1. business and industrial
2. energy
3. electricity

Examination Cover Sheet
Princeton University Undergraduate Honor Committee
January 22, 2007
Course Name & Number: PHY 101
Professors: Galbiati, Kwon, McDonald
Date: January 22, 2007
Time: 9 AM
This examination is administered under the Princeton University Honor Code.
Students should sit one seat apart from each other, if possible, and refrain from
talking to other students during the exam. All suspected violations of the Honor
Code must be reported to the Honor Committee Chair at [email protected].
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that is marked with NO or not mentioned may not be used and should not be
in your working space. Assume items not on this list are not allowed for use on
this examination. Please place items you will not need out of view in your bag or
under your working space at this time. University policy does not allow the use of
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•
•
•
•
Course textbooks: NO
Course Notes: NO
Other books/printed materials: NO
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booklet
• Comments on use of printed aids: NO
• Calculator: YES, but only for standard functions. Cannot use graphing
functions or advanced functions, like equation solving
Students may only leave the examination room for a very brief period without the
explicit permission of the instructor. The exam may not be taken outside of the
examination room. During the examination, the Professor or a preceptor will be
available at the following location: outside the exam room.
This exam is a timed examination. You will have 3 hours and 0 minutes to
complete this exam.
Write your name in capital letters in the appropriate space in the next page. Also,
dont forget to write and sign the Honor Code pledge in the appropriate line on the
next page:
“I pledge my honor that I have not violated the Honor Code during this examination”
1
2
Your Name:
PHYSICS 101 FINAL EXAM
January 22, 2007
1
3
5
3 hours
Please Circle your section
9 am
Galbiati
2
10 am
Kwon
11 am
McDonald
4 12:30 pm McDonald
12:30 pm Kwon
Problem
Score
1
/20
2
/15
3
/12
4
/20
5
/20
6
/20
7
/20
8
/20
9
/15
Total
/162
Instructions: When you are told to begin, check that this examination booklet
contains all the numbered pages from 2 through 30. The exam contains 9 problems.
Read each problem carefully. You must show your work. The grade you get depends
on your solution even when you write down the correct answer. BOX your final
answer. Do not panic or be discouraged if you cannot do every problem; there are
both easy and hard parts in this exam. If a part of a problem depends on
a previous answer you have not obtained, assume it and proceed. Keep
moving and finish as much as you can!
Possibly useful constants and equations are on the last page, which you
may want to tear off and keep handy.
Rewrite and sign the Honor Pledge: I pledge my honor that I have not violated the
Honor Code during this examination.
Signature
3
Problem 1
Miscellaneous Questions
1. (5 points) A marathon runner runs at a steady 15 km/hr. When the runner
is 7.5 km from the finish, a bird begins flying from the runner to the finish at
30 km/hr. When the bird reaches the finish line, it turns around and flies back
to the runner, and then turns around again, repeating the back-and forth trips
until the runner reaches the finish line. How many kilometers does the bird
travel?
• 10 km
• 15 km
• 20 km
• 30 km
It takes 30 minutes for the runner to arrive at the finish line. In this time span, the bird
covers a distance of 15 km .
4
2. (5 points) From what minimum height h does the car need to start to make it
around the loop? Ignore friction.
h
R
•
3
R
2
• 2R
•
5
R
2
• Not enough info
By application of the law of conservation of the total energy, the kinetic energy at the
top of the loop is: 12 mv 2 = mg(h − 2R). If the car barely makes it around the loop, the
contact force vanishes on the very top of the loop. At that point, the centripetal force
2
is provided entirely by gravity. Therefore: mv
R = mg. Combining the two equations:
mg(h − 2R) =
h = 25 R
1
mgR
2
5
3. (5 points) It’s a very hot summer in Princeton and the air conditioner is broken.
Your roommate decides that a possible relief might come from the refrigerator,
and proposes to leave the door of the refrigerator open (with the windows closed,
of course). This will:
• decrease the temperature in the room
• increase the temperature in the room
• will not affect the temperature in the room
It will increase the temperature in the room . Every cycle, the refrigerator removes
from the room a quantity of heat QC , from the chilled side of the fridge which is open
to the room, and exhausts the quantity QH , through the vent and to the same
room. The net balance is that the room gets a positive quantity of heat:
QH − QC = W
where W is the work needed to run the refrigerator through one cycle.
6
4. (5 points) A satellite orbiting around the Earth passes nearby the International
Space Station. Operators from the Space Station double the mass of the satellite
by adding a payload with scientific instruments. The radius of the orbit of the
satellite can stay constant if its speed:
• increases by a factor of 8
• increases by a factor of 4
• increases by a factor of 2
• stays constant
• decreases by a factor of 2
The centripetal force acting on satellites is the gravitational force due to the central
body, around which the satellite is orbiting:
mv 2
mM
=G 2
R
R
where m is the mass of the satellite, M is the mass of the central object, and R is the
radius of the orbit. The equation can be solved for R:
R=G
M
v2
The radius depends on the velocity of the satellite, not on its mass. Therefore the
radius stays constant if the speed stays constant
7
Problem 2
A copper flask (emissivity = 1) of radius 10 cm contains liquid argon at its boiling
point for a pressure of 1 atm, i.e. 87 K. The flask is enclosed in a steel vessel which
has a uniform temperature of 200 K. The flask is held to the top of the vessel by a
teflon neck, whose cross-sectional area if 1 cm2 and whose length is 10 cm. The top of
the teflon neck is in good thermal contact with the steel vessel, and the bottom of the
neck is at the temperature of the copper flask. The flask is open to the atmosphere
outside the steel vessel (pressure 1 atm).
Teflon
Neck
Copper
Flask
Stainless Vessel
a.) (5 points) What is the net thermal power adsorbed by the vessel due to the
radiant heat? You may neglect here the radiant power from the neck.
Here we call the area of the spherical flask A = 4πR2 . The net power adsorbed by the
flask through radiation is given by:
4
4
− Tflask
= 1 · 5.67 × 10−8
Prad = σA Tvessel
W
· 4π(0.1 m)2 · 2004 − 874 K4
2
4
m K
Prad = 10.98 W
8
b.) (5 points) What is the thermal power adsorbed by the vessel due to conduction
along the neck? The thermal conductivity of teflon is 0.2 W m−1 K−1 .
The area through which the heat is conducted is B = 1 cm2 = 10−4 m2 . The thermal
power adsorbed by conduction is:
Pcond =
kB∆T
0.2 W m−1 K−1 · 10−4 m2 · 113 K
=
L
0.1 m
Pcond = 0.023 W
c.) (5 points) What is the rate at which the argon inside the flask is boiling? The
heat of vaporization of argon is 6.43 kJ mol−1 . The molar mass of argon is 40 g.
Quote your answer in units of grams/seconds.
The total thermal power adsorbed is:
P = Prad + Pcond = 11 W
The rate of evaporation is:
11
J
1 mol
g
×
× 40
= 0.068 g/s
s
6430 J
mol
9
Problem 3
a.) (6 points) Two cars with masses of 1,500 kg each, both traveling at a speed of
100 km/hr, collide in a head-on collision and stick together. Find the change in
entropy of the Universe if the temperature is 27 degrees Celsius.
This is the “classic” head-on collision. Due to momentum conservation, the two cars
stop altogether. The final kinetic energy is zero. The initial kinetic energy, suddenly,
becomes “unavailable” for work and is degraded into heat or other forms (work required
for deformation) not available for performing work. The change in entropy is of course
positive, as with any irreversible process.
2
1
km
m 1 hr
Wunavailable = KEi = 2× mv 2 = 1, 500 kg 100
1000
= 1.16×106 J
2
hr
km 3600 s
∆S =
Wunavailable
1.16 × 106 J
=
= 3,860 J/K
T
300 K
10
b.) (6 points) The surface of the Sun has a temperature of 7,500 K and the surface
of the Earth is at 290 K. Find the change in entropy of the Universe per each
1,000 J of energy that are transferred, by radiant heat, from the Sun to the
Earth.
The adsorption of heat by the Earth causes an entropy increase. The release of heat
from the Sun causes an entropy decrease. The net balance is an entropy increase, since
the heat flows from a hot to a colder source. The process, of course, in not reversible.
∆S =
1, 000 J
1, 000 J
−
= 3.31 J/K
290 K
7, 500 K
11
Problem 4
Suppose that there is a cylindrical tunnel bored along the diameter of the Earth
through its center. At one end (entrance) of the tunnel, a small ball of mass m = 1 kg
is dropped from rest. In this problem, assume that the Earth is a solid sphere with
uniform mass density.
r R
a.) (4 points) What is the mass density of the Earth?
ME = 6.0 × 1024 kg
RE = 6.4 × 106 m
Therefore,
ρE =
ME
ME
3
3
=
3 = 5.5 × 10 kg/m
VE
(4π/3)RE
12
b.) (4 points) Find the gravitational force on the ball when it is at a distance r from
the center of the Earth. The gravitational force at this position is determined
as if all the mass inside the sphere of radius r (see the figure) is concentrated at
the center. It is also known that the mass outside the sphere of radius r does
not affect any net gravitational force on the ball.
Mass of the Earth inside the radius r is:
M (r < RE ) =
4 3
πr ρE
3
Therefore the gravitational force at r on a ball of mass m is:
ρE (4π/3)r3 m
= −GρE (4π/3)mr
F (r) = −G
r2
where the (−) sign indicates the direction of the force is toward the center.
The numerical value for GρE (4π/3) is 1.5 × 10−6 N kg−1 m−1 , to be used later.
c.) (4 points) The force found in part (b) implies that the ball moves in simple
harmonic oscillation. How long does it take for the ball to reach the other end
of the tunnel?
Note that the force found in (b) can be written in the form F = −kr, where
k = m · 1.5 × 10−6 N kg−1 m−1 . This is the force for simple harmonic osciallation.
Then, the angular frequency ω for the oscillation can be found as:
r
r
k
m · 1.5 × 10−6 N kg−1 m−1
ω=
=
= 1.2 × 10−3 s−1
m
m
The period T of the oscillation is:
T =
2π
= 5.1 × 103 s
ω
The time ∆t for the ball released from the entrance (r = RE ) to reach the other end
is:
∆t = T /2 = 2.5 × 103 s
13
d.) (4 points) What is the speed of the ball when it passes the center of the Earth?
In a simple harmonic oscillation, the energy E = (1/2)kx2 + (1/2)mv 2 is conserved.
Comparing the energy at the entrance and at the center, we have:
1 2
1
kR = mv 2 ,
2 E
2
where v is the speed of the ball at the center.
Therefore,
r
k
v=
RE = ωRE = 7.9 × 103 m/s
m
e.) (4 points) Now we release the ball from rest at a distance r = R/2 from the
center. (Don’t ask how we could get there to begin with.) For this case, what are
the period of oscillation and the speed of the ball at the center of the Earth?
In a simple harmonic oscillation, the period does not depend on the amplitude of the
motion. Therefore, the period T remains the same:
T = 5.1 × 103 s
On the other hand, reduing the amplitude to half (RE → RE /2) results in the change
in total energy, hence the change in the speed at the center.
r
k RE
RE
v=
=ω
= 4.0 × 103 m/s
m 2
2
14
Problem 5
A ball of mass m = 0.1 kg is attached to one end of a wire (of length ` = 1 m),
the other end being fastened to a fixed point. Initially the ball is released from rest
when the angle between the tightly stretched wire and the vertical is 60◦ . It swings
down and strikes a block (of mass M = 0.2 kg) initially at rest on a horizontal
frictionless surface. Assume that the air resistance is negligible and the collision is
elastic.
m
!
M
a.) (5 points) During the downward swing of the ball (before collision), are any
of the following observables conserved. Circle the observable(s) which are
conserved and provide a very brief explanation for your answer.
• the momentum of the ball
• the kinetic energy of the ball
• the total energy of the ball
The gravitation acts as an external force on the ball. Therefore, the momentum of the
ball is not conserved.
In the downward swing the gravitaional potential energy decreases, resulting in an
increase of the kinetic energy. Therefore, the kinetic energy of the ball is not conserved.
The total energy of the ball is conserved. Ignoring the air resistance (and any other
non-conservative forces), the conserved energy is equal to the sum of the gravitational
potential energy and the kinetic energy.
15
b.) (5 points) Find the velocity (magnitude and direction) of the ball just before
and just after the collision.
• Just before collision: let’s v1 be the velocity of the ball just before collision.
1
mgh = mg` (1 − cos(60◦ )) = mv12
2
p
v1 = 2g`(1 − cos(60◦ )) = 3.13 m/s
The direction is to the right.
• Just after collision:
Since the ball makes an elastic collision with the block, both the total linear
momentum and the total kinetic energy are conserved just before and just after
the collision.
One way to answer this problem is to solve the simultaneous equations arising
from linear momentum and kinetic energy conservations.
Another way is to work in the center-of-mass (CM) frame where the velocity
vectors simply change its sign before and after the collision.
Velocity of the CM:
vCM = mv1 /(m + M ) = 1.04 m/s
Let v1CM and V1CM be the veolcities of the ball and the block, respectively,
measured in the CM frame, just before collision:
v1CM = v1 − vCM = 2.09 m/s
V1CM = V1 − vCM = −1.04 m/s
Just after collision, the velocity vectors flip sign (denoted with a subscript “2”):
v2CM = −v1CM = −2.09 m/s
V2CM = −V1CM = +1.04 m/s
Coming back to the laboratory frame, the velocity of the ball becomes:
v2 = v2CM + vCM = -1.04 m/s
The (−) sign indicates that the ball moves to the left after the collision.
16
c.) (5 points) If the ball is in contact with the block for a time of 0.001 s, what is
the average force exerted on the ball?
Let F be the average force exerted on the ball during collision.
m(v2 − v1 )
0.1 kg · (−1.04 m/s − 3.13 m/s)
∆p
=
=
= - 417 N
∆t
∆t
0.001 s
The (−) sign indicates that the force (on the ball) is directed to the left.
F =
d.) (5 points) How high (measuring from the point of collision) will the ball reach
after the collision?
mgh =
h=
1
mv 2
2 2
v22
= 5.56 × 10−2 m
2g
17
Problem 6
Playing a baseball game, you are catching a ball (of mass m = 0.15 kg) with an
initial speed of 40 m/s (≈ 90 miles per hour) thrown by the pitcher in a horizontal
direction.
a.) (3 points) If the ball moves 18 m horizontally before being caught by you, how
much will it fall down due to vertical gravitational force? You may neglect air
resistance.
Time interval δt for the ball to move 40 m horizontally:
δt =
18 m
= 0.45 s
40 m/s
Vertical fall δy during this time is:
δy =
1
g · δt2 = 0.99 m
2
b.) (3 points) What is the vertical component of its velocity just before it is caught?
vy = −g · δt = 4.41 m/s
18
c.) (4 points) How much work must be done on the ball to stop it? (Do not forget
the vertical component of the velocity.)
The work needed to stop the ball is equal to the kinetic energy of the ball just before
being stopped.
1
1
W = ∆KE = 0 − mv 2 = − (0.15 kg) (40 m/s)2 + (4.41 m/s)2 = - 121 J
2
2
where we ignored the work done by the gravitational force during the stopping process.
The (−) sign indicates that the work was done to decrease the kinetic energy of the
object.
d.) (3 points) If the ball is stopped in 2.0 cm with a constant deceleration, what is
the average force acting on the ball while it is brought at rest?
Suppose that you apply a force in the opposite direction to the final velocity of the ball.
Then, the work (which you exert on the ball) can be written as W = F δs, where F is
the average force you exert on the ball, and δs is the stopping distance.
Then,
W
F =
= −6.07 × 103 N
δs
The (−) sign indicates that the force was applied in the opposite direction to the
displacement of the ball.
19
e.) (3 points) What is the average power transferred to the ball during the stopping
process in part (d)?
Assuming that the ball is stopped in a constant deceleration, the average velocity during
the stopping process is obtained by averaging the initial and final velocity:
v=
p
1
0 + (40 m/s)2 + (4.41 m/s)2 = 20.1 m/s
2
Then the average power P transferred to the ball is:
P = F v = 1.22 × 105 W
f.) (4 points) Suppose that the same amount of power as obtained in part (e) is
applied to 1 kg of water at 20◦ C. How long will it take to heat this water to
90◦ C?
Let M be the amount of water.
Q = CM ∆T = 4186 J kg−1 ◦ C−1 × 1 kg × 70 ◦ C = 2.93 × 105 J
The time ∆t required to transfer this amount of heat with the power obtained in part
(e) is:
Q
∆t =
= 2.40 s
P
20
Problem 7
Water emerges from a vertical faucet, whose diameter is D = 1 cm, with an initial
downwards velocity of v0 = 1 m/s. Thereafter, gravity accelerates the water, and
since water is incompressible the diameter of the water column must decrease so as
to keep the mass flux ρvA constant.
a.) (5 points) What is the diameter d of the column at a distance h = 30 cm below
the outlet of the faucet?
The velocity v of the water after falling through height h is given by:
v 2 = v02 + 2gh = (1 m/s)2 + 2 · 9.8 m s−2 · 0.3 m = 6.88 m2 /s2
so that v = 2.62 m/s.
The flow of incompressible water with mass density ρ has a constant flux of mass, ρvA,
where A is the area of the flow. Hence, v0 Atop = vAbottom , or πv0 D2 /4 = πvd2 /4,
and the diameter d of the flow at the bottom is:
p
p
d = D v0 /v = 1 · 1/2.62 = 0.62 cm
21
b.) (5 points) Water emerges from a hole in the side of a water tank at a height
h = 5 m below the water level. What is the velocity v of the water just after
it exits the hole in the tank? What is the pressure P1 inside the tank at the
height of the hole? You may ignore the viscosity of water in this situation, so
that energy is conserved. The tank is large enough that the vertical velocity v0
of water in the tank is negligible. Also, you may ignore the small difference in
the air pressure at the top of the tank and at the hole.
A solution can be based directly on conservation of energy, or on Bernoulli’s law, which
is based on conservation of energy.
One view is that the water that emerges from the hole has disappeared from the top of
the tank, so that it has gained gravitational potential energy PE = mgh. This energy
is converted to its kinetic energy KE = mv 2 /2. Thus:
v=
√
2gh =
p
2 · 9.8 m/s · 5 m = 9.89 m/s
Alternatively, we can invoke Bernoulli’s law that P +ρgh+ρv 2 /2 is constant for smooth
fluid flow in which viscosity can be ignored. Assuming also that the pressure P is the
same at the top of the tank and just outside the hole and that the velocity of the water
is negligible at the top of the tank, we obtain the relation gh = v 2 /2, which leads to
the boxed answer for v.
The pressure inside the tank at the height of the hole follows from another application
of Bernoulli’s law in which v = 0 in both locations. That is:
P1 = P0 + ρgh = 100, 000 Pa + 1000 kg m−3 · 9.8 m s−2 · 5 m = 149,000 Pa
22
c.) (5 points) The velocity of the water increases as it falls after emerging from the
hole of area A in the tank, and the area of the water jet decreases in a manner
related to part a of this problem. Careful observation reveals that, in addition,
the area of the water jet decreases rapidly for a short time just after it exits the
hole, before gravity has any effect on the velocity. The area of the jet after it
has converged is A0 , as shown in the figure.
To understand this effect, you will make an analysis in parts c and d using
the concept of momentum flux, Φp , i.e. the total momentum carried by all the
water particles that crosses a surface perpendicular to the flow per unit time.
Give an expression for the momentum flux Φp across area A0 in terms of the
velocity v of the jet across that area (which is the velocity that you found in
part b).
Hint: keep in mind that the mass flow is ρAv. What is the relation between
mass and momentum?
Momentum = mass times velocity, so momentum flux = mass flux times velocity, and,
recalling parts a and b:
Φp = ρv 2 A0 = 2ρghA0
23
d.) (5 points) Continue the analysis of the horizontal, converging flow of water out
of a hole of area A by considering a tube of area A within the tank, together
with the volume outside the tank that tapers to an area A0 , as shown in the
figure.
The pressure in the tank at the height of this tube is P1 , while the pressure
outside the tank is P0 . The horizontal force on the water in this tube is its area
times the pressure difference between its left and right ends.
But also, the force equals the rate of change of momentum in the tube, which
equals the momentum flux across the right end of the tube, if we choose the
length of this tube so that its volume equals the volume of water that leaves
the tank in 1 sec.
Combine this fact with your results from parts b and c to find the reduced area
A0 of the flow (a few cm outside the hole).
IF we knew the velocity of the water vH at the hole, then we could use the law of
mass conservation, v0 A0 = vH A. However, we do not know vH , so we need another
approach. As hinted, a second approach is based on momentum flux, rather than energy
conservation.
The horizontal force on tube is F = A(P1 − P0 ) and NOT AP1 − A0 P0 since air
pressure acts on the curved part of the jet as well as on its flat end such that the
total horizontal force is just P0 times the total vertical area of the jet outside the
hole, which area is simply the area of the hole = A.
The horizontal force equals the momentum flux, so A(P1 − P0 ) = ρv 2 A0 , using part
c. Also, from part b we have that P1 = P0 + ρgh and v 2 = 2gh. Altogether,
Aρgh = 2ghA0 , so that:
A0 = A/2
This analysis was first made by Borda in 1766.
24
Problem 8
A hexagonal pencil of mass m = 5 g has edges of width a = 3 mm. When placed
with its axis horizontal on an inclined plane of angle θ to the horizontal, the motion
(if any) of the pencil is in general a combination of rolling and sliding.
a.) (5 points) If the coefficient of static friction between the pencil and the plane is
µ = 0.3, what is the largest angle θmax for which the pencil will remain at rest
if placed on the plane at rest?
The forces on the pencil when at rest are shown in the figure to the right. The force of
gravity has components mg cos θ and mg sin θ perpendicular and parallel to the inclined
plane, respectively. The normal force N balances the perpendicular component of the
force of gravity, so that N = mg cos θ. The force of static friction point up the slope
and has magnitude F = µN = µmg cos θ. The pencil does not slide so long as the
force of friction exceeds the component of the force of gravity parallel to the slope. The
critical condition is therefore, mg sin θmax = F = µmg cos θmax , or µ = tan θmax , and
hence, θmax = tan−1 µ = tan−1 0.3 = 0.29 rad = 16.7◦ .
25
b.) (5 points) In the rest of the problem the angle of the inclined plane is less than
θmax , so the pencil moves only if it is given an initial velocity. The pencil then
rolls in a sequence of 1/6 turns during each of which the edge of the pencil in
contact with the plane serves as the instantaneous axis of rotation.
At the end of each 1/6 turn a face of the pencil collides with the inclined plane.
Energy is lost during this collision, and so there exist a “terminal velocity” of
the pencil in which case the energy lost in each collision just equal the potential
energy change of the pencil during the previous 1/6 turn.
Give an expression for the change in potential energy, ∆PE, during each 1/6
turn of the pencil.
As seen in the figure to the right, during a 1/6 turn, the center of mass of the pencil
moves distance a parallel to the inclined plane, and hence the CM falls by height a sin θ.
Thus, ∆PE = mga sin θ.
26
c.) (5 points) A simple model of the collision at the end of each 1/6 turn of the pencil
between the pencil and the plane is that a fixed fraction of the kinetic energy
gets lost during the collision. For the pencil and the plane under consideration,
the fraction of kinetic energy lost in each collision is 50%. Give an expression
for the change of kinetic energy, ∆KE, during a collision in terms of the velocity
v of the CM just before the collision. I = 17ma2 /12 is the moment of inertia of
a hexagonal pencil about one of its edges.
The angular velocity just before the collision is ω = v/a.
We know the moment of inertia I about the instantaneous axis of rotation, so the total
kinetic energy is simply KE = Iω 2 /2. We do NOT add mv 2 /2 to this; the energy
of the CM motion is already included. If we want to write KE = mv 2 /2 + I 0 ω 2 /2,
then I 0 must be the moment of inertia about the center of mass, which is ICM =
7ma2 /12. Then, since v = aω, we find that KE = ma2 ω 2 /2 + (7ma2 /12)ω 2 /2 =
17ma2 /12)ω 2 /2 = Iω 2 /2.
Since 1/2 of this KE is lost during the collision, the change of kinetic energy during a
collision is:
∆KE = Iω 2 /4 = Iv 2 /4a2 = 17mv 2 /48
27
d.) (5 points) During “steady” rolling of the pencil down the plane the kinetic
energy lost in each collision with the plane equals the change in potential energy
experienced each 1/6 turn. What is the velocity v at the end of a 1/6 turn for
this “steady” motion when the angle θ of the inclined plane is 15◦ ?
From parts b and c,
∆KE = Iv 2 /4a2 = 17mv 2 /48 = ∆PE = mga sin θ
So:
v=
p
p
p
4mga3 sin θ/I = 48ga sin θ/17 = 48 · 9.8 · 0.003 · 0.26/17 = 0.15 m/s = 15 cm/s
28
Problem 9
a.) (9 points) A tube of length L = 1 m is closed at one end, open at the other,
and rotates with uniform angular velocity about an axis perpendicular to the
closed end, as shown in the figure, which motion excites resonant vibrations in
the air inside the tube. What is the lowest frequency heard by an observer in
the plane of the rotation, if the period of rotation is 1 sec?
The lowest frequency vibration of a tube of length L that has one end closed and one
end open has wavelength λ = 4L = 4 m. The frequency f of the vibration is given by
f = vs /λ = (343 m/s)/(4 m) = 85.8 Hz.
However, because the open end of the tube is moving, an observer at rest hears
Doppler-shifted frequencies. The lowest frequency heard occurs when the end of the
tube is moving away from the observer.
The velocity of the end of the tube is vs = ωL = 2πL/T = 2π m/s for length L = 1 m
and period T = 1 s.
The observed frequency fo is given by the appropriate Doppler formula:
fo =
f
1+vs /v
=
85.6
1+2π/343
= 84.1 Hz
29
b.) (6 points) When a guitar string is plucked, its initial waveform is triangular,
as shown in the upper figure below. The subsequent motion of the string is
periodic with a period T . Sketch the waveforms at times T /4 and T /2 in the
blank grids below
Recall that a standing wave is equivalent to the sum of a left-moving traveling
wave and a right moving traveling wave: 2 cos(kx) cos(ωt) = cos(kx − ωt) +
cos(kx + ωt). A similar decomposition holds for the triangular wave on the
guitar string. The spatial period of the wave is 2L if the length of the string is
L, so the left- and right-moving component waves at time t = 0 both have the
form of the dotted curve in the figure.
30
At time t = T /4 the left and right moving triangular waves are as shown by the dotted
lines in the upper figure below. Adding these curves in the physical region of length L,
we obtain the desired waveform. The same goes for the time t = T /2.
31
POSSIBLY USEFUL CONSTANTS AND EQUATIONS
You may want to tear this out to keep at your side
x = x0 + v0 t + 12 at2
F = µFN
s = rθ
p = mv
ω = ω0 + αt
KE = 21 Iω 2
W = F s cos θ
2
ac = vr
τ = F ` sin θ
Ifull cylinder = 21 mr2
Ifull sphere = 52 mr2
F = Y A ∆L
L0
P V = nRT
Q = mL
V
Wisotherm = nRT ln Vfi
Q
= kA
∆T
t
L
Q = σT 4 At
v = λf
v = v0 + at
F = −kx
v = rω
F ∆t = ∆p
P E = 12 kx2
KE = 12 mv 2
L = Iω
ω 2 =p
ω02 + 2α∆θ
ω = k/m√
2πr3/2 = T GM
Ihollow cylinder = mr2
∆L = αL0 ∆T
P1 V1γ = P2 V2γ
P V = nkT
Q = cm∆T
γ =q
CP /CV
v=
v=
√
γkT
m
v2 = v02 + 2ax
F = −G Mr2m
a = rα
P
xcm = M1tot i xi mi
P E = mgh
Wnc = ∆KE + ∆P E
P
τ = Iα
∆θ =pω0 t + 12 αt2
ω = g/l
P
I = mi ri2
Q = πR4 ∆P/8ηL
Q = Av
P + ρgh + 21 ρv 2 = const
∆S = ∆Q
T
∆U = Q − W
U = 32 nRT
q
F
v = m/L
√
v = Yρ
β = (10 dB) log II0
f0 = fs / 1 ∓ vvs
sin θ = Dλ
Bad ρ v
1± 0
f0 = fs 1∓ vvs
v
v
fn = n 2L
sin θ = 1.22 λd
Monatomic:
Diatomic:
CV = 3R/2
CV = 5R/2
CP = 5R/2
CP = 7R/2
R = 8.315 J/K/mol
u = 1.66 × 10−27 kg
σ = 5.67 × 10−8 W/m2 /K4
NA = 6.022 × 1023 mol−1
k = 1.38 × 10−23 J/K
Mearth = 6.0 × 1024 kg
Mmoon = 7.4 × 1022 kg
0◦ C = 273.15 K
Rearth = 6.4 × 106 m
Rmoon = 1.7 × 106 m
1 kcal = 4186 J
GNewton = 6.67 × 10−11 Nm2 /kg2
f0 = fs (1 ± vv0 )
v
fn = n 4L
vsound = 343 m/s