Physics 141.
Review Exam # 3.
Fuel efficient aviation.
(see http://www.treehugger.com/files/2009/02/fuel-efficient-aviation.php)
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Exam # 3.
• Same format as previous exams:
•  10 multiple choice questions
•  3 analytical questions
• The three analytical questions will be distributed as follows:
•  One question will come from the material discussed in Chapter 11.
•  One question will be an equilibrium question.
•  One question from come from the material discussed in Chapter 12.
• Note:
•  Practice exam # 3 covered also the material discussed in Chapter 13
(question 11).
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Midterm Exam # 3.
Chapter 11.
• The focus of this Chapter is rotational motion and angular
momentum.
• Rotational motion and angular momentum is described in
terms of angular variables, such as angular position,
velocity, and acceleration, and torque.
• There is a great deal of symmetry between the way we use
linear variables and the way we use of angular variables.
• We discussed the requirements of conservation of angular
momentum (no external torques)
• We also discussed the concept and consequences of the
quantization of angular momentum.
• Sections excluded: 11.12 (page 453 – 455), and 11.13.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
1
Review Midterm Exam # 3.
Chapter 11.
• Terminology introduced:
•  Angular position, velocity, and acceleration.
•  Rotation axis.
•  Moment of inertia.
•  Rolling motion.
•  Torque.
•  Angular momentum.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11. Rotational variables.
•  The variables that are used to
describe rotational motion are:
•  Angular position θ
•  Angular velocity ω = dθ/dt
•  Angular acceleration α = dω/dt
•  The rotational variables are
related to the linear variables:
•  Linear position l = Rθ
•  Linear velocity v = Rω
•  Linear acceleration a = Rα
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11. Rotational variables.
Angular velocity and acceleration are vectors! They have a magnitude and
a direction. The direction of ω is found using the right-hand rule.
The angular acceleration is parallel or antiparallel to the angular velocity:
If ω increases: parallel
If ω decreases: anti-parallel
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
2
Review Chapter 11. The moment of inertia.
•  The kinetic energy of a rotation
body is equal to
K=
1 2
Iω
2
where I is the moment of inertia
which is defined (for discrete
mass distributions) as
I = ∑ mi ri 2
i
•  For continuous mass distributions
I is defined as
I = ∫ r 2 dm
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11.
Moments of inertia.
•  As part of the exam you will
receive a table of moments of
inertia for various objects (see
Figure on the right).
•  There will be no analytical
questions that require you to
calculate the moment of inertia
for non-uniform objects (like you
had on WeBWorK).
•  But ….. You need to know who
to determine the moment of
inertia using the parallel-axis
theorem.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11. Parallel-axis theorem.
•  Calculating the moment of
inertial with respect to a
symmetry axis of the object is in
general easy.
•  It is much harder to calculate the
moment of inertia with respect to
an axis that is not a symmetry
axis.
•  However, we can make a hard
problem easier by using the
parallel-axis theorem:
I = I cm + Mh2
Frank L. H. Wolfs
Easy
Hard
Icm
I
Department of Physics and Astronomy, University of Rochester
3
Review Chapter 11. Rolling motion.
•  Rolling motion is a combination
of translational and rotational
motion.
•  The kinetic energy of rolling
motion
has
thus
two
contributions:
•  Translational kinetic energy =
(1/2) M vcm2.
•  Rotational kinetic
(1/2) Icm ω 2.
energy
=
•  We assume that the wheel does
not slip: ω = v / R.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11. Torque.
•  The torque τ of the force F is
proportional to the angular
acceleration of the rigid body:
τ = Iα
  
τ =r×F
•  This equation looks similar to
Newton’s second law for linear
motion:
F = ma
F
φ
r
A
•  Note:
linear motion rotational motion
mass m
moment I
force F
torque τ
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11. Angular momentum.
•  The angular momentum is
defined as the vector product
between the position vector and
the linear momentum.
•  Note:
•  Compare this definition with the
definition of the torque.
•  Angular momentum is a vector.
•  The unit of angular momentum is
kg m2/s.
•  The angular momentum depends
on both the magnitude and the
direction of the position and linear
momentum vectors.
•  Under certain circumstances the
angular momentum of a system is
conserved!
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
4
Review Chapter 11. Angular momentum and circular motion.
•  Consider an object carrying out
circular motion.
•  For this type of motion, the position
vector will be perpendicular to the
momentum vector.
•  The magnitude of the angular
momentum is equal to the product
of the magnitude of the radius r and
the linear momentum p:
L = mvr = mr2(v/r) = Iω
•  Note: compare this with p = mv!
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11. Conservation of angular momentum.
• Consider the change in the angular momentum of a particle:



⎛  dv dr  ⎞
dL
d  
   
=
r × p = m⎜ r ×
+
× v⎟ = m r × a + v × v =
dt
dt
dt dt
⎝
⎠



 
= r × ma = r × ∑ F = ∑ τ
(
)
(
)
• When the net torque is equal to 0 Nm:



dL
∑ τ = 0 = dt ⇒ L = constant
• When we take the sum of all torques, the torques due to the
internal forces cancel and the sum is equal to torque due to
all external forces.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11. Quantization of angular momentum.
•  Consider the "classical" picture of the
motion of electrons in atoms.
•  The angular momentum is a integer
multiple of h/2π, the orbit must be
such that
rp = N
h
= N
2π
•  This leads to a quantization of the
orbital radius and energy:
r = 4πε 0
N 22
me2
2
1 ⎛ 1 ⎞ me4
13.6
E = K +U = − ⎜
= − 2 eV
2 ⎝ 4πε 0 ⎟⎠ N 2  2
N
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
5
Review Chapter 11. Quantization of angular momentum.
• The energy levels of an electron in the Hydrogen atom
exactly match the levels predicted using this simple model,
and the quantization of the energy levels is a direct
consequence of the quantization of angular momentum.
• In addition to the orbital angular momentum of the electrons
in the atom, they also poses spin. The projection of the spin
of the electron on a particular axis will be either +(1/2)h/2π
or - (1/2)h/2π. It will never be zero. The electron is said to
have be a spin 1/2 particle.
• Many other particles, such as muons, neutrinos, and quarks,
are spin 1/2 particles.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 11. Quantization of angular momentum.
• Since quarks are the building blocks of hadrons, we also
expect that hadrons have a well defined spin.
•  Hadrons that contain three quarks can either be spin 1/2 or spin 3/2.
•  Hadrons that contain two quarks can either be spin 0 or spin 1.
• The total spin of a particle limits how particles can be
distributed across the various energy levels of the system.
•  If the spin is a half integer, the particle is called a Fermion, and it
must obey the Pauli exclusion principle (two fermions can not be in
the exact same quantum state).
•  If the spin is an integer, the particle is called a Boson, and it is not
subject to the Pauli exclusion principle (there is not limit to the
number of Bosons that can be in the exact same quantum state).
• The spin of macroscopic objects will also be quantized, but
the difference between different spin states is so small that it
is impossible to observed effects of this quantization.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Example Problem: Problem 11.P62.
• A yo-yo is released
from rest with the string
vertical.
• Determine the tension
in the string as the yoyo falls.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
6
Example Problem: Problem 11.P84.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Midterm Exam # 3.
Equilibrium.
• This topics focuses on the conditions for equilibrium.
• The conditions for equilibrium are:
•  First condition: net force = 0 N
•  Second condition: net torque = 0 Nm
• Both conditions must be satisfied for the object to be
equilibrium.
• Static equilibrium:
•  The conditions for equilibrium are met.
•  P = 0 kg m/s
•  L = 0 kg m2/s
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Midterm Exam # 3.
Equilibrium.
•  Equilibrium in 3D:
∑F
∑F
∑F
x
=0
y
= 0 and
z
=0
With respect to every reference point!
∑τ
∑τ
∑τ
x
=0
y
=0
z
=0
•  Equilibrium in 2D:
∑F
∑F
∑τ
x
=0
y
=0
z
=0
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
7
Example Problem
Problem 11.P56
• Calculate the torque due to
the gravitational force of
each person.
• Can you determine
normal force FN?
the
• What is the net torque with
respect to the pivot point?
• When will the seesaw
•  Rotate clockwise?
•  Rotate counter clockwise?
•  No rotate?
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12.
• The focus of this Chapter is a discussion of the applications
of the statistical model of solids and gases. Many important
properties of solids and gases can be derived using the
fundamental assumption of statistical mechanics.
• Using the fundamental assumption of statistical mechanics
we can predict the energy distribution of molecules in a gas
(this distribution is known as the Boltzmann distribution).
The corresponding velocity distribution is known as the
Maxwell-Boltzmann velocity distribution.
• Sections excluded: none (sorry).
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12.
• Terminology introduced:
•  Reversible and irreversible processes.
•  The fundamental assumption of statistical mechanics.
•  Entropy and thermal equilibrium.
•  Temperature.
•  The Boltzmann distribution.
•  The Maxwell-Boltzmann distribution.
•  Most-probable, average, and root-mean-square velocities.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
8
Review Chapter 12. Reversible and irreversible Processes.
•  Many processes in physics are
reversible (e.g. elastic collisions,
projectile motion).
•  Other processes, such as the
process of achieving thermal
equilibrium, are irreversible.
This implies that the process
appears always in one particular
order (e.g. heat flows from hot to
cold).
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12. The fundamental assumption of statistical mechanics.
•  In order to determine the
probability to observe a certain
configuration, we rely on the
fundamental
assumption
of
statistical mechanics to make this
determination:
The fundamental assumption in
statistical mechanics is that in our
state of microscopic ignorance, each
microstate (microscopic distribution
of energy) corresponding to a given
macrostate (total energy) is equally
probable.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12. Microstates vs Macrostates.
•  Each state shown in the Figure on the
right is a micro state.
•  Since our assignment of dof 1, dof 2,
and dof 3 is arbitrary we cannot
distinguish various microstates. For
example, (4,0,0), (0,4,0) and (0,0,4) will
look exactly the same. These three
microstates belong to the same macro
state.
•  There are a total of 4 macro states for
the system shown in the Figure.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
9
Review Chapter 12. Microstates vs Macrostates..
•  In this example, dof 1 has a different
energy level as dof 2 and dof 3. Thus,
dof 1 can be distinguished from dof 2
and dof 3.
•  Consider that the system has 2 quanta
of energy and that both quanta are in
the same dof.
•  There are three ways to do this:
•  (2, 0, 0)
•  (0, 2, 0)
•  (0, 0, 2)
•  (2, 0, 0) is one macro state.
•  (0, 2, 0) and (0, 0, 2) are two
microstates that belong to the same
macro state.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12. Entropy.
•  The evolution of a system depends
on the number of microstates, the
concept of entropy is introduced.
•  The entropy S of a system with Ω
states is defined as
S = k lnΩ
where k is the Boltzmann constant
-24
(1.4 x 10 J/K).
•  The entropy of complex system is the
sum of the entropy of each subsystem that makes up the system.
•  The most probable configuration is
the configuration for which the
entropy has a maximum (this is the
second law of thermodynamics).
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12. Entropy and the temperature.
•  The temperature of a system is
defined as
1
dS
=
T dEint
•  The temperature defined in this
manner is expressed in units of Kelvin
(K).
•  Two objects are in thermal
equilibrium if their temperatures are
the same.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
10
Review Chapter 12. The Boltzmann distribution.
•  If we consider a single atom in contact
with a system, consisting of a large
number of atoms, we can show that the
number of states of the combined
system is proportional to e-ΔE/KT, where
ΔE is the excitation energy of the single
atom.
•  Since the probability is proportional to
the number of states, we conclude that:
•  The probability of finding a microscopic
system to be in a state with energy ΔE
above the ground state of the system is
proportional to e-ΔE/KT.
•  This probability distribution is called
the Boltzmann distribution.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12. The Boltzmann distribution.
•  Consider that the gas molecule is
moving with a velocity vx along
the x axis.
•  The energy of the gas molecule,
associated with its motion along
the x axis, will be (1/2)Mvx2.
•  The probability of finding the gas
molecule with a velocity between
vx and vx + dvx is equal to
( )
P vx dvx ∝ e
⎛1
⎞
− ⎜ Mvx 2 ⎟ / kT
⎝2
⎠
dvx
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12. The Maxwell-Boltzmann speed distribution.
⎛ M ⎞
P v = 4π ⎜
⎝ 2π kT ⎟⎠
()
3/ 2
v 2e
⎛1
⎞
− ⎜ Mv 2 ⎟ / kT
⎝2
⎠
Integral = N
Most probable v
Average of v2
Average v (50% below, 50% above)
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
11
Review Chapter 12. The Boltzmann distribution and internal energy.
• The root-mean-square kinetic energy associated with the three
translational degrees of freedom is equal to
K rms = K x,rms + K y,rms + K z,rms =
3
kT
2
• It turns out that the average energy associated with each degree
of freedom, including vibrational and rotational degrees of
freedom, is (1/2)kT.
• Note:
•  The internal energy only depends on the temperature; it does not depend
on the mass of the gas molecules.
•  At a given temperature, the rms velocity of heavier molecules will be
smaller than the rms velocity of lighter molecules.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12. Other degrees of freedom.
• For
a
diatomic
molecule other degrees
of freedom must be
considered.
There
importance depends on
their energy spacing
compared to kBT.
• Order of importance:
• Rotational motion
• Vibrational motion
• Electronic excitation
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Review Chapter 12. The Boltzmann distribution and internal energy.
CV = ΔQ / ΔT
Remember: Q = C ΔT.
For one degree of freedom (d.o.f):
K = (1/2)NkT and ΔK = (1/2)Nk ΔT
Frank L. H. Wolfs
2 d.o.f.: U = kT
2 d.o.f.: U = kT
3 d.o.f.: U = (3/2)kT
Department of Physics and Astronomy, University of Rochester
12
Example Problem:
Problem 12.P38.
• Calculate the entropy of this system for a total energy of 0,
1, 2, 3, 4, and 5 quanta.
• Calculate the approximate temperature of the system when
the total energy is 4 quanta.
• Calculate the heat capacity when the total energy of the
system is 4 quanta.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Study tips.
• Review the homework assignments related to this material
and look at the solutions that are posted on the WEB.
• Review the end-of-chapter problems, especially those for
which you have received the solutions. However, make sure
you read all other problems and determine if you know what
approach to take to solve them.
• Use the practice exam to determine how well prepared you
are for the exam, but please note that Chapter 13 was also
included on that exam when it was given.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
Good luck preparing for exam # 3.
Frank L. H. Wolfs
Department of Physics and Astronomy, University of Rochester
13