PHY 3221: Mechanics I
Fall Term 2010
Final Exam, December 14, 2010
• This is a closed book exam lasting 90 minutes.
• Since calculators are not allowed on this
√ test, if the problem asks for a numerical answer,
answering 2 + 2 is as good as 4, and 2 is as good as 1.4142....
• There are six equally weighted problems worth a total of 30 points. The problems appear
on the second and third page of this test. Begin each problem on a fresh sheet of paper.
Use only one side of the paper. Avoid microscopic handwriting.
• Put your name, the problem number and the page number in the upper right-hand corner
of each sheet.
• To receive partial credit you must explain what you are doing. Carefully labelled figures
are important! Randomly scrawled equations are not helpful.
• Draw a box around important results (or at least results which you think might be important).
• Good luck!
1
Problem 1. [5 pts] You use a rope to pull a crate across a level floor. The maximum
tension that the rope can have without breaking is Tmax , and the coefficient of kinetic friction
is µk . Suppose that you tug on the rope so that it makes an angle θ with the horizontal (refer
to Fig. 1). Find the maximum possible mass Mmax of the crate that you can pull at constant
speed without breaking the rope, and the optimal angle θopt that will allow you to do that.
Hint: make sure to draw a force diagram and clearly label all forces.
y
~g
x
M
θ
µk
Figure 1: An illustration for the crate pulling problem.
Solution. The fources which act on the crate are: gravity G = Mg, tension T , normal
reaction N and friction F = µk N. When we pull the crate at constant speed, the net force is
zero. In components:
x : −F + Tx = −µk N + T cos θ = 0,
y:
−G + Ty + N = −Mg + T sin θ + N = 0.
Eliminate N
N=
T cos θ
µk
−Mg + T sin θ +
and solve for M:
M=
T cos θ
=0
µk
T
(cos θ + µk sin θ)
µk g
Obviously, the mass will be maximum when T = Tmax . To find the corresponding optimum
angle θopt , find the extremum of M(θ):
dM
=0
dθ
Explicitly:
=⇒
− sin θopt + µk cos θopt = 0
=⇒
µk
sin θopt = q
,
1 + µ2k
tan θopt = µk
1
cos θopt = q
1 + µ2k
2
=⇒
θopt = tan−1 µk
and
Mmax =
Tmax
Tmax q
1 + µ2k
(cos θopt + µk sin θopt ) =
µk g
µk g
Problem 2. [5 pts.] Obtain the Fourier expansion of the periodic function
F (t) =



+1,
−1,
if − π/ω < t < 0,
if 0 < ω < +π/ω.
Solution. (See also Problem 3-28.) This is an odd function: F (−t) = −F (t), therefore all
an = 0. For bn use the formula from the formula sheet:
ω +π/ω
F (t) sin(nωt)dt
π −π/ω
Z
Z
ω +π/ω
ω 0
(+1) sin(nωt)dt +
(−1) sin(nωt)dt
=
π −π/ω
π 0
Z
bn =
0
π/ω
ω − cos(nωt) ω − cos(nωt) −
=
π
nω
π
nω
−π/ω
0

 − 4 , if n odd
2
nπ
n
=
(−1 + (−1) ) =

nπ
0, if n even
Then
F (t) = −
1
1
1
4
sin(ωt) + sin(3ωt) + sin(5ωt) + sin(7ωt) + . . .
π
3
5
7
Problem 3: [5 pts.] A particle with mass m moves in one dimension along the x-axis in
the region x > 0. It is acted upon by two forces, F~1 and F~2 . The first force F~1 is directed away
from the origin, and its magnitude is linearly proportional to x as
x
F1 (x) = C1 ,
L
where C1 is a fixed force constant and L is a given length scale. The second force F~2 is directed
towards the origin, and has magnitude
F2 (x) = C2
L3
,
x3
where C2 is another fixed force constant. Find
(a) the potential energy U(x);
(b) the equilibrium positions (if any) and discuss whether they are stable or unstable.
Solution. (a)
U(x) = −
Z
F (x)dx = −
x
+C1
dx −
L
Z 3
Z
L3
−C2 3 dx
x
!
U(x) = −
C1 x2 C2 L3
−
2L
2x2
(b) The equilibrium position is where F1 = F2 :
L3
C2 1/4
x
C1 = C2 3 =⇒ xeq = L
L
x
C1
It is unstable, which is easily seen as follows: for x < xeq , F2 wins and the net force is towards
the origin, for x > xeq , F1 wins and the net force is away from the origin.
Problem 4. [5 pts] A massless spring of relaxed length L and spring constant k hangs from
the ceiling as shown in Fig. 2(a). At t = 0, a piñata of mass m is attached to the free end of
the spring and released (see Fig. 2(b)).
(a) Describe the subsequent motion of the piñata x(t). You can treat the piñata as a point
particle and ignore air resistance. For grader’s convenience, use the coordinate system shown
in Fig. 2(b).
(b) Suppose that you are now blindfolded, handed a bat and spun around several times.
Because of the blindfold, you cannot see the location of the piñata at any given point. At what
height x (measured from the ceiling, as shown in Fig. 2(b)) should you aim the bat in order to
maximize your chances of hitting the piñata? Explain your reasoning.
x=0
k, L
m
(a)
(b)
x
Figure 2: An illustration for the piñata on a spring problem.
Solution (a) The piñata will behave as an undamped oscillator in the presence of a constant
force (gravity):
mẍ = mg − k(x − L)
or
ẍ + ω02 x = g + ω02 L,
ω0 ≡
s
k
m
The general solution is
g
x(t) = xc (t) + xp (t) = A1 cos ω0 t + A2 sin ω0 t + L + 2
ω0
4
!
The constants A1 and A2 are determined from the initial conditions x(0) = L and ẋ(0) = 0 as
A1 = −g/ω02 and A2 = 0.
x(t) = L +
mg
g
(1 − cos ω0 t) = L +
(1 − cos ω0 t)
2
ω0
k
(b) Recall Problem 3-5. It is best to aim at the location where the oscillator spends the
longest amount of time (alternatively, where the velocity is zero), which is at x = L or x =
L + 2mg/k.
Problem 5. [5 pts] In the Pixar movie “Wall-E” the main character finds himself drifting
away from the mothership “The Axiom” with some velocity V . Luckily, he has picked up a
fire extinguisher which can eject fire-retarding foam. By switching on the fire extinguisher and
spraying foam in the direction away from the ship, Wall-E can propel himself back to the ship.
Let the velocity of the foam with respect to a stationary fire extinguisher be u, the mass of
Wall-E be M, the mass of an empty fire extinguisher be m and the total mass of the foam be
µ. Find the maximum value of V (in terms of u, M, m and µ) for which Wall-E will be able to
return safely back to the Axiom.
Solution. There are no external forces present, thus
F = 0 = mv̇ + uṁ
from where
dm
m
In order to be safe, Wall-E needs to acquire some infinitesimal speed towards the mothership
(i.e. basically he needs to stop):
mdv = −udm
0
=⇒
dv = −u
dm
m
dv = −u
Z
M +m
M +m
V = −u ln
M +m+µ
!
µ
= u ln 1 +
M +m
Z
−V
M +m+µ
Problem 6. [5 pts.] A particle of mass m is constrained to move in one dimension under
the influence of a retarding force whose magnitude is given by F = Ceαv , where C and α are
positive constants. At t = 0 the particle is released from the origin (x = 0) with initial velocity
ẋ = +v0 . Find the time T that it would take for the particle to stop completely.
Solution. Refer to Problem 2-39.
mẍ = m
−αv
e
C
dv = − dt
m
=⇒
Z
0
v0
−αv
e
dv
= −Ceαv
dt
C
dv = −
m
5
Z
0
T
dt
=⇒
T =
m 1 − e−αv0
Cα
Formula sheet
A·(B × C) = B·(C × A) = C·(A × B) ≡ ABC
A×(B × C) = (A · C)B − (A · B)C
(A × B) · (C × D) = A · [B × (C × D)]
= A · [(B · D)C − (B · C)D]
= (A · C)(B · D) − (A · D)(B · C)
(A × B) × (C × D) = [(A × B) · D] C − [(A × B) · C] D
= (ABD)C − (ABC)D = (ACD)B − (BCD)A
v = ṙ er + r θ̇ eθ + r sin θφ̇ eφ
a =
+ 2ṙ φ̇ sin θ + 2r θ̇φ̇ cos θ + r θ̈ sin θ eφ
v = ṙ er + r φ̇ eφ + ż ez
a =
r̈ − r θ̇2 − r φ̇2 sin2 θ er + 2ṙ θ̇ + r θ̈ − r φ̇2 sin θ cos θ eθ
r̈ − r φ̇2 er + r φ̈ + 2ṙ φ̇ eφ + z̈ ez
X
k
εijk εlmk = δil δjm − δim δjl
X
εijk εljk = 2 δil
j,k
X
εijk εijk = 6
i,j,k
Time averages over one period T :
1
T
Z
t+T
1
hcos ωti =
T
Z
t+T
hsin2 ωti =
2
t
t
6
dt sin2 ωt =
1
2
dt cos2 ωt =
1
2
Simple harmonic oscillator:
mẍ + kx = 0
x(t) = A sin(ω0 t − δ)
x(t) = A cos(ω0 t − φ)
s
2π
=
ω0 = 2πν0 =
τ0
Damped oscillator:
k
m
b
ẍ + 2β ẋ + ω02 x = 0, 2β =
m
√ 2 2
√ 2 2 x(t) = e−βt A1 e β −ω0 t + A2 e− β −ω0 t
Underdamped motion
x(t) = Ae−βt cos(ω1 t − δ),
ω1 =
x(t) = e−βt (A1 cos ω1 t + A2 sin ω1 t) ,
q
ω02 − β 2
ω1 =
Critically damped motion
q
ω02 − β 2
x(t) = (A + Bt)e−βt
Overdamped motion
−βt
x(t) = e
Driven oscillator
h
ω2 t
A1 e
−ω2 t
+ A2 e
i
,
ω2 =
q
β 2 − ω02
F0
ẍ + 2β ẋ + ω02 x = A cos ωt, A =
m √
√ 2 2
− β 2 −ω02 t
β −ω0 t
−βt
+ A2 e
A1 e
xc (t) = e
A
xp (t) = q
cos(ωt − δ)
(ω02 − ω 2 )2 + 4ω 2β 2
−1
δ = tan
q
2ωβ
2
ω0 − ω 2
!
ω02 − 2β 2
ωR
Q=
2β
ωR =
RLC circuit
VL = L
Gauss’s law
Z
S
dI
dt
VR = RI
~n · ~g da = −4πG
7
VC =
Z
V
ρ dv
q
C
Fourier series
∞
X
1
(an cos nωt + bn sin nωt) ,
F (t) = a0 +
2
n=1
2Z τ
ω Z +π/ω
F (t′ ) cos nωt′ dt′ =
F (t′ ) cos nωt′ dt′
π −π/ω
τ 0
ω Z +π/ω
2Zτ
=
F (t′ ) sin nωt′ dt′ =
F (t′ ) sin nωt′ dt′
π −π/ω
τ 0
an =
bn
Green’s function
′
G(t, t ) =



1
mω1
0,
′
e−β(t−t ) sin ω1 (t − t′ ),
if t < t′ .
x(t) =
Z
t
F (t′ )G(t, t′ )dt′
−∞
Rocket motion
Fext = mv̇ + uṁ
8
if t ≥ t′ ,