Poisson and Gaussian
Throw N balls at random into B boxes. Let a be the average number of balls, N=B,
in a box. Let P(x) be the probability that a given box has exactly x balls in it.
(a) Show that
P(x) ¼ is approximately equal to [(a^x)(e^-a)]/x!
Certain assumptions are needed for this expression to be valid. What are
they?
(b) Show that if a is large, the above Poisson distribution essentially becomes a
Gaussian distribution,
P(x) = [(a^x)(e^-a)]/x! and is approxiamtely equal to {e^[-(x-a)^2]/2a}/sq.rt. of [2a(pi)]
About Me
- juLxisT
- Quezon City, National Capital Region, Philippines
- a simple guy who loves being himself but sometimes forgets who he really is..
10 May 2008
pang theo mech mga parekoy!! sabi sa source "A boring problem daw!"
Consider a rigid, solid Earth with radial symmetry and outside radius R, rotating with constant angular velocity w. Start at the equator and drill a sufficiently large diameter hole straight through the center of the Earth and out the other side. Never mind that the center of the Earth is pretty hot, and probably flows. Don't worry about ground water, or ocean water, filling the hole, etc., etc. Consider that these don't happen, maintaining the integrity of the hole you've drilled.
Now hold an object above the hole and release it to fall down the hole. Will the object hit a wall? If so, which one, the East, West, North, South, or some other wall of the hole?
Solution. Since the force on the falling object is radial at all times there's no torque on it (about the center of the Earth). Therefore, the angular momentum of the falling object is conserved. Then rw = Rwe where w is the angular velocity of the falling object, r is its distance from the center of the Earth, R is the radius of the Earth and we is the angular velocity of the Earth. Since r decreases with time, w must increase with time, therefore w > we all the way down. The falling object gets "ahead" of the walls of the hole. Since the Earth rotates west to east, the falling object must hit the east wall.
The same argument can be used for an object dropped from a high tower at the equator.
This conclusion is not affected by the radial distribution of mass within the Earth, just so long as it is a solid Earth (as given).
Some books give the problem in a different fashion. They specify that the Earth also be at rest. Kleppner and Kolenkow [An Introduction to Mechanics, McGraw-Hill, 1973] ask students to show that in this case the motion is simple harmonic with a period of 84 minutes, and note that this is the same as the period of an Earth satellite in low orbit with r approximately equal to R. That solution is straightforward.
K&K specifically exclude consideration of hitting the wall, by saying: "In deriving this result, you need to treat the Earth as a uniformly dense sphere, and you must neglect all friction and any effects due to the Earth's rotation." I wonder why they didn't just say, "Assume the Earth doesn't rotate."
They fail to address some obvious questions.
Why should this be the same period as the satellite in low Earth orbit? Is there some insightful-conceptual way we might have expected this?
What if the hole were drilled through the center starting at some other latitude?
What if the hole were drilled perpendicular to the Earth's axis and through the Earth's center, starting at any latitude?
What if the hole were drilled in some other way? If it were drilled pole to pole, the rotation of the Earth wouldn't affect the fall. There's a related problem. In Chapter 7 of Lewis Carroll's 1893 book Sylvie and Bruno. The fictional German professor, Mein Herr, proposes a way to run trains by gravity alone. Dig a straight tunnel between any two points on Earth (it need not go through the Earth's center), and run a rail track through it. With frictionless tracks the energy gained by the train in the first half of the journey is equal to that required in the second half. And also, in the absence of air resistance and friction, the time of the journey is about 42 minutes (84 for a round trip) for any such tunnel, no matter what the tunnel's length.
Martin Gardner discussed this in one of his Scientific American "Mathematical Games" columns.
The period in the frictionless hole:
Restating the question: The period for the circular "skimming" orbit (of radius equal to the Earth's radius, R) is
T = 2p (R/g)1/2
Show that this is the same as the period of motion in frictionless straight hole through the center of the stationary Earth.
Partial Answer:
First let's address the question of the force on an object within a homogenous spherical distribution of mass.
The force on a piece of mass at point P a distance r from the center is due to all mass in the sphere of radius less than r.
image for this problem..
http://www.lhup.edu/~dsimanek/scenario/shell.gif
All mass at distances greater than r contributes nothing. This can be seen by subdividing the mass in the outer shell into many shells of infinitesimal thickness. Then take infinitesimal solid angles with vertices at the point P. The two pieces of mass within the solid shell contribute oppositely directed forces at P, of size proportional to the inverse square of their distances from P. But by geometry, their volume is proportional to the square of their distance from P. Therefore, their forces are equal, and being oppositely directed, add to zero.
The mass within r is proportional to r3. Its gravitational force acts as if it were concentrated at the center of the sphere, and therefore the force it exerts at P is proportional to r-2. Therefore the gravitational force at distance r is proportional to r3r-2 = r.
Now consider the straight hole or tunnel through the Earth, but not through the center of the Earth.
The hole has frictionless walls, or tracks. The gravitational force within the Earth is f = -kd where d is the distance from the center. When d = R the force is mg. At the center of the Earth the force is zero.
At any angle ß, the component of gravitational force along the tunnel is f = F(r)sin(ß). At distance x from the center of the hole an object is distance r from the center of the Earth. x = r[sin(ß)], and we already know that F(r) = -kr and
k = mg/R so
f = -kr sin(ß) = -kx
e2 ung image for that question..
http://www.lhup.edu/~dsimanek/scenario/earthole.gif
Therefore the force is proportional to the distance, meeting the requirements for simple harmonic motion with an effective "spring constant" k.
Now hold an object above the hole and release it to fall down the hole. Will the object hit a wall? If so, which one, the East, West, North, South, or some other wall of the hole?
Solution. Since the force on the falling object is radial at all times there's no torque on it (about the center of the Earth). Therefore, the angular momentum of the falling object is conserved. Then rw = Rwe where w is the angular velocity of the falling object, r is its distance from the center of the Earth, R is the radius of the Earth and we is the angular velocity of the Earth. Since r decreases with time, w must increase with time, therefore w > we all the way down. The falling object gets "ahead" of the walls of the hole. Since the Earth rotates west to east, the falling object must hit the east wall.
The same argument can be used for an object dropped from a high tower at the equator.
This conclusion is not affected by the radial distribution of mass within the Earth, just so long as it is a solid Earth (as given).
Some books give the problem in a different fashion. They specify that the Earth also be at rest. Kleppner and Kolenkow [An Introduction to Mechanics, McGraw-Hill, 1973] ask students to show that in this case the motion is simple harmonic with a period of 84 minutes, and note that this is the same as the period of an Earth satellite in low orbit with r approximately equal to R. That solution is straightforward.
K&K specifically exclude consideration of hitting the wall, by saying: "In deriving this result, you need to treat the Earth as a uniformly dense sphere, and you must neglect all friction and any effects due to the Earth's rotation." I wonder why they didn't just say, "Assume the Earth doesn't rotate."
They fail to address some obvious questions.
Why should this be the same period as the satellite in low Earth orbit? Is there some insightful-conceptual way we might have expected this?
What if the hole were drilled through the center starting at some other latitude?
What if the hole were drilled perpendicular to the Earth's axis and through the Earth's center, starting at any latitude?
What if the hole were drilled in some other way? If it were drilled pole to pole, the rotation of the Earth wouldn't affect the fall. There's a related problem. In Chapter 7 of Lewis Carroll's 1893 book Sylvie and Bruno. The fictional German professor, Mein Herr, proposes a way to run trains by gravity alone. Dig a straight tunnel between any two points on Earth (it need not go through the Earth's center), and run a rail track through it. With frictionless tracks the energy gained by the train in the first half of the journey is equal to that required in the second half. And also, in the absence of air resistance and friction, the time of the journey is about 42 minutes (84 for a round trip) for any such tunnel, no matter what the tunnel's length.
Martin Gardner discussed this in one of his Scientific American "Mathematical Games" columns.
The period in the frictionless hole:
Restating the question: The period for the circular "skimming" orbit (of radius equal to the Earth's radius, R) is
T = 2p (R/g)1/2
Show that this is the same as the period of motion in frictionless straight hole through the center of the stationary Earth.
Partial Answer:
First let's address the question of the force on an object within a homogenous spherical distribution of mass.
The force on a piece of mass at point P a distance r from the center is due to all mass in the sphere of radius less than r.
image for this problem..
http://www.lhup.edu/~dsimanek/scenario/shell.gif
All mass at distances greater than r contributes nothing. This can be seen by subdividing the mass in the outer shell into many shells of infinitesimal thickness. Then take infinitesimal solid angles with vertices at the point P. The two pieces of mass within the solid shell contribute oppositely directed forces at P, of size proportional to the inverse square of their distances from P. But by geometry, their volume is proportional to the square of their distance from P. Therefore, their forces are equal, and being oppositely directed, add to zero.
The mass within r is proportional to r3. Its gravitational force acts as if it were concentrated at the center of the sphere, and therefore the force it exerts at P is proportional to r-2. Therefore the gravitational force at distance r is proportional to r3r-2 = r.
Now consider the straight hole or tunnel through the Earth, but not through the center of the Earth.
The hole has frictionless walls, or tracks. The gravitational force within the Earth is f = -kd where d is the distance from the center. When d = R the force is mg. At the center of the Earth the force is zero.
At any angle ß, the component of gravitational force along the tunnel is f = F(r)sin(ß). At distance x from the center of the hole an object is distance r from the center of the Earth. x = r[sin(ß)], and we already know that F(r) = -kr and
k = mg/R so
f = -kr sin(ß) = -kx
e2 ung image for that question..
http://www.lhup.edu/~dsimanek/scenario/earthole.gif
Therefore the force is proportional to the distance, meeting the requirements for simple harmonic motion with an effective "spring constant" k.
thermody nmn..
Helium (ç=1.67) is adiabaticly expanded (very slowly) in a well insulated container to a volume four times its original volume. It started at room temerature (22 degrees Celsius) and at a pressure of one atmosphere.
a. What is the final pressure?
b. What is the final temperature?
Consider Helium to act as an ideal gas.
eto mga answers and solution..
a. http://zebu.uoregon.edu/~probs/therm/adiabatic/node2.html
b. http://zebu.uoregon.edu/~probs/therm/adiabatic/node3.html
it's better to do some practices para handa agad sa pasukan di ba friends!!
a. What is the final pressure?
b. What is the final temperature?
Consider Helium to act as an ideal gas.
eto mga answers and solution..
a. http://zebu.uoregon.edu/~probs/therm/adiabatic/node2.html
b. http://zebu.uoregon.edu/~probs/therm/adiabatic/node3.html
it's better to do some practices para handa agad sa pasukan di ba friends!!
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