Kinematics : Speed andacceleration. Making time a mechanical quantity.
Dynamics : Conservation ofmomentum (Newton's three laws).
(2011-01-28) 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 toallsentient animals,including human beings: Two people can push or pull twice as hardas just one of them...
Archimedes established the less obvious fact that simple passivemachines (lever, inclined plane, etc.) can allow one man toexert the same force as two men, provided he's willing to do itover a distance twice as large.
The mechanical advantage of a simple machineis 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 anangle with the horizontal) can be used to lift a load vertically. In this function, its mechanical advantage is:
1 / sin
This indicates by how much that "machine" reduces the force required to perform the task,provided there is nofriction 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 mechanicaladvantage equals the ratio of distances fromthe fulcrum axis to the input and output lines, respectively. (The angular displacements being equal, because oftherigidity of the lever, the actual displacementswould be proportional to distances measured along those lines.)
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)
(2011-01-28) 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 scholarGerard 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) byMarshallClagett (1916-2005).
Unlike the ancient Greeks, Gerard of Brussels allowed quotients of dissimilar quantities as legitimate objects of study. (viz. distance over time).
Gérard thus paved the way for the next generation of kinematicists who would further focus on the concept ofacceleration, including the Oxford calculators at Merton College (discussednext) Nicolas Oresme (1323-1382) in France and Casale (Giovanni di Casali, c.1310?-1375) in Italy.
(2011-01-28) The distance traveled at constant acceleration is what would be traveled at asteady 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 wasWalter Burley (c. 1275-1344) nemesis ofWilliam of Ockham (c. 1288-1348).
John Suisset, the Calculator : Richard Swineshead (c. 1323?-1354). He appears in Cardano's famous list of the 12 Greatest Minds.
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. theradius vector):
The velocity is dr/dt = v. The acceleration dv/dt = g is constant. We may, therefore, integrate twicesuccessively to obtain:
v = v0 + tg and r = r0 + tv0 + t 2g
That last relation implies that r remains in a planeparallel to ( g , vo ). 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 usedfor 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 thatfalling objects have a constant acceleration only some 250 years later).
The one-dimensional Merton rule can also be used for angular displacementsor 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).
(2004-10-15) Reviving a dimensionless constant of great historical interest...
Time is defined so that motion looks simple. J.Henri Poincaré,as quoted byJohn A. Wheeler in Gravitation (1973) Misner, Thorne & Wheeler.
Arguably, the scientific revolution in Physics started in 1581when Galileo Galilei (1564-1642) discovered the isochronism of the pendulum, namely the fact that a givenpendulum oscillates with the same period regardless of its amplitude (as long as that amplitude isnot too large). To do that, he had to use his own pulse as a timer, because nothingelse 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 pendulumis proportional to the square of its period of oscillation (at least for small amplitudes).
At first, Galileo had worked without a precise device tomeasure 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 duringeach motion he wanted to time. Assuming the flow had aconstant rate,he would measure the time ellapsed by weighing the water collected. For example, he obtained 903 grains of water during a fall from restthrough 2000 punti (about 1.88 m). He then defined his new unit of time, the tempo, ascorresponding to 16 grains of water on this particular device.
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 thegravitational 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 duringthe time it takes for a body to fall from a height H of2000 punti (about 1.88 m).
The modern answer (underideal conditions) is that R/H is 8/2, so R is about 1621.14 punti,irrespective of the value of a punta (singular ofpunti). The experimental value obtained by Galileo was 1590 punti. This precision does correspond to a timing error of less than 10 ms,as advertised.
Brian Kieffer (2011-05-10,Yahoo!) What is the period of a pendulum for large swings?
AfterGalileo first described the law of the pendulum and thelaw of falling bodies,a century would ellapse before Newtonian dynamics offered both laws a common rational explanation, which is now commonplace.
) forthe constant discussed in theprevious section, whichGalileo could only obtain experimentally.
The motion of a simple pendulum in a vertical planeis always discussed at the introductory level (slightly generalized detailsare presentedelsewhere on this site). The angle between the local vertical (gravitational field g) and a pendulum of length L obeys the followingdifferential equation:
'' + (g / L) sin = 0
For small oscillations, sin is nearly equal to and the followingdifferential equation is thus approximately satisfied:
'' + (g / L) = 0
The solution is a so-called simple harmonic motion expressed by:
= A cos ( ( t-t0) ) [ A and t0 are arbitrary constants ]
The pulsatance = (g/L) does not depend on the amplitude (A) of the oscillation,neither does the period T = 2... 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
L
g
What about large oscillations?
Choosing the unit of time so the small-swing period T0 is 2, we have:
'' + sin = 0
Let's call A the amplitude of the motion, whichwe may define as the positive value of for which ' is zero. By multiplying theabove equation into', we see that the left-handside of the following equation is constant, since its derivative vanishes (it's essentially the total mechanical energy).
(' ) 2 cos = cos A
Introducing the notations k = sin A/2 and x = (sin /2) / k we obtain:
(' ) 2 = 2 (cos cos A) = 4 [ sin2(A/2) sin2(/2) ] = 4 k2(1 x 2)
This gives t as the following indefinite integral of afunction of x, endowed with a sign that changes twice per period (everytime x reaches an extremum, as itoscillates 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 =
dx
( 1 x2)( 1 k2x2)
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 ofelementary functions. The complete elliptic integral of the first kind isthedefinite integral:
K(k) =
1
dx
0
( 1 x2)( 1 k2x2)
Traditionally, k is called the modulus of this elliptic integral or itsrelatives.
As x goes from 0 to 1, the angle goes from 0 to A and the time t increases by one fourth of the true period (measured in our special unit of timewhere the small-swing period is 2). Therefore, going back to arbitrary units of time:
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 , (1-k2 ) ).
Since k = sin(A/2) this translates intothe following robust expression:
True Period of the Pendulum
T = T0 /agm ( 1 , cos A/2 )
That elegant formula is all we really need to computethe 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° = . For example:
A
k = sin A/2
T / 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 = G2, where G is Gauss's constant.
Of course, release angles above 90° are only possible witha rigid pendulum; the usual bob-on-a-string won't do.
We may now investigate therelevant power series, using the expansion (A002894) of K(k) that involves squares of central choice numbers :
T = 2
L
g
2n n
2
(
sin A/2 ) 2n
Expanding each term of that sum as a power series of A and collectinglike 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 = 2
L
1+
A2
+
11 A4
+
173 A6
+
22931 A8
+
...
g
16
3072
737280
1321205760
Here are the next terms of the above bracketed power series:
Although theradius of convergence of thatpower series is [ probably ] , it ought to be shunned for precise numerical computations, in favor of theabove simpler, faster and more robust arithmetic-geometric formula.
(2011-01-28) Smoothly combining a vertical fall and a uniform horizontal motion.
Starting in .../... "law of odd numbers" => "law of falling bodies"
At time t, the projectile's cartesian coordinates are:
x = (v0 cos ) t y = (v0 sin ) t g t 2
The range d is the value of x at the time when y returns to zero:
= (2 v0 / g) sin d = (v02 / g) sin 2
The maximum range (v02 / g) is thus obtainedfor a tilt = 45°.
The top height h (the value of y when t = /2) is one fourth of the height of theright triangle whose base is the range d and whose hypotenuse is in the direction of the initialvelocity (use the above picture to memorize that).
h = d tg = (v0 sin ) 2 / 2g
It wouldn't be fair to Noviomagus and other expert gunners of that era to suppose that they hadn'tnoticed that actual cannonball trajectories didn'tfit Aristotelian triangles at all. The gunners chose to overlook such obvious shortcomings becausetheir beloved Aristotelian theories did providethe correct relation between range and tilt,which is really all you need to operate a cannon, at least when the target islevel with it (Aristotelian physics doesn't give correct predictions in other cases).
(2011-01-28) The key to Newton's three laws of motion.
The notion of inertia was originally not part of Aristotelian physics. It was conceived by Ibn Sena (980-1037) and formally introduced, around 1340, byJean 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).
Motion remains uniform, unless disturbed by a force.
Force is the rate of change of momentum: F = dp/dt withp = mv.
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 vectorialquantity (now called linear momentum) which can be exchanged between separate systems but canbe 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 rate at which they exchange momentum over time.
(2007-10-31) The work done is equal to the change in kinetic energy.
The infinitesimalwork 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 itspoint of application during a time dt. Namely:
F . v dt = m (dv/dt) . v dt = mv .dv = d ( ½ mv 2)
Therefore, the work done over any interval of time is equal tothe 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).
The above computation may serve as an introduction to the modern concept of energy, which is often credited to Thomas Young (1807) and whose relativistic generalization is presented in the next section. The word itself was coined by Aristotle () and had been used by theologers andphilosophers since the fourth century AD.
(2008-09-18) (Einstein, 1905) Based on the relativistic expression of the work done to a point mass.
Are not the gross bodies and light convertible into one another ? Isaac Newton (Query 30, 1717)
In Special Relativity, the dynamics of a point-mass are still described by the Newtonianrelation F = dp/dt provided the momentump is redefined as follows for a nonzero mass m at speed v:
p =
mv
1 -v2/c2
The limiting case allows particles of zero mass traveling (only) at speed c.
In this, c is the speed of light (Einstein's constant). Assuming the rest mass (m) to be constant,we maydifferentiatethis (vectorial) expression:
dp =
m dv
+
m (v.dv/c2)v
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 =
mv . dv
= d
m c2
(1 -v2/c2) 3/2
1 -v2/c2
Theabove expression of the work done is thus only slightly modified:
F .v dt = (dp/dt) .v dt = v . dp
= d
m c2
1 -v2/c2
This yields the following relativistic equivalent of kinetic energy (up to an additive constant) for a pointlike particle of constant mass, first given byAlbert Einstein in 1905:
E =
m c2
1 -v2/c2
Using the above value of p, we get: E2 ( cp ) 2 = ( m c2 ) 2 Thus, E = m c2 (1 + p2/ mc)
So, at low speed, we have E mc 2 + ½ mv 2 m c 2 + ½ p 2/m.
Those correspond to the first two terms in theTaylor expansions of the above relativistic valueof E/mc2 using either x = v/c or y = p/mc :
The constant leading term seems irrelevant at first. However, Einstein postulated that it correspondsto a real energy at rest which anything of mass m always possesses. That's the principle of inertia of energy :
E = m c 2
The more general relation shown above remains valid for photons (m=0):
E2 = (m c2) 2 + ( cp ) 2
When the energy of a system at rest increases, so does its mass. Thenext section investigates the fundamentals of what's involved in such a process.
(2008-09-18) Variable rest mass in the case of a pointlike system.
If a pointlike object is somehow endowed with a structure thatallows it to store energy at rest (e.g., spin or heat) then its rest mass (m) may vary. Redoing theabove computation to allow for that, wefind 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
1 -v2/c2
1 -v2/c2
With the proper algebraic massaging, this differential relation becomes:
d
m c2
=
v .dp
+
1 -v2/c2
d ( mc2)
1 -v2/c2
dE
=
W
+
Q
This is actually one enlightening case of the first principle of thermodynamics : The variation in the energy of this simplesystem is an exact total differential (dE) of the parameters which describe it (m andv). Thisdifferential form is equatedto the sum of two infinitesimal quantities which are not exact total differentials themselves; the "work" (W) and "heat" (Q) received by our point-mass.
rsg160 (2001-04-15) I have been told that a satellite in a circular orbit that startsto 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 slowthe 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 altitudeat 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 2and its potential energy is exactlyMgz/(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:
M ( V V' + gz' ) = FV
Now, we may remark that the quantity(-z' )/ Vis simplysin(),the sine of the angle of reentry (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 sinF/M
As long as the drag force F is less thanMg sin(),V' is therefore a positive quantity,which means that the spacecraft will indeed gain speed initially.
cannot be too small, or else the spacecraft could "bounce" off the atmosphere and be backinto outer space after losing just a little bit of energy. If no action is taken, an orbiting spacecraft could thus keep bouncing back untilenough energy is lost and/or its reentry angle is sufficiently large--possibly dangerously so,since a large 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.
(2002-01-27) Consider an equilibrium realized when a "perfect"rope is passed over a frictionless and massless pulley with a ten-pound weight onone 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 ofAlice in Wonderland,Lewis Carroll(1832-1898,né Charles Dodgson),who agonized over it. He was aprofessor of mathematics at Oxford from 1855 to 1881. The above picture once illustrated a discussion of the puzzle by themathematical columnist Sam Loyd (1841-1911), who called the problem"Lewis Carroll's Monkey Puzzle", while stating that it was not known whetherLewis 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 havethe 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 thebalancing weight are always equal. There are only two such forces for each body; the downward weightand the upward tension of the rope. The weights are equal because the two bodies have the same massand 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 thatthe 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 ofmass for both pulley and rope (otherwise the rope's tension would not be the same oneither side of an accelerating pulley and it would vary along the length of an acceleratingrope).
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 anoriginal 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 higherfrom the ground, but willrequire 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 theprocess (the monkey and the weight went up one foot each), which is just as difficult atask for a 10 lb monkey as lifting his own weight up two feet...
This deceptively simple puzzle is an excellent way to start a healthydiscussionabout the fundamental principles of classical mechanics.
(2007-01-03) A heavy ball is dropped on a hard floor from a heightofone foot with a light ball (ping-pong ball) on top of it. How high could the light ball bounce?
: Up to 9 feet ! :
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 Vto 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 speedreverses itself (since the mass ratio of the two balls is very large). An upward speed of 2Vrelative to the heavy ball translatesinto an upward speed of 3V...
Ideally, the kinetic energy of the ping-pong ball thus goes from ½ mV2before impact to 9 times that muchafter the shock ! This is enough to propel that ballup to 9 times the height it was dropped from.
In practice, a lesser height is reached (because of imperfect elasticity and a finite mass ratio) but the actual demonstration remains impressive.
(2007-10-09) The tangential acceleration is dv/dt. The normal acceleration is v2/R.
Consider a smooth curve in space. Let s be the curvilinearabcissa along that curve and T = T(s) a unit vector tangent to the curve in the direction ofincrease of s. Geometrically, the curvature 1/R at s can be definedas 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 tothe trajectory has magnitude v2/R (speed squared divided by radius of curvature).
(2007-10-09) Harnesses shouldn't be needed to keep riders on their seats.
Traditional roller-coasters rides are designed so that a componentof the acceleration always makes the wheels of the car push against thetracks and/or the bodies of the riders push against the seat cushions.
At the top of avertical 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 byreleasing a car of mass m from a height h above the top ofthe loop is greater than the kinetic energy it retains atthat point (the two would be equal if there was no lossof mechanical energy due tofriction). Therefore:
h > ½ R
Thus, a roller-coaster car must be released froma height above the top of a loop-the-loop (much) greater than half thetop radius of curvature.
(2007-10-18) A hanging bob traveling along an horizontal circle.
Let L be the length of the string and the (constant) angle it makes with the vertical. The radius R of the trajectory is:
R = L sin
The horizontal centripetal acceleration is v2/R = 2 R, as the bob travels around the circle at speed v = R. (This is a special case of the normal accelerationdiscussed 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 theweight (mg) and the string tension (F). The magnitudes of the relevant components of theacceleration are thus obtained as the sides of the right trianglepictured at right (featuring the aforementioned angle ).
Thus, v2 = R g tan = L g sin2 / cos = L g (1-u2) / u which can be recast into the following quadratic equation, withrespect to u = cos :
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 = u = ( 1 + x2)½ - x where x = v2 / 2Lg
To solve this,we put x = sh and find that u =e. In other words:
= arccos (exp (argsh( v2 / 2Lg )))
v 2 = 2 L g sh ( ln cos )
(2011-01-11) Let the above hang from the top of a hemisphereof radius a > L/2
We assume that a large enough orbital speed v makes the (pointlike) bobtouch the hemisphere and that there's nofriction. The trajectory of the bob is a horizontal circle of radius x at a height y > 0 above the horizontal planeof 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 2a y = 2a 2 L2 which gives a value of y that can be plugged into either equation to obtain x > 0.
y = a[ 1 (L/a)2] and x = L[ 1 (L/a)2]
To obtain the tension T of the string, we may project Newton's second law on the direction parallel to (y, -x) since the unknownreaction of the sphere is perpendicular to that (it's normal to the surface and, therefore, it'sin the direction of the (x, y) vector). The mass m multiplied into the centripetal(horizontal) acceleration so projected is equal to the sum of therespective projections of the bob's weight and of the tension force acting on it:
This boils down to the simple expression: T = m [ g + y.v2/x2] L/a Using the angular velocity = v/x and the above value of y, we obtain:
T = m g L/a + [ 1 (L/a)2] m L2
(2008-02-13) 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 itsmoment of inertia is J.
Let O be the center of the spherical bowland C be the center of the small ball. is the (oriented) angle from the vertical to OC and ' is the derivative d/dt.
In a pure roll, the point of contact of the ball with the sphericalsurface has zero speed. Since the speed of C is (R-r) ', we may deduce that the ball spins with angular velocity (1-R/r) '. The potential energy of the ball is -M g (R-r) cos (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(' )2 M g (R-r) cos
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) '' + g sin = 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 restfrom a point 3 inches above the equilibrium position?
An oscillation where the acceleration is proportional to the distance from a fixed pointis 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.Understandard 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 thatMy"=-Mg-y(Mg/L) (Newton's Law).If we consider instead the position x=y-L above the equilibrium position,we havex"+(g/L)x=0 (in the absence of friction or damping).Introducing the quantity such that2=g/L(in other words is(9.80665/0.1016)or about 9.82457 rad/s), x is therefore equal toA cos(t) +B sin(t), 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 simplyx = A cos(t),with A=3"=76.2 mm and=9.82456847...rad/s(corresponding to a period of oscillation T=2of 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 Twould be different under a gravitational field other than the "standard" one.
gatman(Central Florida.2001-03-31) How fast (rpm and mph) are electrons going around a nucleus? What factors affect it and how?
For small objects like electrons,quantum mechanics states thatthe very notion oftrajectory breaks down. You can't measure both the momentum and the position of a particle: The product of the uncertainties in the measurements of suchconjugatequantities cannot be less than the so-calledHeisenberg limit. Thus, the electron does not have a precisespeed in the classical sense.However, you can work out what the expected value of the momentum would beif 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 ofexpected speed...
This being said (and it had to be said),you may work out things numerically using something as crude as theold-fashioned semi-classical Bohr model of the atom(circular "trajectories" --oh, well-- with an angular momentum whichis only allowed to be n times therationalized Planck constant "h-bar" = h/2). The number n is theprincipal quantum numberyou 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 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 ofthe mass of the electron to that of the nucleus). In other words, E(n) = (-2.18 J) / n2.
As is the case for planets around the Sun, it turns out thatthe kinetic energy ½ mv2 is equal to -E(n). The mass of the electron ism = 0.911 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 isv = (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 isaon2,whereao= 0.53 misBohr's radius. The period is equal to 2/v times this,which means it is proportional to n3: T = (1.52 s) n3. The frequency is thus inversely proportional to the cube of n: f = (6.58 Hz)/n3,or if you insist on using rpm's: f = (395 000 000 000 000 000 rpm)/n3,since60 rpm = 1 Hz.
All this pertains to the hydrogen atom (Z = 1). For a lone electron around a nucleus with Z protons (a so-calledhydrogenoid 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 protonsat a speed (Z/n) Vo and a frequency (Z2/n3 ) 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 arounda circuit at frequency f0 is equivalent toa current q f0 of about 1 mA.
The so-called Bohr magneton (q/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 ofvectorial area is = I. The Bohr magneton would be the moment classically associated with anelectron actually moving at the above speed along a circular path witha radius equal to the aforementioned Bohr radius.
(2002-06-10) Is there anything harder than diamond?
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 theNanoScan (NS)scanning force microscope (SFM). Some experimental studies indicate thatultrahard fullerite is abouttwice 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 togetherthe hexagonal carbon planes underlying thestructure of graphite (the density of graphite is 2267 g/L).
However, the polymerization of fullerene which occurs at high temperatureunder 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 structureleaves little room for compression... The material is more than three times stiffer than either diamond or osmium,according to theabove 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'sstiffness[itsbulk modulus], which was thoughtuntil recentlyto be a good indicator of the more elusive quality calledhardness. Noteworthy candidates which did not live up to expectations, according to this model,includedcubic boron nitride. (CBN is the hard form of "BN". It was first synthesized in 1957 by Dr. Robert H. Wentorf oftheGeneral Electric Company. It's known in the trade asborazon.)
On the other hand, Cohen's model clearly indicated that acarbon nitride crystalshould bestiffer (and possiblyharder) than diamond. The race was on to obtain the stuff in crystalline form and measure its properties. Someearly effortsby the team of Yip-Wah Chung (at Northwestern University) resulted in a layered composite oftitanium nitride and carbon nitride (a so-calledsuperlattice)which was, surprisingly,almost as hard as diamond.
Crystals of carbon nitride were apparently first synthesized atLawrence Berkeley Lab (LBL)by Eugene Haller and William Hansen,using an approach similar to the synthesis of industrial diamonds. A 1993 patent for this newsuperhard material was subsequently awarded toCohen, Haller and Hansen. However, the jury seems still out...[SeeMarch 1998APS conference, andMay 2000Nature 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 ofsynthetic diamond was foundto 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) Hardnessis the resistance to permanent surface damage...
It's totally different from elastic resilience, in spite of a loose correlation.
Unlikestiffness, which is anelastic property of a solid[quantitatively,stiffness is simply the solid'sbulk modulus],hardness actually indicates the resistance of a solid's surface topermanent deformation (scratching or indentation) by another solid. Hardness issomewhat correlated with elastic moduli (the larger the moduli,the harder the material isexpected to be). This correlation isfar from perfect,as was spectacularly demonstrated in 2002 by a measurement provingosmium stiffer than diamond, although osmium is not nearly as hard as diamond. (The termindentability is a less ambiguous alternative forhardness.)
Hardness does not have a theoretical definition. Instead, it is evaluated using a number ofpractical scales,which are onlyroughly compatible with each other. Hardness values obtained by conversion between such scales arenotoriously fuzzy and/or unreliable.
The oldest scale of hardness is theMohs' scale,which canbarely 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 materialscratches 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 othersubstances of intermediate hardness: Lead is 1½,a fingernail is 2½ (so aregalena andgold),a knife blade is 5½ (so iswindow 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, becauseall other mineralsare much softer (as of 2002, only one or twosynthetic substances areknown to be harder than diamond, as discussedabove).
Some English speakers have been known tomemorize 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 someplastometric hardness scales (described below),diamond (MH 10) is about 3½ times harder thancorundum (MH 9), whereasfluorite orfluorspar(MH 4; sometimesmisspelled "flourite" or "flourspar")is only 20% harder thancalcite (MH 3).
This problem with Mohs' scale has been somewhat corrected in a so-calledextended scale,which departs from the above beyond MH 6 and assigns a hardness of 15to diamond (instead of 10). This extended scale remains much less popularwith geologists than the aboveoriginal one...
Better quantitative ratings of hardness are mostly obtained withtwo very different kinds of instruments,which may well measure different characteristics of the material under test. One of these is known as adurometer, the other is called aplastometer(of various types, named after the specificindenter used).
Adurometer ["dur" is French for "hard"] is simply a diamond-pointedhammer which slides under its own weight in a glass tube and rebounds off thesurface of the material under test. The height of the rebound is measured and compared with what would be obtainedfor some reference material. If a conventional rating of 100 is assigned tohigh carbon steel,this principle defines the so-calledShore scale,which is divided into overlapping subscales (A, B, C,D, O, OO)covering progressively softer materials with different measurement specifics.
Other hardness scales are based on the size of the indentation left by aplastometer after pressing [usually for 30 seconds]an object (indenter) of known geometry with a calibratedforce against a planar surface of the material under test.
Aplastometric hardness is then defined (in units of pressure)as the ratio of the calibrated force to thetotal surface area of that part ofthe indenterat rest which has the samecross-sectional areaas the observed indentation.
In those cases where the indenter is not much harder than the material under testa theoretical correction may be needed to estimate the size of an indentationthat would have been left by aninfinitely hard indenter. This is also thecorrect way to extend to diamond(and/or substances likeultrahard fullerite, the hardest stuff known to Man)a scale like theVickers scale, which is normally based on direct readingsfrom instruments with diamond indenters(valid for ordinary materials, compared to which diamond may be considered "infinitely hard").
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 closerto 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 locationsat the samelocal solar time, say noon. You may as well compare thesituations of two points B and G on the half meridian directly facing theSun (noon local time) and located at the respective latitudes of Bakersfieldand Guam (the longitudes of Bakersfield and Guam are irrelevant).
Let's do a rough calculation first (always a good idea). The Earth isalmost a perfect sphereof radius R = 6371000 m. Because the sun is so far away, the difference in the solar distances of two pointsA and B on the Earth is accurately estimated as the distance between the twoparallel 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 oftwo 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] lessthan 6371 km. Since the Sun is about 150 000 000 kmaway, 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 theradiant illumination)is inversely proportional to thesquare of the distance to the Sun. The difference in radiant illuminationdue to the difference in solar distance is thus no more than 0.0085%. Obviously, such a minuscule difference could notpossibly account forthe observation concerning sunburns. There's another explanation...
What's important is theangle 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 theSun 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'ssurface "sees" the Sun at a more oblique angle and thus receives a narrowerbeam of sunlight. Also, there's less time available between sunrise and sunset to receive the Sun's energy.
Your observation about sunburns, however, involves yetanother angular aspect.Sunburns are directly related to UV exposure in the middle of the day. It's important to realize that theatmosphere 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) A nice suggestion for an experimental project in physics. [ 12th grade ]
In 1867, Lord Kelvin (1824-1907) devised a simple way togenerate high voltages by harnessing the power of...falling drops of water.
To the best of my knowledge, this experiment has not yet been performedin a high-schoolcontext, but it should make a great student project...
(2007-07-24) A dropped object always fallsto the east of a plumb line.
The Coriolis force (or Coriolis acceleration) is observed only forsomething thatmoves with respect to arotating frame of reference.
Thebeautiful 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/s2 ). Finally, let denote thesidereal angular rate of rotation of the Earth, namely:
= 2 / 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:
(R+h) and 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 anonrotating frame of reference(momentarily) at rest with respect to the beach, the top of the toweris seen to have a velocity h. A marble dropped with zero velocity with respect to the top of thetower is really dropped with an horizontal velocity h with respect to the beach and it willmaintain that horizontal velocity throughout the time (t) it takes to hit the ground. Thus, the marble will land at a distance x = h t to the east of the point marked by the plumb line. Since h = ½ gt,we have:
x = h t = (2/g) h3/2 = 0.00003297554 h3/2
The numerical coefficient applies if x 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 , then the above remains applicable, except thatthe horizontal velocity of the top of the tower with respect to the ground (in an inertial frame "tangent to" the rotating one) is now h cos . Also, the value of g to be used (which determines the duration ofthe fall) is clearly thelocal value, which is greater thanthe above equatorial value.
x = cos(2/g) h3/2
of a location isthe angle from the planeof the equator to the local vertical. This is the only type of latitude used in geography. It differs from thegeocentric latitude (the angle between the equatorial plane and a line from the centerof the Eartn) since the Earth isan oblateellipsoid. The "local vertical" is perpendicular to the "plane of the horizon"(which is tangent to the Earth surface) and it doesn'tgo through the center of the Earth, except for points on theEquator or at the poles.
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 (asmeasured before 1901) is 9.80991 m/s.
Neglecting air drag, a marble dropped from the top floor of the Eiffel Towerwould hit the ground 118 mmeast of the droppingvertical, 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.
x = (14.8548 mm/s) t
The Leaning Tower of Pisa is locatedat a latitude of 43° 43' 23'' N (and a longitude of 10° 23' 47'' E). At that latitude , the rotation of the Earth makes the top of the Tower (h = 55 m above the ground) moves at a speed h cos relative to the ground. This results in a Coriolis deviation x to the east, which is related to the timeof fall t by the equation:
x = ( h cos ) t = (2.89846 mm/s) t
Neglecting air resistance, the time of fall t 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 towerbecause, at a similar latitude, it is proportional to the height (h) raised to the power of 1.5...
(2007-07-26) 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 thosetwo terms, we obtain a differential equation which can besolved analytically.
(2007-09-22) 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 velocityof the infinitesimal element of mass dm):
p = v dm
The angular momentumL of a spinless point-mass about a fixed origin O is defined as the cross-productof the position r into the momentum of that point:
L = rp = m rv
The angular momentumL of an extended collection of suchpoints is the sum of the angular momenta of its components:
L = (rv ) 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 ofthe whole mass concentrated at the center of mass C).
L = S + rp = S + OCp
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).
(2017-06-19) 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 relevant generalized coordinateswhich would be constant over time in any actual motion. Some examples are:
Total mechanical energy in the absence of heat dissipation.
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 quantityexists 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 onecylinder and speed up the other until the two circumferential speeds are equal (so that there's no longer any slipping at the rims).
Express the final rates of rotation () as functions of the initial rates () knowing the radii (R1, R2) of the cylinders and theirmoments 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 totalangular momentum. However, external forces must be applied to theaxes 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...
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 aroundthe central axis is the moment of the friction force applied at its rim:
F . R1 = J1 . d1 / dt (-F) . (-R2) = J2 . d2 / dt
Therefore, the following quantity remains constant over time:
J1 .1 / R1 J2 .2 / R2 = k
In the final state, that equation does hold, along withan additional relation stating that the speeds at the rims are equal at the point of contact:
J1 .1 / R1 J2 .2 / R2 = k R1 .1 = -R2 .2
Eliminating 2 between those two equations, we deduce:
R1 .1 ( 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:
In the special case of two solid cylinders of given thicknessmade from the same homogeneous material, the moment of inertia is proportionalto the fourth power of the radius (it's what the formula ½MR2 implieswhen M is proportional to the square of R). The above formulas become:
(2023-05-05) Juggling doesn't make conserved quantities disappear.
rsg160 (
(2002-01-27) 
(2007-01-03) 
(
= h/2
A and B on the Earth is accurately estimated as the distance between the twoparallel 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 oftwo illuminated points on Earth may not exceed the Earth radius (R). This overestimate is good enough for our next point...
