PHY 3221: Mechanics I
Fall Term 2009
Midterm Exam 3, December 2, 2009
• This is a closed book exam lasting 50 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 three problems and two multiple choice questions worth a total of 20 pts. The
problems and questions 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. [2 pts] An object is dropped from an altitude of one Earth radius above the
Earth’s surface. If M is the mass of Earth and R is its radius, the speed v of the object just
before it hits Earth is given by:
A. v =
GM
R
1/2
B. v =
GM
2R
1/2
C. v =
2GM
R
D. v =
GM
R2
1/2
E. v =
GM
2R2
1/2
1/2
.
Problem 2. [2 pts] A spherical shell has inner radius R1 , outer radius R2 , and mass M ,
distributed uniformly throughout the shell. The magnitude of the gravitational force exerted
on the shell by a point mass m, located a distance d from the center, outside the outer radius,
is:
A. 0
B.
GM m
R21
C.
GM m
d2
D.
GM m
R22 −d2
E.
GM m
.
(R1 −d)2
Problem 3. [4 pts] The speed of sound v in a gas might plausibly depend on the pressure
P , the density ρ and the volume V of the gas. Use dimensional analysis to find a relation
between v, P , ρ and V .
2
Problem 4. [7 pts.] Refer to Fig. 1. Consider a thin semi-circular disk of total mass M
and radius R. The surface mass density of the disk is not uniform, but varies as ρ(r) ∼ r,
where r is the distance to the center (point O). Find the gravitational field vector ~g at point
O. Hint: at some point, you will need to determine the constant of proportionality C in the
relation ρ(r) = Cr, in terms of the given parameters M and R.
M
R
ρ∼r
O
Figure 1: An illustration for the thin semi-circular disk problem.
Problem 5. [5 pts.] Refer to Fig. 2. Suppose that the gravitational acceleration g(B) at the
top of the Earth’s atmosphere (point B) was measured to be exactly equal to the gravitational
acceleration g(A) at the surface (point A). Let the height of the atmosphere be h, the Earth’s
radius be R and the mass of the planet Earth be M . Find the total mass m of all the air in
the atmosphere in terms of M , h and R. (You can assume perfect spherical geometry in this
problem.)
g(A) = g(B)
R
A
h
B
M
m
Figure 2: An illustration for the Earth atmosphere problem.
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 Z t+T
1
hsin ωti =
dt sin2 ωt =
T t
2
2
1
hcos ωti =
T
2
Z
t+T
t
dt cos2 ωt =
Tidal force
2GmMm r cos θ
D3
GmMm r sin θ
= −
D3
FT x =
FT y
4
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
k
m
Damped oscillator:
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