PHY 3221: Mechanics I
Fall Term 2009
Final Exam, December 16, 2009
• 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 problems worth a total of 21 points, out of a maximum of 20 points. Each
problem carries equal weight. 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. [3.5 pts] In Cartesian coordinates, some force is given by F~ = k(y, x, 0), where
k is a constant. Is this a conservative force? If the answer is “yes”, find the corresponding
potential energy U (x, y, z).
Problem 2. [3.5 pts] Refer to Fig. 1. A clumsy skateboarder lets her skateboard roll down
a frictionless “quarter pipe” ramp, which has the profile of a cylinder of radius R. Assume that
the skateboard started rolling from rest at point A, which is a distance H above the ground,
and landed at point B, which is a distance D from the end of the ramp. Find D in terms of
H, R and g.
A
R
~g
H
B
D
Figure 1: An illustration for the skateboarder problem.
Problem 3. [3.5 pts] A particle starting at rest is attracted to a force center according to
the relation F = −mk 2 /x3 . Show that the time required for the particle to reach the force
center from a distance L is L2 /k.
2
Problem 4. [3.5 pts.] The natural frequency of a critically damped oscillator is ω0 . Initially,
at time t = 0, the oscillator is displaced a distance L from its equilibrium position and then
released. Find the velocity v of the oscillator at a later time t.
Problem 5. [3.5 pts.] Refer to Fig. 2. A hole has been drilled straight through the center
of a spherical planet of mass M and radius R. Assume uniform mass density and neglect
rotational effects and friction.
(a) A stone of mass m has been dropped in the hole at point A. Find the time it takes to
arrive at point B. Hint: show that the stone’s motion in the hole is simple harmonic and find
its frequency.
(b) Now suppose the stone is launched into a circular orbit just above the surface of the planet.
How long does it take to reach point B in this case?
A
R
M
B
Figure 2: An illustration for the supertunnel problem.
Problem 6. [3.5 pts.] A point mass m is located a distance D from the nearest end of a
thin rod of mass M and length L, along the axis of the rod. Find the gravitational force exerted
on the point mass by the rod.
3
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 θ φ̇ 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 θ̇ 2 − 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
εijk εlmk = δil δjm − δim δjl
k
X
εijk εljk = 2 δil
j,k
X
εijk εijk = 6
i,j,k
Time averages over one period T :
1
hsin ωti =
T
2
hcos2 ωti =
Tidal force
Z
t+T
dt sin2 ωt =
t
1 Z t+T
1
dt cos2 ωt =
T t
2
2GmMm r cos θ
D3
GmMm r sin θ
= −
D3
FT x =
FT y
1
2
Rocket motion
Fext = mv̇ + uṁ
4
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 =
q
ω02 − β 2
Critically damped motion
x(t) = (A + Bt)e−βt
Overdamped motion
h
i
x(t) = e−βt A1 eω2 t + A2 e−ω2 t ,
Driven oscillator
ω2 =
q
F0
ẍ + 2β ẋ + ω02 x = A cos ωt, A =
m
√ 2 2 √ 2 2
xc (t) = e−βt A1 e β −ω0 t + A2 e− β −ω0 t
xp (t) = q
A
(ω02 − ω 2 )2 + 4ω 2 β 2
δ = tan
−1
q
2ωβ
2
ω0 − ω 2
cos(ωt − δ)
!
ω02 − 2β 2
ωR
Q=
2β
ωR =
RLC circuit
VL = L
Gauss’s law
β 2 − ω02
Z
S
dI
dt
VR = RI
~n · ~g da = −4πG
5
VC =
Z
V
ρ dv
q
C