Midterm Exam on Relative Motion 101 - unicyclist, bohemian duck, ligh

1. A unicyclist is riding his wheel in a straight line on the flatbed of a truck passing you fixed to a point on the road edge. You can see that the unicyclist is moving @ u = 1m/s with respect to a series of 1 meter marks along the length of the truck bed. You measure the truck rolling by @ v = 9 m/s with respect to the embankment.

2. The unicyclist is replaced with a bohemian duck trained to fly flat and straight at a constant velocity in the same direction as the truck motion. When released you measure the duck motion at 2 m/s wrt the 1 meter marks on the truck bed. The truck moves at 9 m/s wrt the embankment as before.

3. the bohemian duck is replaced by with a light emitter at the rear of the truck pointing in the same direction as the unicyclist and bohemian duck motions. The emitter begins to emit short pulses at 1 pulse/s. The truck moves as before at 9 m/s.

The examination is open book, but no drinking, smoking or lacivious activty permitted during the examinaton.

What are the various speeds as indicated?

Code:

1. the unicyclist wrt: 
   a. embankment 
   b. the truck? 
2. the bohemian duck wrt: 
   a. the embankment 
   b. the truck?
3. the light pulse wrt:
   a. embankment 
   b. the truck

 

Log in or Sign up to hide all adverts.

Log in or Sign up to hide all adverts.

James R, You realize you are on the embankment making all these measurements, do you not? If so how is answer to #3 determined? Or, alternatively, from where you considered the obvservations were conducted?

 

Log in or Sign up to hide all adverts.

The exam asked for the speed of the various items with respect to the truck, which is what James answered.
The location and motion of "you" doesn't matter. You can still directly measure speeds wrt to the truck if you use rulers and clocks which are:
- at rest with respect to the truck,
- synchronized wrt the truck, and
- in the same place as the events of interest.

Don't make the mistake of reading too much into the notion of an "observer". The "observer", in special relativity literature, is just shorthand for a reference frame with rulers and clocks. It's supposed to be simpler to say "measured by an observer at rest in reference frame A" than "measured using rulers and clocks at rest in reference frame A", but they mean exactly the same thing.

 

You, does matter because it is you making the determination of the respective velocities wrt to the embankment. I agree with you that observer may be a system of measuring devices that summarily we all refer to a "you".

All of your analyses being a given and that 'You' are the observer and as present as any measuing devices you bring along on the embankment to facilitate your measurement processes. I place one demand that you be as objective as you have ever been in your life, equivalent to the level of a sworn juror as limited by a judge's 'juror instrution' to you as a member of a jury judging a serious matter, that you forego the use of any learned system or theory aquired ante the publication of this thread. The judge orders you to determine the issue based on your analyses restrictive scrutiny restricted to the four corners of the documents in this thread. Is this fair? Or is this unfairly restrictive?

Pete, I truly am unconcerned regarding how you make the measurements, but am crucially interested in being informed, by you, of how the measuring is conducted. Myself, I would respond as you and James R have indicated in your responses to question #2 and #1. However, both of you, are on the embankment observing the motions. James R answered #3 that,

"Speed of light relative to truck = 299792458 m/s.
Speed of light relative to road = 299792458 m/s."

From the perspective of an embankment observer how can such an observer measure a relative velocity of light wrt a parallel moving truck to be equal to the relative velocity of light and a stationary road, especially when all measurements are conducted from the embankment? Is this a newly discovered law of nature?

When I first began posting in this forum (oh yes, back in the olden days) James R made it clear to me the importance of naming, identifying, the reference frame from which measurements are made. While the truck may be moving in Wickenburg, Arizona, and the two human observers, James R and Pete, may be on embankment in Portsmouth, New Hampshire, these conditions are not a satifactory rearrangement of the experimental set up. If you both have other plans then please provide us an adequate electronic-mechanical measurment system as outlined in the opening thread, before you leave the facilities.

The truck and prodigal student have returned, pursuant to a self imposed obligation owed the staff, the particpants in this thread, the members and viewers of sciforums.

 

Geistkiesel, your questions do not specify who does the measuring. If you want to insist that only someone on the truck can make measurements wrt the truck, James's answers are still correct.


But as I said, measurements wrt the truck can also be made by someone on the embankment who can see rulers and clocks on the truck. I don't see why this is a difficulty.

Perhaps you could imagine that the embankment person has instruments on the truck that make measurements and transmit the results. That would have the same effect.

 

Please Register or Log in to view the hidden image!

 

How many times have you posted this now? I've posted it at least twice.

Please Register or Log in to view the hidden image!

 

First time but I think I've seen Ben post it, as well as you. Just shows how predictable cranks are. Geist's scenario is just 'race car on a train' relabelled.

 

Now here's my midterm exam, also open book:



An infinitesimally thin uniform spherical shell of total mass \(M\) and radius \(R\) is placed in deep outer space, far away from any stars and planets.

1. Calculate the gravitational field strength produced by the shell as a function of distance (\(r\)) from the center of the shell:

Code:

a) Inside the shell ([tex]r<R[/tex])
b) Outside the shell ([tex]r>R[/tex])

2. Using your result from Part 1, or any other method you like, calculate the gravitational field strength as a function of \(r\) in both regions, assuming that instead of a uniform spherical shell, we now have a uniform solid sphere, also with total mass \(M\).

3. In 4 lines or less, calculate the electric field strength produced by a uniformly charged solid sphere with total charge \(Q\) and radius \(R\), again as a function of distance from the center (\(r\)), for both \(r>R\) and \(r<R\). You may use the results you found in Part 1 and Part 2.

You may drink whatever you want and smoke whatever you want, just make sure you share with your colleagues. Assume the classical laws of Newton gravitation and Coulomb electrostatics apply here (i.e. ignore any quantum/relativistic corrections).

 

I was careless. I intended to add the constraint in clear enough terms that you, on the embankment are making the measurements as stated.

What are the various speeds as indicated where you are making the measurements.
Code:
1. the unicyclist wrt:
a. embankment
b. the truck?
2. the bohemian duck wrt:
a. the embankment
b. the truck?
3. the light pulse wrt:
a. embankment
b. the truck

Using this criteria:
a. what is the SOL wrt the embankment?
b. what is the SOL wrt the truck?

Lets say you are measuring the the speeds from a fixed points on the road where you distributed detectors, equally spaced, and as the objects go by the measurements are made by recording the time the objects trigger the known location of the switches. Even the light pulses can trigger the counters as the pulses pass by.

Now what are the velocities as described? Arew trhere any ambiguities tghat I haven't corrected for?

 

If you are only using rulers and clocks that are at rest and synchronized in the embankment rest frame, then all speeds you measure will be speeds with respect to the embankment.

 

There are an infinite number of solutions.
The unevaluated result of integrating the shell integral is, \(F = K[-k/S + S]\) evaluated from a to b and where K = GmM/(d^2)(4R) and k = r^2 - R^2.However, there are many results
2. Using your result from Part 1, or any other method you like, calculate the gravitational field strength as a function of \(r\) in both regions, assuming that instead of a uniform spherical shell, we now have a uniform solid sphere, also with total mass \(M\).

3. In 4 lines or less, calculate the electric field strength produced by a uniformly charged solid sphere with total charge \(Q\) and radius \(R\), again as a function of distance from the center (\(r\)), for both \(r>R\) and \(r<R\). You may use the results you found in Part 1 and Part 2.

You may drink whatever you want and smoke whatever you want, just make sure you share with your colleagues. Assume the classical laws of Newton gravitation and Coulomb electrostatics apply here (i.e. ignore any quantum/relativistic corrections).[/QUOTE]

 

What is S, and d?

I also suggest you read the wiki page on Gauss' law for gravity.

 

No there aren't. The mathematical theory of Riemann integration proves that the integral performed over the shell has a unique value, regardless of what coordinate system and integral scheme you use. If you insist math is wrong and you can prove otherwise, then go ahead and do the integral by two different methods.

You haven't specified the values S, a, b, d. In its present form your answer looks like gibberish. There is only one possible value for the shell integral, and there are no undetermined constants remaining after you've performed it.

Now time to grade this, as I would if I were marking a freshman or second year course on the subject.

Question 1: Answer has undetermined constants, is expressed in an unintelligible form, and no work is shown to support any of the student's assertions. 3/33 points.

Question 2: Unanswered, 0/33 points.

Question 3: Unanswered, 0/33 points.

Total: 3/99

Needless to say, we expect far, far more from our students, especially when they're seeking to enroll in courses such as Special Relativity. Please come by my office at your earliest opportunity to discuss the results of your test, and your future plans here as a student.

 

Geistkiesel reporting for the Examination Cap'n and is ready to proceed.

I will start from b.


b. For an isolated shell of mass M = 1, then as \(r -> infinity Fr -> 0\). This tells me that in isolation there is no “gravitational potential field” at points r > R.

a. The force on a test particle inside the shell result in other than popularly believed conditions.

If only points are considered without reference to a test mass it appears that the shell center is a point of instability as all points on the shell are attracting each other symmetrically. With a test mass m, where \(m << M\) located a distance h from the surface along any radial line effectively separates the masses affecting m unequally. The area of the least size is defined in terms of \(M_1 = h\) and \(M_2 = (2 - h)\). All points in \(M_1\) are nearer to m than all points in \(M_2\).

Concentrating the two mass segments to points located along the radial at points coinciding with the centers of mass of each mass segment the distance of the mass center in \(M_1/\) from m is \(\frac{h}{2}\). The point of the COM of \(M_2\) is a distance \(\frac{(2 – h)}{2}\) from m.

From the universal law of gravity

Code:

 [tex]F_r=\frac{GmM}{r^2}[/tex],   with [tex]G = 1; M_1 = h; M_2 = 2 – h [/tex]

then

Code:

[tex]F_1 = \frac{4(2 – h)}{h}[/tex] 

[tex]F_2 =\frac{4h}{(2- h)}[/tex]. 

The ratio of the two forces for [tex] 2 > h > 0[/tex], is
 [tex]\frac{F_1}{F_2} = \frac{(2 - h)^2}{h^2}[/tex].

Rewmember, the above assumes validity of the universal law of gravity (ulg), or as the gravity law of the universe (glu) for which this writer offers no critical scrutiny here.

To get a flavor for the distribution of forces within the shell evaluate the expression \(K[\frac{-k}{X} + X]\) evaluated from \(R - r to 1\), then from \(1 to R + r \) and Using real numbers as R = 1, r = .9then r = .5, and finally r = .1 and the integrated expression, \(F_r = K[\frac{-k}{X} + X] evaluated from a to 1, then from 1 to b, where a = R - r and b = R + r. As [tex]k = R^2 - r^2\) and \(K = \frac{F_r}{4R}\) the successive values for k are .19, .75 and .99 for R - r = .9, .5. and .1 respectively and for \( 0 < r < R\) the calculated \(|F_r| > 0 \). The calulated \(|F_1| = |F_2|\). But, \(F_1 + F_2 = 0\). The shell theorem calculates the total force of two shell halves and without evaluating the integrated function in parts one gets an ambiguous and distracting result. Evaluating from the + side first, the first 1/2 shell segment returns postive calculated forces and after crossing the midpoint of tghe shell negative values indicate a symmetrical result as stated abgove.

For an isolated system the gravitational field strength, depending on the inverse of the distance squared, is zero, as for\( r > R r --> inf\)

As \( r -> infinity F_r - > 0\)

1. The field strength for r > R of a uniformly charged sphere is, in the absence of a test particle, zero. 2. Similarly, for \(r < R \), absent a test charge within the shell cavity, the forms of the expression for gravity are analogus.

A problem with the shell theorem is that using the universal law of gravity as expressed as,

Code:

 [tex]F_r=\frac{GmM}{r^2}[/tex]

is that this expression says noting about concentrating the mass of the shell at the COM. The expression merely locates the center of the shell wrt a test mass located a distance r from m. But such rhetoric is anethma to Newton's faithful who do not realize the shouldefrs they arfe standing on turned to dust many years ago. If these faithful followers believe they are viewing nature from such a preferred and elevated position they are sadly mistaken, they remain faithful to be sure, but mistaken nevertheless.

 

For points r > R;
For an isolated thin shell of mass M = 1, then as r -> ∞ Fr -> 0. This tells me that in isolation there is no “gravitational potential field” at points r > R.

For R > r.
The force on a test particle inside the shell result in other than popularly believed conditions. With a test mass m, where m << M located a distance h from the surface along any radial effectively separates the mass of the shell affecting the test mass m unequally.

The surface area a spherical segment A = 2\pi h for R =1.

The area of the least SHELL segment (M1 ) defined by a line through m perpendicular to the radial through m, \(M_1 = 2\pi h\) and \(M_2 = 2\pi (2 - h)\)for \(M = 1 = M_1 + M_2\). The distances of the centers of mass of each segment are\(h/2\). and \((2 – h)/2\) for \(M_1\) and \(M_2\) respectively.

Using the universal law of gravity,

\(F=\frac{GmM}{r^2}\)
and with ignoring [text2\pi[/tex], \( G = 1; M_1 = h; M_2 = 2 – h\) , then
\(F_1 = \frac{h(2 – h)}{[\frac {h}{2}]^2\)
and,
\(F2 =\frac{(h)(2 – h)}{[(2- h)/2]^2}}\).

The ratio of the forces for \(2 > h > 0\),
\(\frac{F_1}{F_2} = \frac{[h(2 - h)]^2}{h/2)^2}\)

For h = 1 the ratio of the forces
\(\frac{F_1}{F_2}=1\)

The ratio of the forces at h = 1 (the shell center) tells us that the forces on m attributed to each segment are equal, at the shell center.

The heretical result here is that the integrated shell theorem before evaluation for \(K = \frac{GmM}{4Rr^2}\) and \(k = r^2 - R^2\) is,
\(F_r = K[\frac{-k}{x} + x]\)

If one evaluates this expression from R - r to R + r the reult will be zero.

However, evaluating from R - r to 1, then from 1 to R + r the result is not quite what dogmatic followers of Newton's Shell Theorem believe. Newton screwed up and those standing on Isaac's shoulders must bear their own consequences.

Notice that for \(K = \frac{GmM}{4Rr^2}\) the universal law of gravity is embedded in the shell the theorem as a constant multiplied by 1/4R and that the evaluated expression for the shell integral returns a value of 4R, which is in units of distance. The shell theory re the evaluated shell integral is a manipulated and concocted sham.


Using the universal law of gravity question # 2 has a similar result.

Answer to question #4.There is no observable electric field strength in the absence of a test charge at r > R, and similar results are analogous to a point R > r. The expression for electric-forces use the same form of the mass-forces with a change in variable notation only.

The ULG expressed as,

\()F=\frac{GmM}{r^2}\)

says only that the force on a test mass m from a this shell located a distance r from the shell center is F. The expression for F is that analyzed by Newton et al as reflecting that the shell behaves as if all the mass was concentrated at the shell center requires a bit more thought - the development of the shell integral did not address the locations of the location of forces forces attributed to the infitesimal mass segment pairs.
Take any point on the circle of infitesimal mass in the nearest shell segment nearest to m and consider the mirror image infitesimal mass in the segment farthest from m and you will clearly see that all forces attributed to mass points in the nearest and furthest segments that \(F_n\) > \(F_f\). There is no way using ULG that m can presume the forces are centered at the shell center. The forces on m point to the shell along any extended radial (duh), but the mass force center is offset in the direction of m.

 

Evaluate the shell integral from R - r to 1 the from 1 to R + r and see what you get. Try a numerical example for R = 1, r = .9, .5, and .1 in successive evaluations of the integrated expression.

a and b are the limits of evaluation, d the distance of the shell from m. I was giving you the benefit of the doubt that you could figure this out yourself.

Question 1: Answer has undetermined constants, is expressed in an unintelligible form, and no work is shown to support any of the student's assertions. 3/33 points.

Question 2: Unanswered, 0/33 points.

Question 3: Unanswered, 0/33 points.

Total: 3/99

Needless to say, we expect far, far more from our students, especially when they're seeking to enroll in courses such as Special Relativity. Please come by my office at your earliest opportunity to discuss the results of your test, and your future plans here as a student.[/QUOTE]
I demand a total grade of zero and in my evaluation of the tester I determine he manifests an excellent and faithful reproduction of his collegiate mentors and associates and falls within the center of maas of scientists comprising the main stream of scientific absurdities.

Personally, I would remove any accumulated points acquired for the proctor's tenure and would summarily fire him - but then with family and a need to maintain gainful employment, firing may be extreme, but not too extreme - the decision stands, the proctor's appeal denied..

See revised answer, not submitted for grade correction purposes below.