
B(k − ω02 m)
P =
(k − ω02 m)2 + (cω0 )2
B(cω0 )
Q=
(k − ω02 m)2 + (cω0 )2
The first two terms decay with e−rt . The last two terms are oscillatory of constant amplitude and do not
depend on initial conditions. The amplitude of this part is M .
p
M = P 2 + Q2
B
=p
2
(k − ω0 m)2 + (cω0 )2
(c) If c2 ≪ 4mk, show that for large times the oscillation approximately
obeys the following statements:
p
If the forcing frequency is less than the natural frequency (ω0 < k/m), then the mass oscillates “in
phase” with the forcing function (i.e., when the forcing function is a maximum, the stretching of the
spring is a maximum, and
pvice versa). On the other hand, if the forcing frequency is greater than the
natural frequency (ω0 > k/m), then the mass oscillates “180 degrees out of phase” with the forcing
function (i.e., when the forcing function is a maximum, the compression of the spring is a maximum and
vice versa).
If c2 ≪ 4mk then
P ≈
B
m(k/m − ω02 )
Q≈0
p
The forcing contribution
to the solution is P cos ω0 t. If ω0 < k/m, then P > 0 and the oscillation is in
p
phase. If ω0 > k/m, then P < 0 and the oscillation is out of phase.
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(d) In part (b) the ratio of the amplitude of oscillation of the mass to the amplitude of the forcing
function is called the response. If c2 ≪ 4mk, then at what frequency is the response largest?
The response is M/B, which is maximized by minimizing
(M/B)2 = (k − ω02 m)2 + (cω0 )2
∂
((k − ω02 m)2 + (cω0 )2 )
0=
∂ω0
= −4ω0 m(k − ω02 m) + 2c2 ω0
= 2(c2 − 2m(k − ω02 m))ω0
≈ −4m(k − ω02 m)ω0
r
k
ω = 0,
m
Since c is very small, the maximum response will be at
r
ω=
k
m
This is consistent with what would be expected based on resonance.
Problem 13.1
Assume that friction is sufficiently large (c2 > 4mk).
(a) Show that the mass either decays to its equilibrium position (without passing through it), or that
the mass shoots past its equilibrium position exactly once before returning monotonically towards its
equilibrium position.
The general solution is (making r1 and r2 positive rather than negative)
x = c1 e−r1 t + c2 e−r2 t
0 = c1 e−r1 t + c2 e−r2 t
c2
e(r2 −r1 )t = −
c1
c2
(r2 − r1 )t = ln −
c1
1
c2
t=
ln −
r2 − r1
c1
c2 c1 < 0 and t ≥ 0, the mass passes the equilibrium. Otherwise it does not. Since there is at most one
possible solution for t, it cannot cross a second time.
(b) If the initial position x0 of the mass is positive, then show that the mass crosses its equilibrium
position only if the initial velocity is sufficiently negative. What is the value of this critical velocity?
Does it depend in a reasonable way on x0 , c, m, and k?
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Including boundary conditions,
x0 r2 + v0 −r1 t x0 r1 + v0 −r2 t
e
−
e
r2 − r1
r2 − r1
x0 r2 + v0
c1 =
r2 − r1
x0 r1 + v0
c2 = −
r2 − r1
x=
Assume that r2 > r1 > 0. The requirement c2 c1 < 0 leads to v0 > −x0 r1 or v0 < −x0 r2 . t > 0 and r2 > r1
imply that − cc12 > 1.
c2
c1
x0 r1 + v0
=
x0 r2 + v0
1<−
If v0 > −x0 r1 , both the numerator and denominator are positive, but the denominator is larger. That rules
out this option. If v0 < −x0 r2 , both are negative, but the numerator has larger magnitude. This is what is
required. The condition on v0 is v0 < −x0 r2 .
Problem 14.2
Consider a mass m located at x = xi + yj, where x and y are unknown functions of time. The mass is
free to move in the x − y plane without gravity (i.e., it is not connected to the origin via the shaft of a
pendulum) and hence the distance L from the origin may vary with time.
(a) Using polar coordinates as introduced in Sec. 14, what is the velocity vector?
x
y
cos θ
=L
sin θ
cos θ
− sin θ
x ′ = L′
+ Lθ′
sin θ
cos θ
x=
= L′ r + Lθ′ θ
(b) Show that the acceleration vector a = (d2 /dt2 )x is
2
dL dθ
d θ
θ+
a= L 2 +2
dt
dt dt
8
2 !
d2 L
dθ
r
−L
dt2
dt
cos θ
− sin θ
+ Lθ′
sin θ
cos θ
cos θ
− sin θ
− sin θ
cos θ
x′′ = L′′
+ 2L′ θ′
+ Lθ′′
− L(θ′ )2
sin θ
cos θ
cos θ
sin θ
x ′ = L′
= (L′′ − L(θ′ )2 )r + (Lθ′′ + 2L′ θ′ )θ
Problem 14.3
Consider a mass m located at x = xi + yj and only acted upon by a force in the direction of r with
magnitude −g(L) depending only on L ≡ kxk, called a central force.
(a) Derive the differential equations governing the angle θ and the distance L [Hint: Use the equation
for a determined in exercise 14.2b],
mx′ = f
m(L′′ − L(θ′ )2 )r + (Lθ′′ + 2L′ θ′ )θ = −g(L)r
m(L′′ − L(θ′ )2 ) = −g(L)
Lθ′′ + 2L′ θ′ = 0
(b) Show that L2 (dθ/dt) is constant [Hint: Differentiate L2 (dθ/dt) with respect to t]. This law is called
Conservation of Angular Momentum.
The second differential equation from (a) can be partially integrated.
Lθ′′ + 2L′ θ′ = 0
L2 θ′′ + 2LL′ θ′ = 0
(L2 θ′ )′ = 0
L2 θ ′ = k
This can be used to eliminate the θ dependence from the differential equation for L.
Additional Problem
A rigid body can be modeled as a collection of point masses at positions xp with masses mp . Let x(t)
the the position of some fixed point in the rigid body, and let R(t) be a rotation matrix (RRT = I,
det(R) = 1). Relative to this point, we can say that xp = x + Rrp , where rp represent the displacements
of the point masses from the reference point x, when the rigid body is in its reference orientation (R = I).
With this representation, rp is constant for each point mass; only x(t) and R(t) depend on time.
9
˙ T is skew-symmetric (that is, show that
(a) Show that RR
define the angular velocity to be ω, where

 
0
−ω3
ω1
0
ω ∗ =  ω3
ω = ω2 
−ω2 ω1
ω3
˙ T = −(RR
˙ T )T ). Because of this, we can
RR

ω2
−ω1 
0
˙ T
ω ∗ = RR
and ω ∗ is defined so that ω ∗ u = ω × u. Why are we mathematically justified in defining ω in this way?
˙ Note that this also provides the ODE
(It suffices to show that ω can be computed for any R and R.)
˙ = ω ∗ R.
for orientation, R
RRT = I
d
(RRT ) = 0
dt
˙ T + RR
˙T =0
RR
˙ T = −(RR
˙ T )T
RR
A matrix A with AT = −A has the general form

0
A = −a
−b
a
0
−c

b
c ,
0
from which the entries of ω can be read off by setting ω ∗ = A.
˙ angular
(b) Express the velocity vp of each particle p in terms of the rigid body velocity (v = x),
velocity (ω), position (x), and orientation (R).
xp = x + Rrp
˙ p + R˙rp
x˙ p = x˙ + Rr
˙ T )(Rrp )
vp = v + (RR
vp = v + ω ∗ (Rrp )
(c) Express the total mass (m) and momentum (p) of the rigid body in terms of fixed per-particle
states (mi and ri ) and the rigid body state (x, v, R, and ω).
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m=
X
mp
p
p=
X
mp v p
p
=
X
mp (v + ω ∗ (Rrp ))
p
=
X
mp v + ω ∗ R
p
X
mp r p
p
= mv + ω ∗ R
X
mp r p
p
(d) Up to this point, we have not chosen the point in the rigid body where x is computed in any
particular way. Show that the total momentum can be written as p = mv if we choose x to be the
center of mass,
x=
1 X
mp x p
m p
Assume for the rest of this problem that x is at the center of mass.
1 X
mp x p
m p
1 X
mp (x + Rrp )
x=
m p
1 X
1 X
x=
mp x +
mp Rrp
m p
m p
1 X
mp Rrp
x=x+
m p
1 X
0= R
mp r p
m
p
x=
X
mp r p = 0
p
This simplifies our formula for momentum
p = mv + ω ∗ R
X
p
= mv
11
mp r p
(e) Show that the total kinetic energy of the rigid body has the form
KE =
1
m
kvk2 + ω T Jω
2
2
where J is called the inertia tensor. Show that J = RJ0 RT , where J0 is the constant intertia tensor in
the rest configuration (J0 = J when R = I.) Provide a formula for J0 in terms of mp and rp . You may
assume that J0 and J are invertible.
This becomes easier once we work out some properties of the cross product matrix u∗ .
ω∗ u = ω × u
= u∗T ω
(Ru)∗ (Rv) = (Ru) × (Rv)
= R(u × v)
= Ru∗ v
= Ru∗ RT (Rv)
(Ru)∗ = Ru∗ RT
With these,
KE =
X 1
2
p
mp v p · v p
X
1
mp (v + ω ∗ (Rrp )) · (v + ω ∗ (Rrp ))
2
p
1X
1X
1X
1X
mp v · v +
mp (ω ∗ (Rrp )) · v +
mp v · (ω ∗ (Rrp )) +
mp (ω ∗ (Rrp )) · (ω ∗ (Rrp ))
=
2 p
2 p
2 p
2 p
X
1X
1X
mp kvk2 + vT ω ∗ R
mp (ω ∗ (Rrp ))T (ω ∗ (Rrp ))
mp r p +
=
2 p
2
p
p
1X
1X
2
∗T
T
=
mp kvk +
mp ((Rrp ) ω) ((Rrp )∗T ω)
2 p
2 p
!
1 T X
1X
2
∗
∗T
mp kvk + ω
mp (Rrp ) (Rrp )
ω
=
2 p
2
p
X
J=
mp (Rrp )∗ (Rrp )∗T
=
p
J=
X
mp (Rr∗p RT )(Rr∗p RT )T
p
J=
X
T
mp Rr∗p r∗T
p R
p
J=R
X
mp r∗p r∗T
p
p
J0 =
X
!
RT
mp r∗p r∗T
p
p
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(f ) Define the angular momentum of a rigid body to be
X
L≡
(xp − x) × mp vp
p
Show that L = Jω.
This becomes easier once we work out some properties of the cross product matrix u∗ .
X
L=
(xp − x) × mp vp
p
L=
X
p
L=
X
p
(Rrp ) × mp (v + ω ∗ (Rrp ))
(Rrp ) × mp v +
X
L=
R
L=
X
p
mp r p
!
X
p
×v+
∗
mp (Rrp ) (Rrp )
(Rrp ) × mp (ω ∗ (Rrp ))
X
mp (Rrp )∗ ((Rrp )∗T ω)
p
∗T
p
!
ω
L = Jω
(g) Define the force f and torque τ to be the change in momentum and angular momentum (f = p˙
˙ Show that in the absence of forces (f = 0) or torques (τ = 0), mass, momentum, angular
and τ = L).
˙ = 0. Conclude
momentum, and kinetic energy are conserved. That is, m
˙ = 0, p˙ = 0, L˙ = 0, and KE
that an isolated rigid body translates uniformly (v˙ = 0), and its rotation evolves according to
˙ = (RJ−1 RT L)∗ R,
R
0
where J0 and L are constant.
dm
d X
=
mp = 0
dt
dt p
p˙ = f = 0
L˙ = τ = 0
JJ−1 = I
˙ −1 + J d (J−1 ) = 0
JJ
dt
d −1
˙ −1
(J ) = −J−1 JJ
dt
13
d
J˙ = (RJ0 RT )
dt
˙ 0 RT + RJ0 R
˙T
= RJ
˙ T RJ0 RT + RJ0 RT RR
˙T
= RR
= ω ∗ RJ0 RT + RJ0 RT Rω ∗T
d m
1 T
2
˙
KE =
kvk + ω Jω
dt 2
2
1
d
1 T −1
2
=
kpk + L J L
dt 2m
2
d −1
1
J
L
= LT
2
dt
1
˙ −1 L
= − LT J−1 JJ
2
1
˙
= − ω T Jω
2
1
= − ω T (ω ∗ RJ0 RT + RJ0 RT Rω ∗T )ω
2
=0
since ω ∗T ω = ω × ω = 0.
d
dt
J−1 is obtained by differentiating JJ−1 = I.
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