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Final Answers
© 2000-2017   Gérard P. Michon, Ph.D.    


Physics is much too hard for physicists
 David Hilbert  (1862-1943)
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 Never express yourself more
clearly than you're able to think

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The Physics of Baseball:   Pitching  |  Flight of the Ball  |  Hitting  |  Fielding

Classical Mechanics and Beyond

In order of historical appearance, the three branches of mechanics are:

  • Statics :   Forces and geometry.
  • Kinematics :   Speed and acceleration.  Making time a mechanical quantity.
  • Dynamics :   Conservation of momentum  (Newton's three laws).

The introduction below follows roughly the same order of presentation.  For example, we'll avoid "clarifying" the  law of the lever  with references to such things as speed and power.  Those concepts were foreign to Archimedes and they would only contribute to the puzzlement of a beginner  (if one is bold enough to read this).
However, it's occasionally a very good thing to "cheat" by synthesizing a historical development more efficiently with the benefit of hindsight.  Archimedes certainly knew, at least in special cases, that the product of the force by the displacement was the same for the input and the output of mechanical machines.  He didn't give that quantity a name like  virtual work,  but that's a relatively insignificant detail.  He would certainly have grasped that idea instantly.  Not so with time-dependent mechanical quantities like speed of power, which required a major conceptual breakthrough, 15 centuries later  (c. 1200 AD).
Likewise, summarizing Newton's three laws by the sole conservation of momentum may seem bold, but Newton would certainly have understood that.  He would probably have approved that synthetic view.
When it comes to clarifying what momentum really is, we'll examine the concept beyond the historical context of classical mechanics...  Let's not get ahead of ourselves, though.  As always, the beginning is the proper place to start.  Enjoy the ride.

(2011-01-28)  Mechanical Advantage  &  Statics.   (Archimedes)
Applied forces are inversely proportional to their virtual displacements.

Give me a fulcrum and I shall move the World.
Archimedes of Syracuse  (c. 287-212 BC)

The intuitive notion of  force  (in the modern sense of the word)  is well known to all sentient animals, including human beings:  Two people can push or pull twice as hard as just one of them...

Archimedes  established the less obvious fact that simple passive machines  (lever, inclined plane, etc.)  can allow one man to exert the same force as two men, provided he's willing to do it over a distance twice as large.

The  mechanical advantage  of a simple machine is the ratio of the distance traveled by the exerted force  (along its own direction)  to the corresponding displacement of the load  (along the desired direction).

For example, an  inclined plane  (at an angle  q  with the horizontal)  can be used to lift a load vertically.  In this function, its mechanical advantage is:

1 / sin q

This indicates by how much that "machine" reduces the force required to perform the task, provided there is no friction at all  (in practice, the friction could be reduced by rollers or lubricants).

Friction is a lesser issue in the case of a  lever  (resting on a fixed point, or axis of rotation, traditionally called a  fulcrum)  whose  mechanical advantage  equals the ratio of distances from the fulcrum axis to the input and output lines, respectively.  (The angular displacements being equal, because of the rigidity of the lever, the actual displacements would be proportional to distances measured along those lines.)

 Come back later, we're
 still working on this one...

The ratio of the weight moved to the weight moving it
is the inverse ratio of the distances from the fulcrum.

"Mechanica"   Peripatetic School   (c. 300 BC)
Masses are in equilibrium at distances
inversely proportional to their weights.

"On the Equilibrium of Planes"   Archimedes   (c. 250 BC)

Law of the Lever (History)   |   Gear Ratio   |   Mechanical Advantage (Wikipedia)

(2011-01-28)   Speed and velocity.  Birth of Kinematics  (c. 1200)
Space traveled per unit of time, a notion due to  Gérard de Bruxelles.

The formal notion of  speed  is due to the early 13th century scholar Gerard of Brussels  (Gerardus Bruxellensis, fl. 1240 in Brussels)  whose life is almost undocumented. 

Gérard de Bruxelles  founded Western  kinematics  by authoring  Liber de motu  (Book on Motion)  a treatise which was composed sometime between 1187 and 1260,  according to the definitive work on the subject:  The Science of Mechanics in the Middle Ages  (1959)  by Marshall Clagett (1916-2005).

Unlike the ancient Greeks,  Gerard of Brussels  allowed  quotients  of dissimilar quantities as legitimate objects of study.  (viz. distance over time).

In English and French,  the word  "ratio"  is best reserved for the dimensionless result which comes from dividing two quantities of the same nature.  When there's no such restriction, the word "quotient" is preferred.  (Example.)

Gérard  thus paved the way for the next generation of  kinematicists  who would further focus on the concept of acceleration,  including the  Oxford calculators  at  Merton College  (discussed nextNicolas Oresme (1323-1382)  in France and  Casale  (Giovanni di Casali, c.1310?-1375)  in Italy.

National Geographic Video :   Speed of a Peregrine Falcon.  "Frightful" was clocked diving at 242 mph.

(2011-01-28)   The  Mean-Speed Theorem  or  Merton Rule   (c. 1340)
The distance traveled at constant acceleration is what would be traveled at a steady speed equal to half the sum of the initial and final velocities.

The four  Oxford calculators  listed below made good use of that  theorem  (credited to  Bradwardine).  They were all associated with  Merton College, after which the result is named.  The godfather of the group was Walter Burley (c. 1275-1344)  nemesis of William of Ockham (c. 1288-1348).

The theorem is actually a vectorial relation  (call it the  mean-velocity theorem,  if you must)  which is easy to prove  anachronistically  using the tools of  calculus:  Let's call  r  the  position  (viz. the radius vector):

The velocity is  dr/dt = v.  The acceleration  dv/dt = g  is constant.  We may, therefore, integrate twice successively to obtain:

v   =   v0  +  t g         and         r   =   r0  +  t v0  +  ½ t 2 g

That last relation implies that  r  remains in a plane parallel to  ( g , v).  Eliminating  t g  between the two equations, we obtain the advertised result:

The Mean-Speed Theorem
r   =   r0  +  t (v0 + v) / 2

If anything, this vectorial relation is most commonly stated and/or used for a single component.  The result does apply to the vertical position of a falling object  (near the surface of the Earth)  but it was obtained  well before  that application was known to be a valid one  (Galileo established experimentally that falling objects have a constant acceleration only some 250 years later).

The one-dimensional Merton rule can also be used for angular displacements or any other type of quantity known to undergo a constant acceleration.

The rate of change in acceleration is called  jerk.  The Merton rule can be said to apply to  jerkless  motion  (although that modern term is rarely used).

Nonuniform Motion and the Merton College Mean-Speed Theorem  by  William Heytesbury (c. 1350)
Oresme's Mean-Speed Theorem:  "History of Science" exhibit at the University of Oklahoma.
The Oxford Calculators  by  Mark Thakkar  (Oxford Today, Trinity Issue 2007).

 Coat-of-arms of 
 Galileo Galilei (1564-1642) (2004-10-15)   The first scientific unit of time:  Galileo's  tempo.
Reviving a dimensionless constant of great historical interest...

 Time is defined so that motion looks simple.
 J. Henri Poincaré, as quoted by John A. Wheeler
in  Gravitation  (1973)  Misner, Thorne & Wheeler.

Arguably, the  scientific revolution  in Physics started in 1581 when  Galileo Galilei (1564-1642)  discovered the  isochronism  of the pendulum, namely the fact that a given pendulum oscillates with the same period regardless of its amplitude  (as long as that amplitude is not too large).  To do that, he had to use his own pulse as a timer, because nothing else was yet available...

The earliest unit of time used in experimental physics was the  tempo  (plural: tempi)  devised by  Galileo  to establish experimentally the  pendulum law,  which says that the length of a pendulum is proportional to the square of its period of oscillation  (at least for small amplitudes).

At first, Galileo had worked without a precise device to measure time intervals and he relied on his own pulse  (and/or on his musical instincts, as discussed below).  Next, he invented a device which let water flow during each motion he wanted to time.  Assuming the flow had a constant rate, he would measure the time ellapsed by weighing the water collected.  For example, he obtained 903 grains of water during a fall from rest through  2000 punti  (about 1.88 m).  He then defined his new unit of time, the  tempo, as corresponding to 16 grains of water on this particular device.

Thus, Galileo's  tempo  was roughly equal to  11 ms.

The precision of Galileo's experiments can be established from one number, which he derived experimentally, although it happens to be a  pure number  (depending neither on the local value of the gravitational field nor on the unit of length he used):  Galileo determined the length  R  (in  punti)  of a pendulum which swings from rest  (at a small angle)  to its vertical position during the time it takes for a body to fall from a height  H  of 2000 punti  (about 1.88 m).

He did that by releasing the pendulum's bob and the falling body simultaneously and listening to their respective bangs upon arrival.  He changed the length of the pendulum until those arrivals sounded simultaneous, which is "musically" so when the discrepancy is below 10 milliseconds or so  (this is, most probably, why Galileo chose his basic unit of time to have the aforementioned order of magnitude).

The modern answer  (neglecting air resistance)  is that  R/H  is  8/p2,  so  R  is about  1621.14 punti, irrespective of the value of a  punta  (singular of punti).  The experimental value obtained by Galileo was 1590 punti.  This precision does correspond to a timing error of less than 10 ms, as advertised.

Stillman Drake: "The Rôle of Music in Galileo's Experiments"  Scientific American, 232 (June 1975) pp. 98-104.
F. Riess, P. Heering, D. Nawrath: Reconstructing Galileo's Inclined Plane Experiments for Teaching Purposes

Brian Kieffer  (2011-05-10, Yahoo!)   True period of a  pendulum.
What is the period of a pendulum for  large  swings?

After Galileo first described the  law of the pendulum  and the law of falling bodies, a century would ellapse before  Newtonian dynamics  offered both laws a common rational explanation, which is now commonplace.

For small oscillations of the pendulum, this has provided a precise theoretical value  (8/p)  for the constant discussed in the previous section, which Galileo could only obtain experimentally.

The motion of a simple pendulum in a vertical plane is always discussed at the introductory level  (slightly generalized details are presented elsewhere on this site).  The angle  q between the local vertical  (gravitational field  g)  and a pendulum of length  L  obeys the following differential equation:

q''  +  (g / L)  sin q   =   0

For small oscillations,  sin q  is nearly equal to  q  and the following differential equation is thus  approximately  satisfied:

q''  +  (g / L)  q   =   0

The solution is a so-called  simple harmonic motion  expressed by:

q   =   A cos ( w ( t-t0 ) )       [ A  and  t are arbitrary constants ]

The  pulsatance  w = Ö(g/L)  does  not  depend on the amplitude  (A)  of the oscillation, neither does the  period  T = 2p/w...  That statement is the so-called  isochronism of small swings  observed by  Galileo  in 1581.

Period of a Pendulum of Length  L  (small swings)
T0   =   2 p   Ö   L

In older texts, the quantity most often quoted was the  time of swing  (equal to half the period)  which is the time it takes for the pendulum to swing from its leftmost to its rightmost position.  Before another view prevailed, it was suggested that the then-to-be-defined meter unit ought to be the length of a pendulum  beating one second.  Under the current definition, a one-meter pendulum beats about  1.0032 s  at sea-level and mid-latitude, which means that its period is about  2.0064 s.

What about  large  oscillations?

Choosing the unit of time so the  small-swing period  T0 is  2p,  we have:

q''  +  sin q   =   0

Let's call  A  the  amplitude  of the motion,  which we may define as the positive value of  q  for which  q'  is zero.  By multiplying the above equation into q',  we see that the left-hand side of the following equation is constant, since its derivative vanishes  (it's essentially the total mechanical energy).

½ ( q' ) 2  -  cos q   =   - cos A

Introducing the notations  k = sin A/2   and   x = (sin q/2) / k   we obtain:

( q' ) 2  =  2 (cos q - cos A)  =  4 [ sin2(A/2) - sin2(q/2) ]  =  4 k2 (1 - x 2 )

By the chain rule,   dx/dt   =   x'  =  q' cos(q/2) / 2k.  Therefore:

( dx/dt ) 2   =   ( q' ) 2  cos2(q/2) / 4k2   =   ( 1 - x2 ) ( 1 - k2 x2 )

This gives  t  as the following indefinite integral of a function of  x,  endowed with a sign that changes twice per period  (every time  x  reaches an extremum, as it oscillates back and forth between  -1  and  +1).  I wouldn't dare write  |dx|  instead of the less objectionable ± dx  but that's the idea...

t   =   
ò   ± dx
Ö ( 1 - x2 ) ( 1 - k2 x2 )

That integral is known as the  elliptic integral of the first kind.  It can pop up in various guises; this particular one is called the  Legendre normal form.  Elliptic integrals can't be expressed in terms of elementary functions.  The  complete elliptic integral of the first kind  is the definite integral:

K(k)   =   
ò  1   dx
0 Vinculum
Ö ( 1 - x2 ) ( 1 - k2 x2 )

Traditionally,  k  is called the  modulus  of this elliptic integral or its relatives.

As  x  goes from 0 to 1, the angle  q  goes from 0 to A and the time  t  increases by one fourth of the true period  (measured in our special unit of time where the  small-swing  period is  2p).  Therefore, going back to arbitrary units of time:

T   =   4 K(k) T0 / 2p   =   T0 [ K(k) / (p/2) ]   =   T0 / agm (1-k , 1+k)

The  arithmetic-geometric mean (AGM)  of two numbers is recursively defined as the AGM of their arithmetic mean  (half-sum)  and  geometric mean  (the square root of their product).  In particular:

agm (1-k , 1+k)  =  agm (1 , Ö(1-k) ).

Since  k = sin(A/2)  this translates into the following  robust  expression:

True Period of the Pendulum
T   =   T0 / agm ( 1 , cos A/2 )

That  elegant formula  is all we really need to compute the true period  T  very efficiently  (with or without a computer)  since the recursive way to compute an AGM converges  quadratically  (i.e., the number of correct digits roughly doubles with each iteration).  That is so even when  A  is close to the  disallowed  maximum of  180° = p.   For example:

Ak = sin A/2T / T0
0.01°    0.0000872664625    1.00000000190  
0.1°    0.0008726645152  1.00000019039
1°    0.0087265354983  1.00001903892
5°    0.0436193873653  1.00047617249
10°    0.0871557427477  1.00190718814
30°    0.2588190451025  1.01740879760
60°    0.5  1.07318200715
90°     0.7071067811866  1.18034059902
120°    0.8660254037844  1.28434402576
150°    0.9659258262891  1.76220372950
170°    0.9961946980917  2.43936271967
175°    0.9990482215819  2.87766351235
179°    0.9999619230642  3.90106516039
179.9°    0.9999996192282  5.36686710903
 179.99°    0.9999999961923  6.83273733801
180°    1¥

For a  release angle  of  90°,  T/T0 = GÖ2,  where  G  is  Gauss's constant.

Of course, release angles above  90°  are only possible with a  rigid pendulum;  the usual bob-on-a-string won't do.

We may now investigate the relevant  power series, using the expansion  (A002894)  of  K(k)  that involves  squares  of central  choice numbers :

T   =   2p
( 2n
) 2  (  ¼   sin A/2 ) 2n

Expanding each term of that sum as a power series of  A  and collecting like terms, we obtain a power series of  A  with rational coefficients  (since every coefficient is obtained as a sum of finitely many rationals):

True Period of a Pendulum of Length  L   (swings of finite amplitude  A )
T   =   2p Ö L  [ 1  A2  +  11 A4  +  173 A6  +  22931 A8  +   ... ] 
vinculum vinculum vinculum vinculum vinculum
g 16 3072 737280 1321205760

Here are the next terms of the above bracketed power series:

...   +   ( 1319183 / 951268147200 )  A10
+   ( 233526463 / 2009078326886400 )  A12
+   ( 2673857519 / 265928913086054400 )  A14
+   ( 39959591850371 / 44931349155019751424000 )  A16
+   ( 8797116290975003 / 109991942731488351485952000 )  A18
+   ( 4872532317019728133 / 668751011807449177034588160000 )  A20

Although the radius of convergence of that power series is  [ probably ]  p,  it ought to be shunned for precise numerical computations, in favor of the above simpler, faster and more robust  arithmetic-geometric formula.

Pendulum  (Wikipedia)   |   The Nonlinear Pendulum  by  D.G. Simpson, Ph.D.   (2010)
The Pendulum:  Rich Physics from a Simple System   by  Robert A. Nelson  &  M. G. Olsson   (1984)
Exact solution for the nonlinear pendulum  by  A. Belendez, C. Pascual, D.I. Mendez, T. Belendez, C. Neipp  (2007)
Approximations for the period of the simple pendulum based on the arithmetic-geometric mean
by  Claudio G. Carvalhaes  &  Patrick Suppes  (2008)
Half-baked attempts:  TSR Forum   |   Connection between those numbers?  (NRICH Forum)

(2011-01-28)   The trajectory of a cannonball is a  parabola.
Smoothly  combining a vertical fall and a uniform horizontal motion.

Starting in .../... "law of odd numbers" => "law of falling bodies"

 Parabolic trajectory inscribed 
 in an Aristotelian triangle

 Come back later, we're
 still working on this one...

At time  t,  the projectile's cartesian coordinates are:

x   =   (v0 cos j) t
y   =   (v0 sin j) t  -  ½ g t 2

The range  d  is the value of  x  at the time  t  when  y  returns to zero:

t   =   (2 v0 / g) sin j
d   =   (v02 / g) sin 2j

The maximum range  (v02 / g)  is thus obtained for a tilt  j = 45°.

The top height  h  (the value of  y  when  t = t/2)  is one fourth of the height of the right triangle whose base is the range  d  and whose hypotenuse is in the direction of the initial velocity  (use the above picture to memorize that).

h   =   ¼ d  tg j   =   (v sin j) 2 / 2g

 Gunner Firing a Cannon - 1561 
 (woodcut, artist unknown)
Gunner Firing a Cannon  (1561),   Unknown artist.
In   "Problematum Astronomicorum et Geometricorum"
by  Noviomagus = Daniel Santbech  (Basel, 1561).
    Before  the scientific method  required theories to be matched with observations,  Aristotelian concepts did not allow for several simultaneous natural motions.  As illustrated by the woodcut at left, it was thus taught  (not thought)  that an object thrown up in the air was endowed with something called an impetus which made it move in a straight line  (what else?)  until that impetus wore off, at which point the object fell vertically in a straight line !
Aristotelian physics predicted that the high point (F) in the trajectory of a cannonball was four times higher (!!) than the top of the Galilean parabola.

It wouldn't be fair to  Noviomagus  and other expert gunners of that era to suppose that they hadn't noticed that actual cannonball trajectories didn't fit Aristotelian triangles at all.  The gunners chose to overlook such obvious shortcomings because their beloved Aristotelian theories did provide the correct relation between range and tilt, which is really all you need to operate a cannon, at least when the target is level with it  (Aristotelian physics doesn't give correct predictions in other cases).

 Coat-of-arms of 
 Sir Isaac Newton (1643-1727) (2011-01-28)   Conservation of Momentum  (1666, 1687)
The key to Newton's  three laws of motion.

The notion of  inertia  was originally not part of Aristotelian physics.  It was formally introduced, around 1340, by Jean Buridan  (Johannes Buridanus)  who proposed that a moving body is endowed with a so-called  impetus  directly proportional to its mass and its velocity.

Three centuries later, Isaac Newton  (1643-1727)  figured out for himself the fundamental principles of dynamics in 1666, at the age of 23.  He published them formally as a set of  three laws  more than 20 years later  (in 1687).

  1. Motion remains uniform, unless disturbed by a force.
  2. Force is the rate of change of momentum:  F = dp/dt   with p = mv.
  3. Every force is opposed to another force of equal magnitude.

In retrospect, Newton simply equated Buridan's  impetus  to the product  (p)  of the velocity  (v)  of an object into a constant, called its  mass  (m).  He realized that this is a  conserved  vectorial quantity  (now called  linear momentum)  which can be exchanged between separate systems but can be neither created nor destroyed.  Ever.  So say the three laws.

The linear momentum of a system is the sum of the linear momenta of its components.  The force between two systems is the time rate at which they exchange momentum.

Ancient Theory of Impetus

 Gustave Gaspard   
 de Coriolis (1792-1843) (2007-10-31)   The Work-Energy Theorem:
The  work done  is equal to the change in  kinetic energy.

The infinitesimal work done by the resultant  F  of  all  the forces applied  (including conservative forces and  dissipative  friction forces)  to a point-mass of velocity  v  is equal to the  dot product  of that force into the displacement  (v dt)  of its point of application  during a time  dt.  Namely:

F . v dt   =   m (dv/dt) . v dt   =   m v . dv   =   d ( ½ m v 2 )

Therefore, the  work done  over any interval of time is equal to the variation of what's called the  kinetic energy  of the point-mass, namely  ½ m v2.

E   =   1/2  m v 2

The modern nomenclature  (French:  travail  &  énergie cinétique)  for that  clean  result is due to  Gustave Gaspard de Coriolis  (1792-1843).

 Coat of arms of  
 Emilie du Chatelet (1706-1749)
The concept of kinetic energy had been advocated by  Emilie du Châtelet (1706-1749)  in 1740.  She combined the ideas of Leibniz with the observations of Willem's Gravesande (1688-1742).

The above computation can be seen as an introduction to the concept of  energy, whose  relativistic  generalization is presented in the next section:

(2008-09-18)   Relativistic Kinetic Energy   (Einstein, 1905)
Relativistic expression of the  work done  to a point mass.
The following two articles clarify the above notion of energy by putting it in the less forgiving context of Special Relativity.  This is primarily meant for people who have already wrestled with the concept for a while, although any reader familiar with calculus and basic Newtonian mechanics  (F = m a)  needs only accept the single statement of the following paragraph to benefit from the rest of the discussion...

In  Special Relativity,  the dynamics of a point-mass are still described by the Newtonian relation  F = dp/dt  provided the  momentum  is redefined:

p   =   m v
Ö 1 - v2/c2
An erroneous principle  F = m d(p/m)/dt  was once proposed by some authors, including Heinrich Ott (1962) and Henri Arzeliès (1965) in the context of relativistic thermodynamics where  m  varies  (see below).  The issue is a delicate one and the controversy is often revived by newcomers.  The Ott-Arzelies formulation was rebutted by Guessous (1967) and ruled out experimentally by Pemmaraju Ammiraju (1984).  The fundamental issues of "dissipative mechanics", involving variable rest masses, were carefully discussed by Jean Fronteau (1973).

In this,  c  is the speed of light  (Einstein's constant).  Assuming the  rest mass  (m)  to be constant, we may differentiate this (vectorial) expression:

dp   =   m dv     +     m (v.dv/c2 ) v
vinculum vinculum
Ö 1 - v2/c2
(1 - v2/c2 ) 3/2

Dotting that into  v,  we obtain an expression of  v.dp  which  boils down  to:

v . dp   =   m v . dv     =     d (  m c2  )
vinculum vinculum
(1 - v2/c2 ) 3/2
Ö 1 - v2/c2

The above expression of the  work done  is thus only slightly modified:

F . v dt   =   (dp/dt) . v dt   =   v . dp     =     d (  m c2  )
Ö 1 - v2/c2

This yields the following  relativistic  equivalent of kinetic energy for a point-mass  (up to an additive constant)  first given by Albert Einstein in 1905:

E   =     m c2
Ö 1 - v2/c2

This implies that:
E2 - ( c p ) 2   =   ( m c2 ) 2

At low speed, therefore, we have   E   »   mc 2  +  ½ mv 2.  In the above context, the leading term is constant and seems irrelevant at first.  However, Einstein  postulated  that it corresponds to a real energy  at rest  which something of mass  m  is always endowed with.  This is known as the principle of  inertia of energy.

When the energy of a system at rest increases, so does its mass.  The next section investigates the fundamentals of what's involved in such a process.

(2008-09-18)   Basic Relativistic Thermodynamics
Variable rest mass  in the case of a pointlike system.

If a pointlike object is somehow endowed with a structure that allows it to store energy at rest  (e.g., spin or heat)  then its  rest mass  (m)  may vary.  Redoing the above computation to allow for that, we find that  v.dp  is the sum of two terms.  The first one can be deduced from our previous computation  (at constant m)  whereas the second one comes from the fact that  p  is proportional to  m:

v . dp   =    m d (  c2  )   +    v2 dm
vinculum vinculum
Ö 1 - v2/c2
Ö 1 - v2/c2

With the proper algebraic massaging, this differential relation becomes:

d (  m c2  )   =   v . dp   +  
Ö 1 - v2/c2    d ( mc2 )
Ö 1 - v2/c2
dE=dW+   dQ

This is actually one enlightening case of the  first principle of thermodynamics :  The variation in the energy of this simple system is an exact total differential  (dE)  of the parameters which describe it  (m and v).  This differential form is equated to the sum of two infinitesimal quantities which are  not  exact total differentials themselves; the "work"  (dW)  and "heat"  (dQ)  received  by our point-mass.

Relativistic thermodynamics  is discussed at length  elsewhere on this site.

 Spacecraft rsg160 (2001-04-15)
I have been told that a satellite in a circular orbit that starts to enter the atmosphere actually speeds up, at first. When the satellite is outside the atmosphere there are only conservative forces (gravity) acting and, if the satellite is in a circular orbit, its [speed] is constant. When it starts to enter the atmosphere there is a small drag force, since the atmosphere is thin high up. This force always opposes the motion and I would have guessed that it would slow the satellite down. Can you help me in figuring out why the satellite actually speeds up?

Indeed, as long as the drag force remains small, the satellite will gain speed, because speed increases as altitude decreases.  A drag force cannot change that trend unless it's large enough.  The key point is that a spacecraft loses altitude at a steady rate during reentry.  It's fairly easy to work this out quantitatively:

Let's call M the mass of the satellite, V its speed and z its altitude. Let R be the radius of the Earth (assuming its mass distribution has perfect spherical symmetry) and let's call g the gravitational field at z = 0. 

The kinetic energy of the spacecraft is ½MV 2 and its potential energy is exactly Mgz/(1+z/R)  (normalized to 0 when z = 0; it's about Mgz when z is small).

The sum of these two terms is the total energy E, which is constant in the absence of drag.  In the presence of a drag force  F  (in a direction opposite to that of the velocity), there's a loss of total energy equal to the power FV of the drag force.  In other words, E' = - FV.  We may use the above expression of E in this, neglecting the 1/(1+z/R) correction factor for the potential energy (thus making an error on the order of 1%) and obtain the relation:  Angle of 

M ( V V' + gz' )   =   - FV

Now, we may remark that the quantity (-z' )/ V is simply sin(a), the sine of the angle of reentry a (this is a positive quantity since z is decreasing, it would be zero for horizontal flight).  Therefore, the above relation translates into:

dV/dt   =   V'   =   g sin(a) - F/M

As long as the drag force F is less than Mg sin(a), V' is therefore a positive quantity, which means that the spacecraft will indeed gain speed initially.

For proper re-entry the angle of reentry a cannot be too small, or else the spacecraft could "bounce" off the atmosphere and be back into outer space after losing just a little bit of energy.  If no action is taken, an orbiting spacecraft could thus keep bouncing back until enough energy is lost and/or its reentry angle is sufficiently large --possibly dangerously so, since a large a means fast reentry and a lot of heat!

As the atmosphere becomes denser at lower altitudes, the drag force F will eventually exceed the above threshold and the spacecraft will slow down.

 Monkey and 
 weight (2002-01-27)   Lewis Carroll's Monkey
Consider an equilibrium realized when a "perfect" rope is passed over a frictionless and massless pulley with a ten-pound weight on one side and a ten-pound monkey on the other...  What happens when the monkey decides to climb up the rope?

This problem was popularized by the author of Alice in Wonderland, Lewis Carroll (1832-1898, Charles Dodgson), who agonized over it.  He was a professor of mathematics at Oxford from 1855 to 1881.  The above picture once illustrated a discussion of the puzzle by the mathematical columnist Sam Loyd (1841-1911), who called the problem "Lewis Carroll's Monkey Puzzle", while stating that it was not known whether Lewis Carroll originated the question.  (Unfortunately, the solution given by Loyd happens to be erroneous.)

The answer is that the centers of inertia of the weight and the monkey will have the same vertical motion (we assume, of course, that the monkey only goes up or down but does not swing the rope). Thus, if the monkey and the weight are initially motionless at the same height, they will always face each other no matter what the monkey does.  For example, they will both be in free fall if the monkey lets go of the rope, and both falls stop when the monkey grabs the rope again.

The reason for this is simply that all the forces that are acting on either the monkey or the balancing weight are always equal.  There are only two such forces for each body; the downward weight and the upward tension of the rope.  The weights are equal because the two bodies have the same mass and the rope also exerts the same force on either body because of the numerous "ideal" assumptions made here, including the absence of swinging on the monkey's side (so that the rope exerts only a vertical force in either case).  It's also essential to assume not only the lack of any friction, but also the absence of mass for both pulley and rope (otherwise the rope's tension would not be the same on either side of an accelerating pulley and it would vary along the length of an accelerating rope).

Note also that a "perfect" rope retains its length and transmits instantly any change in its tension.  This is clearly unrealistic but it's logically consistent with the axioms of classical  (non-relativistic)  mechanics.

When the same forces act on bodies of equal masses their speeds change in the same way, so that the speeds remain equal if they are originally so (and we're told here of an original equilibrium where both speeds are zero).  Both motions will therefore mirror each other.

From the monkey's perspective, pulling 2 feet of rope will get him only 1 foot higher from the ground, but will require as much effort  (work)  as would be necessary to climb 2 feet on a stationary rope. That's not surprising in view of the fact that 20 lb were lifted one foot in the process (the monkey and the weight went up one foot each), which is just as difficult a task for a 10 lb monkey as lifting his own weight up two feet...

This deceptively simple puzzle is an excellent way to start a healthy discussion about the fundamental principles of classical mechanics.

 Two-ball drop
A heavy ball is dropped on a hard floor from a height of one foot with a light ball (ping-pong ball) on top of it.  How high could the light ball bounce?

Answer Up to 9 feet !       Explanation

Shortly before impact, both balls have the same downward speed V.  An elastic bounce off the floor makes the heavy ball go upward at speed V to meet the light ball still going downward at speed V.  The speed of the ping-pong ball  relative  to the heavy ball is thus 2V.  Upon impact of the two balls that speed reverses itself (since the mass ratio of the two balls is very large).  An upward speed of 2V relative to the heavy ball translates into an upward speed of 3V...

Ideally, the  kinetic energy  of the ping-pong ball thus goes from  ½ mV2  before impact to 9 times that much after the shock !  This is enough to propel that ball, up to 9 times the height it was dropped from.   QED

In practice, a lesser height is reached  (because of imperfect elasticity and a finite mass ratio)  but the actual demonstration remains impressive.

Video :   Stacked Ball Drop  by  Dianna Cowern  (Physics Girl,  2015-01-20).

(2007-10-09)   Tangential and Normal Accelerations
The tangential acceleration is  dv/dt.  The normal acceleration is  v2/R.

Consider a smooth curve in space.  Let  s  be the  curvilinear abcissa  along that curve and  T = T(s)  a  unit vector  tangent to the curve in the direction of increase of  s.  Geometrically, the  curvature  1/R  at  s  can be defined as the  length  of the vector  dT/ds = N/R  (which is perpendicular to  T, since the length of  T  remains constant).  R  is called the  radius of curvature  at that point.

When such a curve is the trajectory of a point-mass traveling at speed  v = ds/dt  its [vectorial] velocity  is  v T.  The  acceleration  of that point-mass is thus:

a   =   d(vT)/dt   =   (dv/dt) T  +  v dT/dt   =   (dv/dt) T  +  (v2/R) N

This means that the component of the acceleration which is perpendicular to the trajectory has magnitude  v2/R  (speed squared divided by radius of curvature).

(2007-10-09)   Loop-the-Loop in a Roller Coaster   (teardrop  shaped)
Harnesses shouldn't be  needed  to keep riders on their seats.

Traditional roller-coasters rides are designed so that a component of the acceleration always makes the wheels of the car push against the tracks and/or the bodies of the riders push against the seat cushions.

 'Loop The Loop'
 at Coney Island (1901-1910)
Loop-the-Loop  at Coney Island  (1901-1910)
This was the first noncircular roller-coaster loop  (the smaller radius of curvature at the top makes it safe at lower speeds)  based on a patent awarded to Edwin Prescott (1841-1931) on Feb. 5, 1901.

At the top of a vertical loop (where riders are upside down) either condition means that the  normal acceleration  exceeds the vertical acceleration of gravity.

v2 / R   >   g

On the other hand, we have:

m g h  >  ½ m v2

This expresses that the gravitational energy obtained by releasing a car of mass  m  from a height  h  above the top of the loop is greater than the kinetic energy it retains at that point  (the two would be equal if there was no loss of mechanical energy due to friction).  Therefore:

h   >   ½ R

Thus, a roller-coaster car must be released from a height above the top of a loop-the-loop (much) greater than half the top radius of curvature.

 Conical pendulum (2007-10-18)   The Conical Pendulum
A hanging bob traveling along an  horizontal  circle.

Let  L  be the length of the string and  q  the (constant) angle it makes with the vertical.  The radius  R  of the trajectory is:

R   =   L  sin q

 Acceleration decomposition

The horizontal centripetal acceleration is  v2/R = wR,  as the bob travels around the circle at speed  v = wR.  (This is a special case of the  normal acceleration discussed above.)  Newton's law says that this acceleration multiplied by the mass  m  of the bob is is the  vector sum  of all the applied forces, namely the weight  (mg)  and the string tension  (F).  The magnitudes of the relevant components of the acceleration are thus obtained as the sides of the right triangle pictured at right  (featuring the aforementioned angle  q ).

Thus,  v2  =  R g tan q  =  L g sin2 q / cos q  =  L g (1-u2 ) / u   which can be recast into the following quadratic equation, with respect to  u = cos q :

u 2  +  ( v 2 / Lg )  u  -  1   =   0

Solving for u, we retain only the solution that's between 0 and 1.  We obtain:

cos q   =   u   =   ( 1 + x2 )½ - x         where   x  =  v2 / 2Lg

To solve this, we put  x = sh j  and find that  u = e j.  In other words:

q   =    arccos (exp (argsh ( v2 / 2Lg )))

v 2   =   2 L g  sh ( ln cos q )

(2011-01-11)   String tension  T  of a  constrained  conical pendulum.
Let  the above  hang from the top of a hemisphere of radius  a > L/Ö2

We assume that a large enough orbital speed  v  makes the (pointlike) bob touch the hemisphere and that there's no friction.  The trajectory of the bob is a horizontal circle of radius  x  at a height  y > 0  above the horizontal plane of the constraining hemisphere's equator.

To obtain  x & y  in terms of  L & a  we use Pythagoras' theorem twice:

x 2  +  y 2   =   a 2               x 2  +  (a-y)2   =   L2

By subtraction, we have   2 a y   =   2 a 2  -  L2   which gives a value of  y  that can be plugged into either equation to obtain  x > 0.

y   =   a [ 1 - ½ (L/a)2 ]       and       x   =   L [ 1 - ¼ (L/a)2 ] ½

To obtain the tension  T  of the string, we may project  Newton's second law  on the direction parallel to  (y, -x)  since the unknown reaction of the sphere is perpendicular to that  (it's normal to the surface and, therefore, it's in the direction of the (x, y) vector).  The mass  m  multiplied into the centripetal (horizontal) acceleration so projected is equal to the sum of the respective projections of the bob's weight and of the tension force acting on it:

m [y.(-v2/x) + (-x).0]   =   [y.0 + (-x).(-m.g)] + [y.(-x)T/L + (-x).(a-y)T/L]

This boils down to the simple expression:   T  =  m  [ g + y.v2/x2 ] L/a
Using the angular velocity  w = v/x  and the above value of  y,  we obtain:

    T   =   m g L/a  +  [ 1 - ½ (L/a)2 ]  m L w2    

(2008-02-13)   Ball in a Bowl
The motion of a small ball  rolling  at the bottom of a spherical bowl.

A ball released at zero speed on a spherical surface moves in a vertical plane.

Let  R  be the radius of that spherical bowl.  The radius of the ball is  r < R.  The mass of the ball is  M  and its moment of inertia  is  J.

Let  O  be the center of the spherical bowl and  C  be the center of the small ball.  q  is the (oriented) angle from the vertical to  OC  and  q'  is the derivative  dq/dt.

In a  pure roll,  the point of contact of the ball with the spherical surface has zero speed.  Since the speed of  C  is  (R-r) q',  we may deduce that the ball spins with angular velocity  (1-R/r) q'.  The potential energy of the ball is  -M g (R-r) cos q  (up to an irrelevant additive constant).  Therefore, the following expression is equal to the  total mechanical energy,  which remains constant:

½ [ M + J/r2 ]  (R-r)2 (q' )2  -  M g (R-r) cos q

For an homogeneous ball,  J = 2/5 M r 2  (therefore, 2/7 of the kinetic energy is rotational energy).  The derivative of the above boils down to:

7/5 (R-r)  q''   +  g  sin q     =   0

A simple pendulum of length  40%  longer than  OC = R-r  would obey the same equation  (a  40%  longer length means an  18.3%  longer period).

brentw (Brent Watts of Hickory, NC. 2001-05-04)
A 24 lb weight, attached to the end of a spring, stretches it 4 inches... What is the equation of motion if the mass is released from rest from a point 3 inches above the equilibrium position?

An oscillation where the acceleration is proportional to the distance from a fixed point is a sinusoidal motion about that point, commonly known as a  simple harmonic motion  (SHM).

Hooke's law  states that the force exerted by a spring is proportional to its elongation. Under standard gravity (g=9.80665 m/s2), the weight (force) Mg of a mass M of 24 lb (10.88621688 kg) is 24 lbf (106.757318766 N). If this corresponds to an elongation of magnitude L=4 inches (0.1016 m), the force corresponding to an elongation y is therefore -y(Mg/L).

Now, if we decide to count upward elongations positively, the acceleration y" is therefore such that My"=-Mg-y(Mg/L) (Newton's Law). If we consider instead the position x=y-L above the equilibrium position, we have x"+(g/L)x=0 (in the absence of friction or damping). Introducing the quantity w such that w2=g/L (in other words w is Ö(9.80665/0.1016) or about 9.82457 rad/s), x is therefore equal to A cos(wt) + B sin(wt), for some constants A and B. Since we are told that, at t=0, x is 3 inches and the speed x' is zero (the mass is just "released from rest"), we know that B is zero and A is 3 inches (76.2 mm). Therefore, the equation of motion is simply x = A cos(wt), with A=3"=76.2 mm and w=9.82456847...rad/s (corresponding to a period of oscillation T=2p/w of about 0.64 seconds).

Note that the value of the mass M turns out to be irrelevant here (the only thing that matters is the magnitude L of the elongation at rest), but the value of the gravitational field g does matter: The value of the period of oscillation T would be different under a gravitational field other than the "standard" one.

 Hydrogen gatman (Central Florida. 2001-03-31)
How fast (rpm and mph) are
electrons going around a nucleus?
What factors affect it and how?

Arms of Niels Bohr (1947) For small objects like electrons, quantum mechanics states that the very notion of trajectory breaks down.  You can't measure both the momentum and the position of a particle:  The product of the uncertainties in the measurements of such conjugate quantities cannot be less than the so-called Heisenberg limit.  Thus, the electron does not have a precise speed in the classical sense. However, you can work out what the expected value of the momentum would be if you were to measure it with infinite precision (which would mean, then, that you would not know at all where the electron is).  From that momentum, you may derive some kind of expected speed...

This being said (and it had to be said), you may work out things numerically using something as crude as the old-fashioned semi-classical Bohr model of the atom (circular "trajectories" --oh, well-- with an angular momentum which is only allowed to be n times the rationalized Planck constant "h-bar"  h-bar = h/2p).  The number n is the principal quantum number you may find listed in chemistry books; it's normally equal to 1 for the electron around an hydrogen nucleus.

All told, you'll find that the binding energy of the electron (a negative quantity) in the Bohr model is  E(n) = -chR/n2 where  c = 299792458 m/s  is the speed of light  (Einstein's constant),  h = 6.626 10-34 J.s  is the  (unrationalized)  Planck constant,  and  R = 10973731 m-1 is Rydberg's constant.  (I'll neglect the so-called  normal mass shift  correcting factor of  1/(1+m/M),  where  m/M  is the ratio of the mass of the electron to that of the nucleus).  In other words,  E(n) = (-2.18 10-18 J) / n2.

As is the case for planets around the Sun, it turns out that the kinetic energy  ½ mv2  is equal to -E(n).  The mass of the electron is m = 0.911 10-30 kg.  So, what you get is a speed of the electron which is inversely proportional to n, namely:  v = (2190 km/s) / n.  The largest speed is for n = 1 and corresponds to about 0.73% of the speed of light, which means it was OK to neglect relativistic effects at our casual level of precision.  If you insist on having the speed expressed in mph, the formula is v = (4900000 mph)/n.

To obtain the period of rotation around the nucleus, you need to know the radius of the "orbit" (again, this is not to be seriously taken as a real trajectory), which is aon2, where ao= 0.53 10-10 m is Bohr's radius.  The period is equal to 2p/v times this, which means it is proportional to n3T = (1.52 10-16 s) n3.  The frequency is thus inversely proportional to the cube of n:  f = (6.58 1015 Hz)/n3, or if you insist on using rpm's:  f = (395 000 000 000 000 000 rpm)/n3, since 60 rpm = 1 Hz.

All this pertains to the hydrogen atom (Z = 1).  For a lone electron around a nucleus with Z protons (a so-called hydrogenoid ion), the energy gets multiplied by Z2 and the speed is therefore multiplied by Z.  Since the size of the "orbit" is divided by Z, the frequency is multiplied by Z2.

To summarize, a lone electron would be expected to go around a nucleus with Z protons at a speed  (Z/n) Vo  and a frequency  (Z2/n) fo , where :

  • Vo is about 0.0073 c, 2190 km/s, 7900 000 km/h, or 4900 000 mph.
  • fo is about 6.58 1015 Hz or 3.95 1017 rpm.

Note that a single electron  (of charge  q = 1.6 10-19 C )  going around a circuit at frequency  f0  is equivalent to a current  q f0  of about  1 mA.

The so-called  Bohr magneton  (qh-bar/2m  =  9.274 10-24 A.m2)  is an important quantity which serves as a unit for whatever  magnetic moment  may be associated with the orbiting electron.  Classically, the magnetic moment associated with a current  I  circulating in a loop of vectorial area  S  is  m = IS.  The Bohr magneton would be the moment classically associated with an electron actually moving at the above speed along a circular path with a radius equal to the aforementioned  Bohr radius.

 Diamond (2002-06-10)
Is there anything harder than diamond?

Several substances are harder than diamond :

Ultrahard Fullerite

Ultrahard fullerite is a polymerized phase of fullerene discovered in 1995, which is the object of several patents awarded to Dr. Michael Yu. Popov.  It is currently used in the NanoScan (NS) scanning force microscope (SFM).  Some experimental studies indicate that ultrahard fullerite is about twice as hard as diamond (approximately 300 GPa, vs. 150 GPa for diamond).

Normally, fullerene crystallizes as a fairly soft yellow solid of low density (1680 g/L) where buckyballs are held together by  van der Waals  forces, similar to what holds together the hexagonal carbon planes underlying the structure of graphite  (the density of graphite is  2267 g/L).

However, the polymerization of fullerene which occurs at high temperature under gigapascals of pressure yields an ultrahard phase whose density  (3170 g/L)  compares to that of diamond  (3516 g/L).  Under normal conditions, the resulting stable structure leaves little room for compression...  The material is more than three times stiffer than either diamond or osmium, according to the above table  (itself based on published  acoustic  properties of ultrahard fullerite).

From borazon (1957) to carbon nitride and beyond...

In 1989, Marvin Cohen and his graduate student Amy Liu (then at UC Berkeley) devised a theoretical model to predict a crystal's stiffness [its bulk modulus], which was thought until recently to be a good indicator of the more elusive quality called hardness.  Noteworthy candidates which did not live up to expectations, according to this model, included cubic boron nitride.  (CBN is the hard form of "BN".  It was first synthesized in 1957 by Dr. Robert H. Wentorf of the General Electric Company.  It's known in the trade as borazon.)

On the other hand, Cohen's model clearly indicated that a carbon nitride crystal should be stiffer (and possibly harder) than diamond.  The race was on to obtain the stuff in crystalline form and measure its properties.  Some early efforts by the team of Yip-Wah Chung (at Northwestern University) resulted in a layered composite of titanium nitride and carbon nitride (a so-called superlattice) which was, surprisingly, almost as hard as diamond.

Crystals of carbon nitride  were apparently first synthesized at Lawrence Berkeley Lab (LBL) by Eugene Haller and William Hansen, using an approach similar to the synthesis of industrial diamonds.  A 1993 patent for this new superhard material was subsequently awarded to Cohen, Haller and Hansen.  However, the jury seems still out... [See March 1998  APS conference, and May 2000  Nature article.]

Because  ultrahard fullerite  is so much harder than diamond, we may guess that it's harder than  carbon nitride  as well, but we won't know for sure until someone makes a  carbon nitride crystal  big enough to test...

Recently (February 2004) a new kind of synthetic diamond was found to be at least 50% harder than  natural  diamond.  This was obtained at the Carnegie Institution's  Geophysical Laboratory  (Washington, DC) by submitting to extreme temperatures and pressures (2000°C,  5-7 GPa) crystals synthesized  [much faster than before]  by a new  chemical vapor deposition  (CVD)  process.

(2002-06-10)   How is Hardness Defined ?
Hardness is the resistance to permanent surface damage...

It's totally different from elastic resilience, in spite of a loose correlation.

Unlike stiffness, which is an elastic property of a solid [quantitatively, stiffness is simply the solid's bulk modulus], hardness actually indicates the resistance of a solid's surface to permanent deformation (scratching or indentation) by another solid.  Hardness is somewhat correlated with elastic moduli (the larger the moduli, the harder the material is expected to be).  This correlation is far from perfect, as was spectacularly demonstrated in 2002 by a measurement proving osmium stiffer than diamond, although osmium is not nearly as hard as diamond.  (The term indentability is a less ambiguous alternative for hardness.)

Some news reports have wrongly described osmium as a "soft" metal, whereas it is one of the hardest metals, with a Mohs hardness of 7...

Hardness does not have a theoretical definition.  Instead, it is evaluated using a number of practical scales, which are only roughly compatible with each other.  Hardness values obtained by conversion between such scales are notoriously fuzzy and/or unreliable.

Friedrich Mohs

The oldest scale of hardness is the Mohs' scale, which can barely be called quantitative.  It was first devised in 1812 by the German mineralogist Friedrich Mohs (1773-1839) who published it in 1822.  This scale is based on comparisons with the materials below, which are assigned the values listed.  If a material scratches another, it's said to be of equal or greater hardness.  Mohs' scale is popular with geologists in the field, who can use it by carrying a kit containing the standard Mohs minerals and/or other substances of intermediate hardness:  Lead is 1½, a fingernail is 2½ (so are galena and gold), a knife blade is 5½ (so is window glass), a steel file is 6½, tungsten is 7½ (tungsten carbide is almost 9).  It's not necessary to have an expensive diamond in such a kit, except to "identify" other diamonds, because all other minerals are much softer (as of 2002, only one or two synthetic substances are known to be harder than diamond, as discussed above).

MH 1MH 2MH 3MH 4MH 5
Basic Mohs Hardness (MH) Scale
MH 6MH 7MH 8MH 9MH 10
Al2SiO4 X2

Some English speakers have been known to memorize this sequence using an infamous  (politically incorrect)  mnemonic sentence:

Those Girls Can Flirt And Other Queer Things Can Do.

The rungs in this hardness "ladder" are uneven:  According to some plastometric hardness scales (described below), diamond (MH 10) is about 3½ times harder than corundum (MH 9), whereas fluorite or fluorspar (MH 4; sometimes misspelled  "flourite" or "flourspar") is only 20% harder than calcite (MH 3). 

Pure corundum is colorless.  Colored corundum gemstones are called either ruby if red (because of the presence of chromium) or sapphire otherwise (mostly blue, in the presence of titanium).  Widely used for grinding and polishing, the emery abrasive contains » 60% corundum (Al2O), mixed with magnetite (Fe3O) and spinel (MgAl2O; MH 8)

This problem with Mohs' scale has been somewhat corrected in a so-called extended scale, which departs from the above beyond MH 6 and assigns a hardness of 15 to diamond (instead of 10).  This extended scale remains much less popular with geologists than the above original one...

Better quantitative ratings of hardness are mostly obtained with two very different kinds of instruments, which may well measure different characteristics of the material under test.  One of these is known as a durometer, the other is called a plastometer (of various types, named after the specific indenter used).

A durometer ["dur" is French for "hard"] is simply a diamond-pointed hammer which slides under its own weight in a glass tube and rebounds off the surface of the material under test.  The height of the rebound is measured and compared with what would be obtained for some reference material.  If a conventional rating of 100 is assigned to high carbon steel, this principle defines the so-called Shore scale, which is divided into overlapping subscales (A, B, C, D, O, OO) covering progressively softer materials with different measurement specifics.  Indentation by a Diamond 
 Pyramid on Silicon Nitride 
 (Vickers Indenter, 10 kgf)

Other hardness scales are based on the size of the indentation left by a plastometer after pressing [usually for 30 seconds] an object (indenter) of known geometry with a calibrated force against a planar surface of the material under test.

The above picture shows an oblique view of the imprint obtained on a silicon nitride surface by applying a force of 10 kgf (about 98 N) to a so-called "Vickers indenter" (a diamond square pyramid with an angle of 136° between opposing faces).  Such a picture offers clues that the observed remnant is smaller than what the indentation used to be when the indenter was still in it.  The elastic recovery which took place is properly disregarded in the evaluation of hardness, which is supposed to be a measure of how difficult it is to inflict permanent indentations (or scratches) on the surface of a solid.

A plastometric hardness is then defined (in units of pressure) as the ratio of the calibrated force to the total surface area of that part of the indenter at rest which has the same cross-sectional area as the observed indentation.

In those cases where the indenter is not much harder than the material under test a theoretical correction may be needed to estimate the size of an indentation that would have been left by an infinitely hard indenter.  This is also the correct way to extend to diamond (and/or substances like ultrahard fullerite, the hardest stuff known to Man) a scale like the Vickers scale, which is normally based on direct readings from instruments with diamond indenters (valid for ordinary materials, compared to which diamond may be considered "infinitely hard").

Video :   This is How Diamonds are Made  (2013).

Kelly (Bakersfield, CA. 2001-08-29; e-mail)
I am from Guam, but now live in Bakersfield, California. [...] I get a sunburn quicker in Guam than in Bakersfield.
I say that it's because being in Guam [latitude 13.5°N] puts me closer to the Sun than being in Bakersfield [35.4°N].  Am I correct?   [...]

Not quite so.  What you want to compare is what happens in the two locations at the same local solar time, say noon.  You may as well compare the situations of two points B and G on the half meridian directly facing the Sun (noon local time) and located at the respective latitudes of Bakersfield and Guam (the longitudes of Bakersfield and Guam are irrelevant).

Let's do a rough calculation first (always a good idea).  The Earth is almost a perfect sphere of radius R = 6371000 m.  Because the sun is so far away, the difference in the solar distances of two points  The Earth lighted by the Sun, 
 between the Spring equinox
 and the Summer Solstice. A and B on the Earth is accurately estimated as the distance between the two parallel planes containing A and B that are perpendicular to the rays from the Sun.  Therefore, you may observe that the difference between the solar distances of two illuminated points on Earth may not exceed the Earth radius (R).  This overestimate is good enough for our next point...

The difference in solar distances at noon is thus [much] less than 6371 km.  Since the Sun is about 150 000 000 km away, this amounts to less than 0.0000425 of the distance to the Sun.  Now, the energy received from the Sun per unit of area (physicists call it the radiant illumination) is inversely proportional to the square of the distance to the Sun.  The difference in radiant illumination due to the difference in solar distance is thus no more than 0.0085%.  Obviously, such a minuscule difference could not possibly account for the observation concerning sunburns.  There's another explanation...

What's important is the angle of the Sun's rays, not the distance to the Sun! This matter of angles also explain why summers are warmer than winters [in the Northern Hemisphere], in spite of the fact that the Earth happens to be closer to the Sun in Winter than in Summer. [Believe it or not.]  The basic reason why it's colder in Winter is that each square mile of the Earth's surface "sees" the Sun at a more oblique angle and thus receives a narrower beam of sunlight.  Also, there's less time available between sunrise and sunset to receive the Sun's energy.

Your observation about sunburns, however, involves yet another angular aspect. Sunburns are directly related to UV exposure in the middle of the day.  It's important to realize that the atmosphere is a natural UV filter.  The lower you see the Sun on the horizon, the thicker the filter.  At noon, the Sun is higher in Guam and will therefore burn you faster.  At sunrise or at sunset, you can't possibly get sunburned.  In Guam or in Bakersfield...

master_at_games_not  (Yahoo! 2007-07-22)   Kelvin's Thunderstorm
A nice suggestion for an experimental project in physics.   [ 12th grade ]

 Coat of arms of  
 Lord Kelvin (1824-1907) In 1867, Lord Kelvin (1824-1907) devised a simple way to generate high voltages by harnessing the power of... falling drops of water.

To the best of my knowledge, this experiment has not yet been performed in a high-school context, but it should make a great student project...

You may want to watch a video demonstration by Walter Lewin at MIT.

Kelvin Water Dropper (Wikipedia)   |   Kelvin's Thunderstorm by Bill Beaty (1995)
Lord Kelvin's Electrostatic Water Drop Experiment   |   Spark-Museum

 Gustave Gaspard de Coriolis   
 (1792-1843) X 1808 (2007-07-24)   The Coriolis Effect:  A Simple Case
A dropped object always falls to the east of a plumb line.

The Coriolis force (or Coriolis acceleration) is observed only for something that moves with respect to a rotating frame of reference.

The beautiful mathematics involved repays study, but it's nice to demonstrate the essentials by analyzing a simple case, using only  elementary methods :

You're on a beach near the Equator, on top of a tower (or a palm tree) of height h.  First, you draw a plumb line to make a small mark in the sand below.  Then, you drop a steel marble.  If there's absolutely no wind, you might expect that marble to fall  exactly  on the mark you just made.  Well, it will fall a few millimeters  east  of the mark...  Why?  Let's be quantitative:

Let h be the height of your tower and R be the equatorial radius of the Earth  (it's equal to 6378137 m, but we won't need the exact value).  Let g be the normal acceleration of gravity at the Equator  (g = 9.780327 m/s).  Finally, let w denote the sidereal angular rate of rotation of the Earth, namely:

w   =   2p / 86164.09   =   7.2921159 ´ 10-5 rad/s

Looking from the south in a  nonrotating  frame (momentarily) at rest with respect to the center of the Earth,  the key observation is that the top of the tower and the beach have  different horizontal velocities,  namely:

w (R+h)     and     w R

What matters is not the huge value of those speeds  (both are about  465.1 m/s,  1674.4 km/h  or  1040.4 mph)  but their tiny  difference.

Another way to put it is that, in a nonrotating frame of reference (momentarily) at rest with respect to the beach, the top of the tower is seen to have a velocity  w h.  A marble dropped with zero velocity with respect to the top of the tower is really dropped with an horizontal velocity  w h  with respect to the beach and it will maintain that horizontal velocity throughout the time  (t)  it takes to hit the ground.  Thus, the marble will land at a distance   Dx  =  w h t   to the  east  of the point marked by the plumb line.  Since  h = ½ gt2, we have:

Dx   =   w h t   =   w Ö(2/g)  h3/2   =   0.00003297554  h3/2

The numerical coefficient applies if  Dx  and  h  are expressed in meters.

Coriolis effect at the Equator  (without air drag)
  Dropping Height    Deviation to the East  
10 m  1.04 mm  
20 m  2.95 mm  
50 m  11.66 mm  
100 m  32.98 mm  
200 m  93.27 mm  
300 m  171.35 mm  

If the experiment is carried out at a nonzero  geodetic  latitude  j,  then the above remains applicable, except that the horizontal velocity of the top of the tower with respect to the ground  (in an inertial frame "tangent to" the rotating one)  is now  w h cos j.  Also, the value of  g  to be used (which determines the duration of the fall)  is clearly the local value, which is greater than the above equatorial value.

Dx   =   cos(j) w Ö(2/g)  h3/2

The  geodetic latitude  j  of a location is the angle from the plane of the equator to the local vertical.  This is the only type of latitude used in geography.  It differs from the geocentric latitude  (the angle between the equatorial plane and a line from the center of the Eartn)  since the Earth is an oblate ellipsoid.  The "local vertical" is perpendicular to the "plane of the horizon" (which is tangent to the Earth surface) and it doesn't go through the center of the Earth, except for points on the Equator or at the poles.    Geodetic latitude and elevation

 Eiffel Tower
 (center view from ground)
Eiffel Tower  (from below)

The  Eiffel Tower  is at latitude  48° 51' 32''.  Its top  floor  is  309.63 m  above surrounding grounds.  The local value of  g  in Paris  (as measured before 1901)  is  9.80991 m/s2.

Neglecting air drag, a marble dropped from the top floor of the Eiffel Tower would hit the ground  118 mm  east  of the dropping vertical,  after a fall lasting about 7.9452 s.

Air resistance (no wind!) would only slow down the fall, thus  increasing  the deviation to the east in  direct proportion  to the increase in duration.

Dx   =   (14.8548 mm/s)  Dt

The Eiffel Tower has long been linked with the study of air resistance, starting in 1903  with the experiments of  Gustave Eiffel (1832-1923) himself:  By dropping objects along a vertical cable hanging from the second floor  (149.23 m  above the ground)  Eiffel obtained the most accurate aerodynamical data of that era.  He improved on this setup by building a wind tunnel next to the Tower in 1909.  The larger of the two wind tunnels he build in nearby Auteuil (in 1912) is still operational today  (Eiffel's is the oldest aeronautical laboratory in the World).

The  Leaning Tower of Pisa  is located at a latitude of  43° 43' 23'' N  (and a longitude of  10° 23' 47'' E).  At that latitude  j,  the rotation of the Earth makes the top of the Tower  (h = 55 m above the ground)  moves at a speed  wh cos j  relative to the ground.  This results in a Coriolis deviation  Dx  to the east, which is related to the time of fall  Dt  by the equation:

Dx   =   (w h cos jDt   =   (2.89846 mm/s)  Dt

Neglecting air resistance, the time of fall  Dt  from a height of  55 m  is about  3.35 s.  This yields a Coriolis deviation of about  9.7 mm.  Air resistance (no horizontal wind) would only increase that Coriolis deviation a bit.  The effect is roughly 12 times less than for the Eiffel tower because, at a similar latitude, it is proportional to the height  (h)  raised to the power of 1.5...

Gustave Gaspard de Coriolis (1792-1843; X1808)

(2007-07-26)   Terminal Velocity   (Settling Velocity)
In the air, the velocity of a falling object has an upper limit.

In the main, the dissipative forces which oppose the motion of a nonrotating smooth  sphere  in a fluid are forces which are opposite to the sphere's velocity  (relative to the fluid).  They are essentially of two distinct types:

  • Viscous resistance, proportional to the speed  v  (mostly for liquids).
  • Quadratic drag, proportional to  v2  (mostly for gases).

If the resistive forces are (somewhat artificially) limited to the sum of those two terms, we obtain a differential equation which can be solved analytically.

 Come back later, we're
 still working on this one...

Vertical fall against air resistance   |   Optimal gear ratio for a car

(2007-09-22)   Angular Momentum & Torque
Spin and  orbital  angular momentum are conceptually distinct.

A body of mass  M,  and center of mass  C  has a total (linear) momentum  p  which is obtained by adding the elementary contributions  v dm  of its massive elements  (v is the vectorial velocity of the infinitesimal element of mass  dm):

p   =   ò  v dm

The  angular momentum  L  of a  spinless  point-mass about a fixed origin  O  is defined as the cross-product of the position  r  into the momentum of that point:

L   =   r ´ p   =   m  r ´ v

The  angular momentum  L of an extended collection of such points is the sum of the angular momenta of its components:

L   =   ò  ( r ´ v )  dm

This breaks down into the sum of two terms:  The  spin  (or  intrinsic  angular momentum)  and the  orbital  angular momentum, (defined as the angular momentum about O of the whole mass concentrated at the center of mass  C).

L   =   S  +  r ´ p   =   S  +  OC ´ p

If the center of mass is motionless,  the linear momentum  p  is zero and the second term vanishes.  In that case,  the angular momentum  L  is equal to the spin  S.

Torque :

Torque  is defined as the moment of an applied force.  Only an external torque can change the angular momentum of a system.  Otherwise,  total angular momentum is always conserved  (even when mechanical energy isn't, as happens when there are internal friction forces).

Until 1924, it was thought that all angular momenta were of an orbital nature, which is to say that a pointlike particle could not have a spin.  Paradoxically,  modern quantum mechanics  allows such a thing.  However, the average  density of spin  in matter is normally zero at the macroscopic level, so the angular momentum of a large object can be obtained as the sum of the orbital momenta of tiny  spinless  constituents.

(2017-06-19)   Example of an  Ad Hoc  Conservation Law
A case where neither mechanical energy nor momentum are conserved.

In a given physical system,  the French call  intégrale première (du mouvement)  any function of the generalized coordinates describing it which would be constant over time in any actual motion.  Some examples:

  • Linear momentum  in the absence of external forces.
  • Angular momentum  in the absence of external torque.
  • Total mechanical energy  in the absence of heat dissipation.
It's always a good idea to try very hard to find what remains the same in a transformation.  This may well turn difficult questions into easy ones.  The  Chameleon Problem  is a nice recreational example.

The concept is mostly used in the context of  analytical mechanics  where all forces are conservative.  Below is a rare example where such a conserved quantity exists in the presence of dissipative forces  (friction).

Consider two cylinders which can rotate freely  (without friction)  at different rates around two distant parallel axes.  If the rims of the cylinders are brought into contact, friction will slow down one cylinder and speed up the other until the two circumferential speeds are equal  (so that there's no longer any slipping at the rims).

Question :   Express the final rates of rotation  (W1, W2)  as functions of the initial rates  (w1, w2)  knowing the radii (R1, R2)  of the cylinders and their moments of inertia (J1, J2)  around their respective axes of revolution.

Note that the friction forces are internal forces which do not modify the system's total angular momentum.  However,  external forces must be applied to the axes to keep them in place and  those  do modify the total angular momentum.  Thus, neither the mechanical energy nor the angular momentum are conserved.  We need another  ad hoc  conserved quantity...

Answer :   By  Newton's third law,  the two friction forces exerted by one rim upon the other are opposite to each other.  Let's call them F and -F.  For either cylinder,  the rate of change of the angular momentum around the central axis is the moment of the friction force applied at its rim:

F . R1   =   J1 . dw1 / dt             (-F) . (-R2)   =   J2 . dw2 / dt

Therefore,  the following quantity remains constant over time:

J1 . w1 / R1  -  J2 . w2 / R2   =   k

In the final state,  that equation does hold,  along with an additional relation stating that the speeds at the rims are equal at the point of contact:

J1 . W1 / R1  -  J2 . W2 / R2   =   k
R1 . W1   =   -R2 . W2

Eliminating  W2  between those two equations,  we deduce:

R1 . W1  ( J1 / R12  +  J2 / R22 )   =   k

A similar relation holds for the other cylinder and we may use the defining expression of  k  (in term of the previous rates)  to obtain the solution:

 Come back later, we're
 still working on this one...

In the special case of two solid cylinders of given thickness made from the same homogeneous material,  the moment of inertia is proportional to the fourth power of the radius  (it's what the formula ½MR2 implies when M is proportional to the square of R).  The above formulas become:

 Come back later, we're
 still working on this one...

A Real Nice Physics Problem (8:54)  by  Walter Lewin   (2017-06-19).
For once, the pedestrian solution of Pr. Lewin may not do justice to this:   15:45  |  2:45  |  20:39

visits since Dec. 6, 2000
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