LECTURE NOTES FOR PHYSICS 307: CLASSICAL MECHANICS
TEXT: THORNTON AND MARION'S CLASSICAL MECHANICS OF PARTICLES AND SYSTEMS
SOME USEFUL REFERENCE STUFF: Greek alphabet, metric prefixes, conversion factors
ASIGNMENTS: (Subject to change: check back often.)
HW #2: Due Friday, Sept. 12: Problems 2.6, 16, 17, 32. (If you can show 1∓2μcosθ - 2sinθ = 0 for prob. 32, that
will be good enough. The upper sign in the ∓ is for the case of the block moving up the slope. If it moves
downward, the friction has the opposite direction. I'll consider extra points if you want to come up with an
expression for θ.)
HW #3: Due Friday, Sept. 19: Problems 2.24, 27*, 47; 9.55, 62.
*Consider the center of mass of the rope!
"Dailies":
Tuesday, Sept. 9: Draw a FBD for Problem 2.16.
Thursday, Sept. 10: Draw a FBD for Problem 2.32.
r  r0

  2 ( rb r0 )

 2e b  , show that the equilibrium is at
Tuesday, September 15: For the Morse potential, V ( r )  V0 e


r=r0 and is stable.
WEEK 3: GO TO LECTURE 4, 5
LECTURE 4: SECTION 2.6, these notes
INCLINED PLANES
We'll start by playing catch up on what we missed in class Thursday.
PULLEYS
OK, here comes something to complicate your life. What happens if a cord connecting two blocks in a problem
passes around a pulley? The effect of the pulley is to change the direction of the tension in the rope. If the pulley
has no mass (the physicist's ideal of the "massless pulley") then the total force on the pulley has to equal zero,
whether or not the pulley is accelerating. In statics, the net force on the pulley is zero as well. If we have a pulley
connected to one edge of a table, or the top of an inclined plane, the support of the pulley will provide the force
that causes the total to equal zero. (But, since we hardly ever are interested in knowing the size of this force, we
tend to ignore it.)
Usually, though, the magnitude of the tension on two parts of a single string passing over a pulley is the same,
only the direction has been changed. This makes two-body problems easier. (See below)
TWO-BODY PROBLEMS
Consider two blocks sitting on top of each other. Let's let the top block weigh 5 lbs and the bottom block weigh 10
lbs. With what force must the floor push up in order to support these two blocks? Your intuition tells you 15 lbs.
But what if we didn't have an intuitive answer? Or, how do we approach problems involving two objects that aren't
as intuitive, such as the same problem in an accelerating elevator?
Consider the problem we just posed. Draw a picture of the problem now. The next task is to isolate the two
objects and consider each one independently. The best way to do that is to draw a dashed circle around each
object. Then draw a free body diagram for each object. Don't be afraid to use up more space on your page
redrawing your figures.
What will your free body diagrams look like for the problem which I proposed? Draw them first then read the rest
of this paragraph. The box on top will have only two forces acting on it: a weight, W1, pulling down, and a normal
force, N1, pushing up from the lower block. The bottom block will have two forces acting on it: N1, pointing down, a
force exerted by contact with the block above; W2, the weight of this block, pulling down, and N2, the contact force
exerted by the floor. Now this is the trick to dealing with more than one object. Isolate each object, and then
consider each object as if it were the only object in the universe. Later, the mathematics will allow you to show
that the two objects are connected, but for now pretend that they aren't.
In this problem, we quickly see that W1 = N1 = 5 lb, and N2 = N1 + W2 = W1 + W2 = 15 lb. In most other problems
involving two objects, the math will be more involved. I simply use this problem to show the basic principles of
solving such a problem.
OK, let's go to a more difficult problem. These same two blocks are connected to a single piece of rope which
hangs over a pulley. Draw this picture, isolate the two blocks, and draw a free body diagram for each. Do it now:
don't peek at the following paragraph.
Did you draw any normal forces? If so, shame on you. The only thing touching either block is the rope, and so
there will be an equal tension, T1, pulling up on each block. In addition to that, each block will have a weight
pulling down, W1 or W2. Now, the system will accelerate. The 5 lb block will accelerate upwards and the 10 lb.
block will accelerate downwards. Draw a bent arrow (“”) --- we will reserve straight arrows for forces --- pointing
up for the small block and another one pointing down for the big block. Label these as accelerations. Write down
the sum of the forces equations for each block. For the small block, you should get ΣFy= +ma = T1-W1. For the
large block, you should get ΣFy =-ma = T1-W2.
To solve an equation like this, an equation for two unknowns -- T and a -- in two equations, first solve one of the
equations for T and then plug this into the other equation. Then you can solve the other equation for a, which is
the only one of these two unknowns that you really care about. We will obviously do other problems, including
obviously more difficult problems involving two or more objects in class.
PROJECTILE MOTION
Projectile motion problems are a class of problems in two dimensions that are characterized by the acceleration
having only a negative-y component, namely
ax  0
ay  g
.
The mathematics of these problems is that you need to separate the x motion and the y motion and treat them as
if they are independent of each other. These equations are not strictly independent because both sets of
equations share one identical quantity, namely time. As far as the techniques for analyzing such problems,
whenever possible you should draw two diagrams for each problems. Label these "before" and "after". Label each
of these diagrams with all of the quantities relevant to the before and after pictures. Between the two pictures, you
might want to label the acceleration.
One more thing that often comes up in projectile motion problems is that the initial velocity components may be
given in terms of magnitude and direction rather then x and y components. For this you can simply write
vox  vo cos 
v oy  v o sin 
An interesting example that shows that you can separate out the x and y motions is the case of a football kickoff
or punt. There are two statistics of interest for each kick. One is the "hang time" and the other is the distance that
the ball travels down the field. If we consider the x motion and the y motion separately, we can show that knowing
only the hang time, we can calculate the y component of the initial velocity, and knowing the distance that the ball
travels down the field and the hang time, we can directly calculate the horizontal component of velocity. We shall
do this in class. So, if you've ever wondered how hard the kicker needs to kick the ball, you can get all the
relevant data from sitting in front of the tube this weekend. (We are obviously ignoring wind resistance (drag).
We'll introduce this in Section 2.4, about a week from now.)
As far as the mathematics involved in simple projectile problems (no drag), if the projectile begins and ends at the
same height, you have a linear equation. If the initial and final heights are different, you need to solve a quadratic
formula for time. Review how to solve quadratic equations now, if you are rusty.
Aside: If asked for the final velocity in a projectile problem, please do not say "zero". Okay, sure, an object ends
up with zero velocity when it slams into the pavement, but that's true for all projectiles: it's not a quantity worth
asking about. We will define "final" as the instant just before impact, when the conditions of free fall are still in
effect (ax=0, ay=-g)
SECTION 2.5 and 2.6: CONSERVATION THEOREMS and ENERGY
I like to ask my students what the two most important terms in Physics are. My choices are "force" and "energy".
Each of these two words describes a process for analyzing physical motion. Mechanics is the central topic in
Physics: its concepts and its methods reappear in fields as diverse as thermodynamics, electricity and
magnetism, relativity.
We have just completed a weeklong or so tour of the "Force Approach" for dealing with motion. We will return to it
again in a bit, but first we will re-introduce the "Energy Approach". These two approaches are complementary, so,
if you solve a problem using each approach, the answers should be consistent, and you can decide to use
whichever one is more convenient. "Energy" has the edge in my book because it is not a vector, and so the
mathematics is much simpler.
One way to approach the concept of energy is by looking at symmetries. Physicists love to talk about symmetries
because, if we know the symmetry of a problem, it allows you to more easily guess the form of the solution. Let's
start with a few symmetries. In each case, the symmetry of the potential energy function (righthandmost column)
is equivalent to some quantity being zero.
IF......
*

F
 0

N  0*
....THEN
SYMMETRY TYPE:

p is conserved
Translational symmetry: U(x)=U(x+R)

L * is conserved
Rotational symmetry: U(x,y,z)=U(r)
There are no "nonconservative"
forces
E=U+T (total mechanical energy) is
conserved
Time-inversion symmetry:
U(x,y,z,t)=U(x,y,z,-t)
  
  
N  r  F and L  r  p .
The relation between zero force or zero torque, on the one hand, and conservation of either momentum or
angular momentum on the other hand, is a very important pattern to be aware of in physics. Memorize these
connections.
Notice that the last relation tells us that, under certain circumstances, the total mechanical energy is conserved.
This is the principle of conservation of energy.
I said that the Force and Energy Approaches yield consistent results: they never contradict each other. This is
unlikely to be due to just chance. The two approaches are fundamentally the same, and the quantity that ties
them together is "work". Two things to beware of with work. First, this is one example where our common sense of
what the English word work means is not very useful in Physics: here work is defined by the integral
 
W   F  ds
Physicists' definition of work
Second, we will use a capital "W" to stand for it: it is easy to confuse this with the "W" for weight. Be careful.
By the way, you may have noticed that Quantum Mechanics is based on energy, not force. We personalize
Schrödinger’s Equation by inserting the appropriate potential energy for a given problem, not the force.
LECTURE 5: SECTIONS 2.6, 9.11 (Return to top.)
WORK THEOREMS:
The definition above tells us how work is related to force. As far as its relation to energy, here goes. First of all, if
we do work against gravity by lifting an object, gravity does negative work. If we then release the object, gravity
will do positive work in returning it to where it started. We can say that we have stored the energy of the gravity in
the object, or that we have increased its gravitational potential energy. Notice that gravity did negative work, so
we can define potential energy in terms of the work done on the object:
ΔU = -W
WARNING: This (defining the potential energy) works only for conservative forces, forces for which the work done
in going from one point to the other does not depend on the specific path taken. What happens if nonconservative forces act? Answer: The total mechanical energy is no longer conserved, and so it no longer makes
sense to talk about some potential energy that we bank and can get back at some later time. In fact, the amount
of energy it loses is equal to the magnitude of the [negative] work done by the nonconservative forces:
Wnc  Emech
More generally, the total work done by all the forces is related to the change in kinetic energy by the Work-Energy
Theorem:
Wnet  KE
Work-Energy Theorem
Let’s focus on systems in which total mechanical energy is conserved. Given how the kinetic energy is related to
the [1D] velocity, v, we can come up for an expression for the velocity as a function of position, as long as we
know how the potential energy varies with position (U(x)):
v( x)  
2
m
E  U ( x ) 
By rewriting the velocity as the derivative of position with respect to time, rearranging things, and integrating, we
can even calculate the time required to travel from one position, x, to another:
t  t0   
dx
2
m E  U ( x ) 
Isn’t calculus wonderful? One of the best applications for this formula is to calculate how the period of motion of a
pendulum depends on the amplitude. To zeroth order, it doesn’t (which is what you were told in Freshman
Physics). In fact, it does depend on amplitude, but not very dramatically until the amplitude gets close to 180º.
We’ll get back to this in Chapter 3.
SECTION 2.6: EQUILIBRIUM
One of the notable features of the harmonic oscillator is that it has an equilibrium state around which it oscillates.
Why is this? The gravitational force near the surface of the Earth doesn’t have such an equilibrium. The second
most notable feature is that this is a stable equilibrium, which means that the motion returns to this position again
and again. An unstable equilibrium, on the other hand is one that is not revisited because once you depart from it,
the force points away from it. Take as an example a pencil (or egg) balanced on its point (or smaller end), or a
salt shaker balanced at an angle on a table. You can picture these two conditions (equilibrium and stable
equilibrium) graphically by plotting the potential energy vs position: equilibrium occurs whenever U(x) has a zero
tangent, and that equilibrium is stable if the function curves upward at that point.
Now let’s look at the relation between force and potential energy. In a one-dimensional system:
dU
; U    Fdx
Relation between U(x) and Force
dx
 dU 
U ( x )  U 0  x
  ... Expansion of potential energy around a point x0
 dx  x 0
F 
So
At equilibrium, the force on a particle is zero, so the expansion of U(x) above must have zero as its second term
when expanded at x0 equal to the equilibrium point.
F  x
In three dimensions, we can generalize to
0
 dU 
 
 0
 dx  x0


U
, etc.
F  U or Fx  
x
 
U    F  ds

with equilibrium occurring when all components of F are zero: Fx r  0  Fy   Fz r
r
0
0
0
The condition for stability can be expressed by the direction of the force vector close to equilibrium:
  dFy 

  dFx 

  dFz 

  0 
 
  dx    0 
  dz    0 
 r0
 r0




  dy  r0

 2 
 and  2
 . Condition for stable equilibrium
 and  2 
 d U   0
 d U   0
 d U   0
 dx 2  r

 dz 2  r

 dy 2  

0
0


r0




SECTION 9.11: ROCKETS
So why do we spend any time on rocket motion? What's important/useful/interesting about it? Mostly, on the
conceptual level, it's that it is an application of Newton's Second Law in which the mass changes, so that you
have to use the momentum formalism, rather than F=ma. It is also the first example of non-uniform acceleration
that we will deal with using the Force Approach. On the mathematical side, it gives us more experience with
differential equations than we get from simple UAM (uniformly accelerated motion) problems, tools which we will
hone in the next class, on drag.
The first thing to pull out of this material is how to deal with the derivative of p=mv. The simple answer is that we
apply the product rule: d/dt(mv)=m(dv/dt)+(dm/dt)v. In free space, where the net external force is zero, the total
momentum of the rocket plus its exhaust is conserved. We use u to denote the velocity of the exhaust. This is an
important short-term-memory item. Memorize this for the following material, and next to any equations you feel
like circling, mark in what the quantities stand for, especially that u is the fuel velocity. At any rate, we find the
following is true in the case of a rocket firing in free space: (a theoretical ideal which never occurs, except
inasmuch as any external forces are much smaller than the force ("thrust") due to the exhaust escaping)
Rocket in free space:
v  v0  u ln
m0
m
In general, it is easier to calculate a rocket's motion numerically rather than analytically. Physlet #3 from this site,
from Evelyn Patterson at the US Air Force Academy, allows you to model the motion of a rocket experiencing
vertical motion. Try an initial mass of fuel of 20000kg, and see what happens.
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