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no1 got that one.. so let's do this again:

... the next person to post (who hasn't already) gets to ask a question.. I'm hoping this will encourage more people to play:)

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No, you have the wrong answer. If you said "pretend that people and the cart are point-masses 13 m from the center of the circle," you'd have the answer, but the effective radius of the cart and people would be approximately 14 m.

The level of zero potential is always and is always assumed to be the lowest possibility in the situation.

Now that's just not true. I don't think you've seen every physics problem in the world. You're just being full of yourself if you think that everybody automatically follows the same implicit rules as your high school physics teacher uses. Zero is often the highest potential in a situation and can also be something arbitrary.

And since the effective radius of the cart's center of mass is really more like 14 m, your choice of zero is about a meter below what you supposed it to be.

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No, you have the wrong answer. If you said "pretend that people and the cart are point-masses 13 m from the center of the circle," you'd have the answer, but the effective radius of the cart and people would be approximately 14 m.

Now that's just not true. I don't think you've seen every physics problem in the world. You're just being full of yourself if you think that everybody automatically follows the same implicit rules as your high school physics teacher uses. Zero is often the highest potential in a situation and can also be something arbitrary.

And since the effective radius of the cart's center of mass is really more like 14 m, your choice of zero is about a meter below what you supposed it to be.

I see your point..

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You're just being full of yourself if

Also, don't think I'm trying to insult you or call you names; I'm just a bit flamboyant in my metaphors, that's all.

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No, you have the wrong answer. If you said "pretend that people and the cart are point-masses 13 m from the center of the circle," you'd have the answer, but the effective radius of the cart and people would be approximately 14 m.

Now that's just not true. I don't think you've seen every physics problem in the world. You're just being full of yourself if you think that everybody automatically follows the same implicit rules as your high school physics teacher uses. Zero is often the highest potential in a situation and can also be something arbitrary.

And since the effective radius of the cart's center of mass is really more like 14 m, your choice of zero is about a meter below what you supposed it to be.

It's important to remember that just about everything in Physics is relative. If I tell you the arm's length is 13m, then obviously that is what is expected to be used as the complete length, especially because the length of the car wasn't given. I also didn't tell was zero potential energy is, so exactly like you said it's either not possible or minimum possibility is assumed to be zero. This is the only possible way to answer the question and since I asked it and have an answer myself, there must be some answer. If we had to make every little specification than something as simple as "What's 2m/s East plus 3m/s East" cannot be answered because no reference point is given. It is assumed that the Earth is the reference point, but it rotates on an axis and revolves around the sun. Who's to say it wasn't 2m/s east relative to the earth and 3m/s east relative to the sun? I also didn't mention what planet this was on which means gravity could be completely different. I also didn't even mention air resistance. Since every little detail could never possibly be given, many details are assumed or negligible (air resistance) in any practicle physics problem. It's better to give an answer based on information given then just to say, "Look, I've found an unimportant piece of information." Give me any physics problem and I can analyze it and scrutinize it as you have. Any normal physicist (excuse my spelling) could still give an answer to my question. Although they might point out some of the facts you did, they could still give me the same answer as I did and most likely would not say it is absolutely impossible. But, can we stop over-analyzing the minor details of this question and move on to another one?

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It's important to remember that just about everything in Physics is relative. If I tell you the arm's length is 13m, then obviously that is what is expected to be used as the complete length, especially because the length of the car wasn't given. I also didn't tell was zero potential energy is, so exactly like you said it's either not possible or minimum possibility is assumed to be zero. This is the only possible way to answer the question and since I asked it and have an answer myself, there must be some answer. If we had to make every little specification than something as simple as "What's 2m/s East plus 3m/s East" cannot be answered because no reference point is given. It is assumed that the Earth is the reference point, but it rotates on an axis and revolves around the sun. Who's to say it wasn't 2m/s east relative to the earth and 3m/s east relative to the sun? I also didn't mention what planet this was on which means gravity could be completely different. I also didn't even mention air resistance. Since every little detail could never possibly be given, many details are assumed or negligible (air resistance) in any practicle physics problem. It's better to give an answer based on information given then just to say, "Look, I've found an unimportant piece of information." Give me any physics problem and I can analyze it and scrutinize it as you have. Any normal physicist (excuse my spelling) could still give an answer to my question. Although they might point out some of the facts you did, they could still give me the same answer as I did and most likely would not say it is absolutely impossible. But, can we stop over-analyzing the minor details of this question and move on to another one?

I agree, as I said earlier just assume 'all else equal'. If something is not stated in the problem then it should be assumed negligible or constant.. depending on the situation.

Now since no one else has appeared interested in this game, I am going to go ahead and ask another question (hope yall don't mind):

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This is actually part of a free response question from a recent exam:

A disk with moment of inertia 3.00 kg*m^2 is rotating with a speed of 7.00 rad/s. A second identical disk that is not rotating is dropped onto the first disk. After 2.00 s, friction forces (assumed to be constant) between the two disks give rise to a torque that causes the two disks to come to a single common final speed.
Calculate the change in kinetic energy of each of the two disks. (Note: this will be two numbers)

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Any1 know? Or just to lazy to work it out lol..
Rashakil ?

lazy to work it out

I have done somewhat similer question, its terrible!!

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Ok, new question:

A massless, rigid board is placed across two bathroom scales that are separated by a distance of 2.00m. A person lies on the board. The scale under his head reads 425 N, and the scale under his feet reads 315 N. Find the weight of the person and locate the center of gravity of the person relative to the scale beneath his head.

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Ok, new question:

A massless, rigid board is placed across two bathroom scales that are separated by a distance of 2.00m. A person lies on the board. The scale under his head reads 425 N, and the scale under his feet reads 315 N. Find the weight of the person and locate the center of gravity of the person relative to the scale beneath his head.

Is the weight of the person 740N, and the center of gravity located at a point 63/74 (~.851)m from the person's head..

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haha, yea I think so.. It was difficult though, I don't think any1 in my class got that question right on the exam. Maybe just a couple of people..

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This is actually part of a free response question from a recent exam:

A disk with moment of inertia 3.00 kg*m^2 is rotating with a speed of 7.00 rad/s. A second identical disk that is not rotating is dropped onto the first disk. After 2.00 s, friction forces (assumed to be constant) between the two disks give rise to a torque that causes the two disks to come to a single common final speed.
Calculate the change in kinetic energy of each of the two disks. (Note: this will be two numbers)

Would it not be 36.25 J? Assuming no heat or sound is generated...

That's a fake answer if it is right, since the two disks have non-rotational energy relative to one another, too.

Edit: it's a wrong answer too; that's just impossible -- the disks would glance off one another, with one disk ejecting the other in the direction of its spin at the point of contact. So something's got to be holding the disks in place, stealing their kinetic energy. In that case the second disk will steal half the first's velocity, and the kinetic energy will thus divide in half, since 2*(1/2)^2 = 1/2. So 18.125 J increase for the first disk, -54.375 J for the second. The apparatus that is apparently holding the disks in place acquires the angular momentum of the entire system (which is rotating in the first disk's direction a little bit). So Earth's rotation changes slightly.

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Is the weight of the person 740N, and the center of gravity located at a point 63/74 (~.851)m from the person's head..

Correct.

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Would it not be 36.25 J? Assuming no heat or sound is generated...

That's a fake answer if it is right, since the two disks have non-rotational energy relative to one another, too.

Edit: it's a wrong answer too; that's just impossible -- the disks would glance off one another, with one disk ejecting the other in the direction of its spin at the point of contact. So something's got to be holding the disks in place, stealing their kinetic energy. In that case the second disk will steal half the first's velocity, and the kinetic energy will thus divide in half, since 2*(1/2)^2 = 1/2. So 18.125 J increase for the first disk, -54.375 J for the second. The apparatus that is apparently holding the disks in place acquires the angular momentum of the entire system (which is rotating in the first disk's direction a little bit). So Earth's rotation changes slightly.

Heres the question again:

This is actually part of a free response question from a recent exam:

A disk with moment of inertia 3.00 kg*m^2 is rotating with a speed of 7.00 rad/s. A second identical disk that is not rotating is dropped onto the first disk. After 2.00 s, friction forces (assumed to be constant) between the two disks give rise to a torque that causes the two disks to come to a single common final speed.
Calculate the change in kinetic energy of each of the two disks. (Note: this will be two numbers)

Hmm.. I think you did it correctly. This is how the solution is supposed to work out:
wf=final speed of rotation for the entire system
I=moment of inertia

You know the initial speed of rotation =
[tex](3.00kg*m^2)*(7.00 rad/s) = (2*(3.00kg*m^2))*wf [/tex]
Therefore, wf= 3.50 rad/s.

Next, to calculate the total change in kinetic energy of each of the disks, you:

disk 1: change in KE=[tex] (1/2)*(I)*wf^2-(1/2)*(I)*(initial speed of disk)^2[/tex]

=[tex](1/2)*(3*kgm^2)*(3.50 rad/s)^2-(1/2)*(3*kgm^2)*(7 rad/s)^2= -55.1J[/tex]

disk 2: change in KE=[tex](1/2)*(I)*wf^2-0[/tex]

=[tex](1/2)*(3*kgm^2)*(3.50 rad/s)^2-0 = 18.4 J[/tex]

As a side note, you will notice that the two change in kinetic energies are of different magnitude. If Newton's third Law is correct, how can equal and opposite torques give rise to changes in kinetic energy that are not equal and opposite? Answer- They rotate through different angles.. One rotates through an angle of 10.5 rad, and the other rotates through an angle of 3.5 rad.

Just in case you were confused like I was about how Newton's third law worked in that case.

And Rashakil Fol- No complaining about the question or saying that the answer isn't a true value.. b/c my professor actually wrote this question himself.. And yes, he has his master's, and doctoral degrees... And he used to be head of the physics department at a large D1 University.. So I think he knows what hes doing when he writes up exams.

Your turn, Rashakil Fol

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And Rashakil Fol- No complaining about the question or saying that the answer isn't a true value.. b/c my professor actually wrote this question himself.. And yes, he has his master's, and doctoral degrees... And he used to be head of the physics department at a large D1 University.. So I think he knows what hes doing when he writes up exams.

Why? He would agree that the kinetic energy of the falling disk is not being taken into account.

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Why? He would agree that the kinetic energy of the falling disk is not being taken into account.

Well you don't know how far it fell.. If it was like placed on top of the other disk, then its kinetic energy would be irrelevant.. Since you don't know the distance the disk fell, then you are to assume that it is negligible. You need to stop looking at the 'big' picture.. Physics is about taking small phenomena and being able to explain what happens

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A cylindrical colony is orbiting the Moon 700 kilometers above the Moon's surface. Its interior radius is 500 meters, and its inhabitants live under Earth-like gravity (10 N).

The air has been sucked out of the colony due to a political emergency, and Brian wants Daniel to receive a message, delivered in the form of a ten-pound brick that smashes through Dan's skylight.

Dan lives 1000 meters away (measuring along the surface of the ground) from Brian's projectile launcher, and the shortest walking path from Brian's projectile launcher to the point underneath Dan's skylight is the arc of a circle, in the direction contrary to the colony's rotation. Dan's skylight is 20 meters above the ground, and Brian's projectile launcher releases projectiles from ground level. His launcher works by imparting a constant force for two seconds on an object, along a magnetic track that can be pointed in any direction.

Brian has a five second window during which the projectile can be in the air before it is detected by the colony's projectile detection system and zapped out of the sky.

If Brian wants to fire the projectile thirty minutes from now, what should the magnitude of the force be, and in which direction should the launcher be pointed, assuming Brian wants to waste as little energy as possible?

P.S. The colony itself weighs 1.00*10^16 kg, not including the projectile.

Votes + Comments
your signature is stupid
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Yeah, we've been through this. Enough with the overanalyzing and "That's an impossible question" statements. It's just annoying, especially when the question can be answered and agreed upon the same by everyone.

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lol.. hmm, I guess this question is unanswerable.. There is no real answer if g= 10N.. The question doesn't even make any sense..

.. just a very small taste of your own jackass medicine

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Last time i studioed physics gravity was a force?

Weight = mass x gravity

So if i i have a mass of 8 N and gravity is 10 N then i have a weight of 80kg

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The strength of a classical gravitational field is measured in N kg ^ -1 (Newtons per kilogram). On the surface of the earth this is measured to be about 9.8 N kg ^ -1. Apologies if this has already been said in the thread, but I'm in an internet cafe so in a rush. If no one minds I'll set another question seeing as the last one apparently went a bit "stale":

An imaginary new HD video system (called Purple - Ray) has been launched and the laser it uses to read the disks has a wavelength of 400 nm (4 * 10 ^ -9 m). The lens used has a numerical appature of 1.5. What is the minimum size the markings on the disk (that represent the data) could have and still be resolved by the laser?

Note: numerical aperture is a dimensionless quantity, hence the lack of a unit.

Steven.

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I've always known gravity to be measured in m/s^2. Of course the force of gravity could be 10N but that is completely dependent on the mass of the object that gravity acts on. If you use the equation F = (m1*m2*G)/(d^2) then G isn't measured in newtons or m/s^2. You get (Nm^2)/(kg^2). Unless I did something wrong.

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