PHY 3221: Mechanics I
Fall Term 2010
Midterm Exam 3, December 1, 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. 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] Which of the figures below depicts the response of a damped harmonic
oscillator to a forcing function which is a single spike (delta function) applied at t = 0?
A. Figure (a)
B. Figure (b)
C. Figure (c)
D. Figure (d)
Figure HbL
2
2
1
1
xHtL
xHtL
Figure HaL
0
0
-1
-1
-2
-2
0
-30 -20 -10
10
20
30
-30 -20 -10
t
Figure HcL
10
20
30
40
50
Figure HdL
xHtL
1.0
xHtL
0
t
1.5
0.5
0.0
-0.5
0
-20
E. Figure (e)
20
40
t
30
20
10
0
-10
-20
-30
-10
0
10
20
30
t
xHtL
Figure HeL
30
20
10
0
-10
-20
-30
-10
0
10
20
30
40
50
t
Solution. D. See Figure 3-24 in the textbook.
Problem 2. [2 pts] A satellite of mass m orbits a planet of mass M in a circular orbit of
radius R. The orbital period is:
A.
B.
C.
D.
E.
independent of M
√
proportional to m
proportional to R
proportional to R3/2
independent of R.
Solution. D. See problem 2-49 (solved in class and in the Student’s manual).
2
Problem 3. [6 pts] Consider an underdamped oscillator with damping parameter β and
characteristic angular frequency ω0 . The oscillator is initially at rest at its equilibrium position
x = 0. Find the response of this oscillator to a forcing function of the form
F (t) =



0,
if t < 0,
if t ≥ 0.
F0 e−βt ,
Solution. Method I. Use Green’s method since the initial conditions are the same as in the
example in the book:
x(t) =
Z
t
F (t′ )G(t, t′ )dt′ =
−∞
Z
0
t
′
F0 e−βt
1 −β(t−t′ )
e
sin ω1 (t − t′ )dt′
mω1
which gives
F0 e−βt
x(t) =
mω1
Z
0
t
"
F0 e−βt cos ω1 (t − t′ )
sin ω1 (t − t )dt =
mω1
ω1
′
′
#t
=
0
F0 e−βt
(1 − cos ω1 t)
mω12
Method II. Write down the solution to the differential equation
ẍ + 2β ẋ + ω02 x =
F0 −βt
e
m
as a sum of the general solution to the homogeneous equation
xc (t) = e−βt (A1 cos ω1 t + A2 sin ω1 t)
plus a particular solution to the inhomogeneous equation, for which you first guess
xp (t) = C e−βt
where C is a constant which you determine by plugging into the inhomogeneous differential
equation:
F0 −βt
e
β 2 C e−βt + 2β(−β)C e−βt + ω02C e−βt =
m
from where
F0
F0 −βt
F0
=
=⇒
x
(t)
=
e
C=
p
2
2
m(ω0 − β 2 )
mω1
mω12
Then
!
F0 −βt
F0
x(t) = e−βt (A1 cos ω1 t + A2 sin ω1 t) +
e
= e−βt
+ A1 cos ω1 t + A2 sin ω1 t
2
mω1
mω12
Now apply the initial conditions ẋ(0) = x(0) = 0 to determine the constants
A1 = −
F0
,
mω12
3
A2 = 0.
Problem 4. [5 pts.] Find the gravitational force of attraction between an infinite wire of
and a point mass m situated a distance R away from the wire.
linear mass density λ = M
L
Solution. Method I. Use Gauss’ law to find the gravitational field of the wire at the location
of the point mass. Due to the cylindrical symmetry, the gravitational field of the wire has the
form ~g (r) = gr (r)~er in cylindrical coordinates (r, ϕ, z). Therefore, consider a cylinder S of
radius R and height H, whose axis is along the wire. There is no gravitational flux through
the top lid or the bottom lid, since there the field is parallel to the surface. Therefore, the only
flux is through the side surface:
Z
S
~n · ~g da = gr (R)
Z
S
da = gr (R)2πRH = −4πG
from where
gr (R) = −
and the force is
Z
V
ρdv = −4πGλH
2Gλ
R
2Gλm
F~ = m~g = −
~er
R
Method II. Repeat the solution to Problem 5-7 from the homework and in the end take the
limit ℓ → ∞.
Problem 5. [5 pts.] In the 1989 animated movie “A Grand Day Out” Wallace and Gromit
run out of cheese and build a home-made rocket to fly to the Moon and replenish their cheese
supply. Let the mass of the Earth be M, the mass of the Moon be m, the radius of the Earth
be R and the distance between Earth and Moon (center to center) be d (refer to the figure
below). Compute the escape velocity V which Wallace’s rocket needs to have on the surface of
the Earth, in order to be able to reach the Moon. You can ignore the effect of friction within
the atmosphere. Hint: think how far the rocket needs to get before the Moon’s gravity is able to
pull it to the Moon’s surface.
M
R
m
V~
x
d−x
d
Figure 1: An illustration for the Wallace and Gromit problem.
4
Solution. Let’s first find the distance x where the Earth’s gravity equals the Moon’s gravity:
G
M
m
=
G
x2
(d − x)2
=⇒
d−x
=
x
r
m
M
=⇒
x=
d
1+
q
m
M
The rocket only needs to reach to the point x on its own, afterwards the Moon’s gravity will
do the job.
Since we are neglecting the effects of friction, energy is conserved and one can write
1 2
Mµ
mµ
Mµ
mµ
µV − G
−G
= 0−G
−G
2
R
d−R
x
d−x
where µ is the mass of the rocket. Solving for V,
V =
√
2G
√
√
m
( M + m)2
+
−
R
d−R
d
v
u
uM
t
5