Just some secondary school concept I need help understanding

[SEPKA]

Sorry for asking these easy question, but I tried to research on these and I get even more confused, so I wish someone can straighten it out for me. Anyway, these questions are about concepts in Maths and Physics:

1. Randomly select:
What is the meaning of this sentence "X objects is randomly selected from Y objects"?
Is it "Each object have a chance of X/Y being selected"?
Is it "Each set of X objects have a chance of 1/X C Y being selected"? (X C Y is read as "X choose Y", you know what it means)
Or something else?
My maths teacher tell me it is the first one which does SOUNDS intuitive, but on the closer examination turn out to be counter-intuitive, but since my teacher have been wrong a lot before, I might as well ask for clarification from you all.

2. Rotational equilibrium:
What is the definition of rotational equilibrium?
Is it "the object have a constant angular velocity"?
Is it "the total torque around any axis is 0"?
Or something else?
The first one seems closer to the truth, yet it is ambiguous (what axis is it spinning around?) while the second one is what my school teach me. Too bad the physics teacher I approach are unable to confirm when I confront them with the first one.

3. Translation (in physics):
Which point on a rigid body define its translational velocity (especially if it does a lot of thing at once like moving/rotating/vibrating/etc.)

4. Conservation of momentum:
If momentum are conserved, then that means in an isolated system less mass->higher velocity. So if I put a radioactive material in a lead box then throw it into space, the material will lose *EDIT* mass overtime, then the velocity of the box will increase. But it sounds rather strange and counter-intuitive, so will it really act like that or am I wrong somewhere?

 

[walfin]

1. Each set's selection probability is 1/Y C X, X/Y chance of each object being selected is also true
2. Doesn't constant angular velocity imply no torque since torque=mass*angular acceleration? (let's not consider no mass)
3. Centroid, I should think?
4. Don't really understand. Are you confusing weight and mass?

Forgive me if it's wrong, physics was years ago. Seems like English isn't your native tongue too; I might've misunderstood your questions.

 

[SEPKA]

1. Each set's selection probability is 1/Y C X, X/Y chance of each object being selected is also true
2. Doesn't constant angular velocity imply no torque since torque=mass*angular acceleration? (let's not consider no mass)
3. Centroid, I should think?
4. Don't really understand. Are you confusing weight and mass?

Forgive me if it's wrong, physics was years ago. Seems like English isn't your native tongue too; I might've misunderstood your questions.

English is not my native tongue but I also live in an English speaking country for 2 years so I hope that you understood what I asked.
1. Yes set's selection probability is 1/(Y C X) also imply that each object is X/Y chance, but the reverse is not true, so I want to clarify which of them is the correct one (or if the correct one is totally different?).
2. Not necessary, it depend on your axis of rotation, and I assume that the object do have a mass.
3. That certainly sounds intuitive. But is there a theoretical explanation of why it is so?
4. I don't think I'm confused. I'm referring to the equation F=dm*dv/dt (Newton's 2nd law) and E=mc^2 (derived from special relativity).

 

[GarmGarf]

4. Don't really understand. Are you confusing weight and mass?

What Walfin was confused about there was that SEPKA stated that "the material will lose weight overtime".

If one threw a body into space, it could have a steadfast mass, but it would still lose weight overtime as it travels further away from the centre of the Earth's gravitational force field (weight is the "force" experienced by objects due to gravitation).

Whether the body in question would be losing mass or not is another issue (in this case: radioactive decay).

 

[fullerene]

1. If english isn't your native tongue, this might be tricky, cause the question you're asking sounds really sensitive to phrasing differences. Statistics is also not my strongest field... though I did have a stats theory course last semester.

"X objects is randomly chosen from Y objects" can produce different math depending on the definition of "chosen".

If repeats are not allowed (after you pick one, it comes out of the Y objects and doesn't go back into the set) and order is important (picking 1, 4, 5 is a "different" set than 1, 5, 4) and you only take X objects (that is, you don't take X+5 objects and say "what's the probability of getting these X somewhere in that X+5), it means (1/Y)*(1/(Y-1))*(1/(Y-2))*...*(1/(Y-(X-1))). This is assuming every object has an equal chance of being chosen (which is usually how "random" is used, but not always).

If order does not matter (choosing "1, 4, 5" or "4, 1, 5" are considered "the same"), then it means Y C X. "Choose" is a convenient word because "12 choose 3" is like saying "the probability of choosing 3 specific objects from a group of 12." That's how I've always remembered it, anyway. This also assumes that all the choices are equally likely, as well as that you're only taking X objects (not taking X+5 and seeing if you got the X you want).

Those are probably the most common/easiest two... but if the problems are coming in with different assumptions about the things being chosen, let me know.

2. The important origin for rotation is the center of mass. If the sum of the torques is 0 around this point, then the object has constant angular momentum. There's an equation very similar to dp/dt = m*(dv/dt) that's written dL/dt = I*(dω/dt), where ω is the angular velocity, L is the angular momentum, and I is the moment of ineria (constant for a given object under normal circumstances... unless it loses mass or changes dimensions somehow).

It's strange, because I could swear I took an engineering class where they said that the sum of the torques computed around any axis is 0 if the object is in equilibrium... but I don't see how that's possible anymore. Obviously, torque around a free-falling object off-axis is not zero, yet it doesn't rotate about that point. So my best intuition is saying that the center of mass has to be the place where you put the origin.

3. I'm not sure what the centroid is, but you measure translational motion about the center of mass too. This one I'm fairly confident of. I think that proving this was actually part of the reason why Newton invented calculus, just so that he could be sure you could treat everything as if it were its center of mass.

4. um... well, radioactivity doesn't just "get rid of" mass. Depending on the type of radioactivity, the atom either shoots off electrons, Helum nuclei, or photons--all of which have momentum, and momentum is conserved in the reaction. If you stick a radioactive atom in a box, it will fire off a photon/helium nucleus/electron in the opposite direction that the "main" particle accelerates, but the center of mass will stay in that same translational equilibrium (like before, this is why I think you always measure translational motion from the center of mass. If the "object" splits into two discontinuous particles, that doesn't really do anything except change your center of mass calculation into a sum of two point masses... and whatever the eratic motion of the two particles, the COM keeps plodding along as it did before) .


You're right that less mass does lead to higher velocity, though. If you haven't done them yet, you'll see this when you study things like rocket ships, which burn off a non-negligible amount of gas. This would be equivalent to the case where you wanted to redefine your system to include only the "main" particle. Instead of dp = m*dv (which leads to newton's law, dp/dt=m*(dv/dt)), you have to do a product rule, which gives dp = dm*v + m*dv, or dp/dt = (dm/dt)*v + m*(dv/dt). The extra (dm/dt)*v term causes a higher acceleration than if the mass were simply constant (although, as is the case with radioactive decay, we don't really know which direction that acceleration will be), but if you want to do it that way, then your "box" can't include the helium/photon/electron that the radioactive decay emits, so it's not all that odd of a picture anymore anyway.

 

[SEPKA]

@walfin & GG: sorry I was not confused, I only use the wrong word. I was talking about mass. I edited the first post.
@ cryptonia:
1. I was talking about situation where no repeat, no identical objects and order not important.
My confusion arise when I suddenly think about this situation:
Let's say you have to pick RANDOMLY 5 ball out of 10 ball all with different colour. So the normal and valid method is to put all into a bag and pick out 5 without looking.
However, when I examine the definition that my school teach me (both from books and teachers) which said that "pick randomly 5 ball out of 10 ball" means "each ball have the probability of 5/10=1/2 of being chosen", I realize another method: take 5 bag, put 2 balls in each bag (the choice of 2 ball are determined non-randomly), then pick 1 ball from each bag without looking; each ball would still have the chance of 1/2 of getting picked. This method run counter-intuitive to the notion of "randomly chosen", yet it held true to the definition given. So I get very confused here.

2. Yes my physics class also said "total torque is 0 around ALL axis". That also don't make sense to me that's why I asked. But if the angular velocity is taken around the centre of mass then it makes sense, although I'm not sure why it have to be centre of mass and not some other point (btw, centroid=centre of mass).

3. That makes sense considering what you said in #2.

4. I learned that nuclear fission process have to break the binding energy between the 2 part of the atom. This increase in energy is provided by a decrease in mass using mass-energy conservation principle by the equation E=mc^2 (it is actually a bit more complicated than this but you get my point). So that meant radioactive material lose mass overtime, and the mass don't go anywhere, it just disappear. Yes they do emit particle, but the total mass of all these particle and the radioactive material will still be less than original.
In the hypothetical situation above I put the radioactive material in a lead box to make sure the system will remains as a whole instead of spliting.

 

[walfin]

OK

1. Yes, the reverse is not true because you could have only 2 objects, one with X/Y chance of being selected, and one with (Y-X)/Y chance of being selected. I don't know what's this thing about "which is the correct one". Why is there a "correct" one? Use the right equation for the right purpose. Is this in context of a problem, or something?

2. If there's no torque it must be about ALL axes. Total sum of moments about every point is 0.

It's strange, because I could swear I took an engineering class where they said that the sum of the torques computed around any axis is 0 if the object is in equilibrium... but I don't see how that's possible anymore.

Don't think of falling objects, think of objects floating around in space and you'll realise it's possible. (constant v is kind of like stationary since all motion is relative)

3. So that you don't have to keep taking weight into account? I suppose you could use some other point and still predict the same things (although the numbers might be different).

4. Await clarifications Re what GarmGarf said.

order is important

Lol, you coulda just said Y P X.

Centroid==centre of mass

 

[SEPKA]

@ walfin: guess you was typing when I post a reply, because I clarify some point there.
1. Simply question: What does it means by "X objects are chosen randomly from Y objects"? (no repeat, order not important, no identical objects). The rest are just the list of the different answer I received so far which I do not know which one is correct. But cryptonia answered this question.
2. Yes it is possible for objects to have no torque around all axis, but I think what cryptonia meant is that it is possible for an object to be NOT rotating and yet still have torque around some axis, which sound counter-intuitive (the example was a falling object: taking any axis far from the object and you will definitely have a torque, yet the object was not rotating, only falling).
3. You do not have to assume that there is weight, but there is definitely mass.
4. As clarified and edited above.

 

[fullerene]

I realized I said something very wrong, in my last post, on probability. I said that the probability of choosing X things from Y total objects was Y choose X. That was pretty stupid, as Y choose X can be > 1, and probability is always < 1. Sorry! It seems like you knew that, given your posts above mine, though.

However, when I examine the definition that my school teach me (both from books and teachers) which said that "pick randomly 5 ball out of 10 ball" means "each ball have the probability of 5/10=1/2 of being chosen"

ok... that's just false, then. Each ball has a 1/10 probability of being chosen, not a 1/2. If you say "what's the probability that the ball chosen will be 1 of these special 5", then the probability is 5/10... but only because there are 5 "special" balls, each with probability 1/10. You add the probabilities, 1/10 + 1/10 + 1/10 + 1/10 + 1/10, one for each ball in the "special" group, to get the 5/10 probability that the ball chosen belongs to the special group.

Due to the conditions like "order not important," and "no repeats", though, the probability of getting 5 specific balls out of 10 is not intuitive. The reason for this is because you accidentally double-count situations. If you couldn't tell from above, the "base" concept in probability theory (as best as I know) is the fraction, (number of successes)/(number of possibilities). Looking at the balls, the "success" is "picking this one particular ball", and there are 10 possibilities of balls that you could get, if you reached in and grabbed one (leading to the 1/10 thing).

If the order is not important, though it throws a wrench into the problem. Suppose you label the balls with letters, and you say "I define 'success' to be 'picking balls a, b, and c out of these 10 balls, in my first three picks'. I will not put the balls back in the bag after picking."

Well, first you ask "how many possibilities are there?" On my first pick, I could get ball a, b, c, ... j, which is 10 possibilities. Since I'm not replacing the balls, my second pick has only 9 possibilities (note: you have no idea which ball you picked on the first try... but you do know that one of them is out of the bag. Since you're only counting generic possibilities, it doesn't matter which one you took). On my third, there are 8 possibilities. Multiplying these together gives 10*9*8 = 720 possibilities of ball combinations. Another way of looking at this: if there are n (= 10) choices to begin the experiment with, and you're choosing k (= 3) things out of them, there are n!/(n-k)! (10!/7!) choices, where 10! = 10*9*8*...*1.

Now you have to count the "number of successes." The only thing defining a success is that you pick 3 particular balls on your first 3 choices. To be defined as a success, your firt three choices must include those. Remember, we said that you want balls a, b, and c. So the different "success" trials could be listed in the order you draw them:

a,b,c
a,c,b
b,a,c
b,c,a
c,a,b
c,b,a

for 6 different "successes". If you listed this out for higher numbers, you would find that there are always k! successes (it's just the number of ways to permute the successes: to remove "order chosen" from the math).

putting this into the (#successes)/(#possibilities) formula gives k!(n-k)!/n! for sets where order doesn't matter.

As you probably already know, (n C k) = n!/k!(n-k)!... so the correct solution for that scenario is 1/(n C k).

So if you want to compute the probability for any other similar situation, just change the required part of the derivation. Order doesn't matter? Simply don't divide by k! at the end. Replacing the ball after each choice? Change 10*9*8 (= n!/(n-k)!) with 10*10*10 (= 10^k). If you're really careful with your logic, you can pretty much blow through elementary statistics using this idea.

I realize another method: take 5 bag, put 2 balls in each bag (the choice of 2 ball are determined non-randomly), then pick 1 ball from each bag without looking; each ball would still have the chance of 1/2 of getting picked. This method run counter-intuitive to the notion of "randomly chosen", yet it held true to the definition given. So I get very confused here.

*nods*, it is different to the definition of randomly chosen. What it sounds like you're doing, though, is designing a new experiment unrelated to the old one, seeing that the probability was the same, and saying that the situation must be the same too? I think they just gave you a shitty definition for probability. Some situation has a probability associated with its success/trial ratio, but it doesn't mean that you can tell very much about situation given only its probability.

hmm... probably good for Physicists to know that, before taking Thermal and Quantum .

2. Yes my physics class also said "total torque is 0 around ALL axis". That also don't make sense to me that's why I asked. But if the angular velocity is taken around the centre of mass then it makes sense, although I'm not sure why it have to be centre of mass and not some other point (btw, centroid=centre of mass).

I'd ask them what they would do with a free-falling object, then, as that seems the clearest counter-example to me.

4. I learned that nuclear fission process have to break the binding energy between the 2 part of the atom. This increase in energy is provided by a decrease in mass using mass-energy conservation principle by the equation E=mc^2 (it is actually a bit more complicated than this but you get my point). So that meant radioactive material lose mass overtime, and the mass don't go anywhere, it just disappear. Yes they do emit particle, but the total mass of all these particle and the radioactive material will still be less than original.
In the hypothetical situation above I put the radioactive material in a lead box to make sure the system will remains as a whole instead of spliting.

I'm not actually sure what to do about this. I mean photons have energy (E=hv) and momentum (p = hv/c), but 0 mass... so maybe(?) the mass that the atom loses turns into photons, so momentum is still conserved? But then the center of mass argument wouldn't work, so you might want to just ignore that part.

I'm also not sure about whether the energy could turn into heat. I know that mass can't just suddenly "poof" and disappear (under any circumstances as normal as radioactive decay, at least), but there aren't many places for it to go. Is it sensible to talk about the temperature of a lone atom? If so, that wouldn't really effect momentum, but would be the closest thing to mass just "poof"ing and vanishing. If I had to guess, I'd say most of the excess mass is emitted as photons, but that's mostly just my guessing.

 

[walfin]

Yeah. Sorry. Posted around same time.

If you sum up all the sets in which that ball will be chosen, it will be half the number of possible sets.

It should intuitively make sense that if you have 10 balls, 5 chosen and 5 not, then there's a 50% chance that a ball will be chosen (i.e. 1/2). Yes, this is just statistics.

(the example was a falling object: taking any axis far from the object and you will definitely have a torque, yet the object was not rotating, only falling).

Falling objects are usually not in equilibrium (acceleration due to gravity)

They can hit equilibrium in a planet with an atmosphere (e.g. this one) due to atmospheric resistance in which case there are 2 forces passing through the same point (gravity & drag; same point: for simplicity's sake), and consequently no torque.

If it is falling and there is torque around some point, the object is rotating relative to that point.

the mass that the atom loses turns into photons, so momentum is still conserved

Yes, I think this is it.
Heat can come in the form of photons too (infrared).
And there's no "principle of conservation of matter" (mass is kind of conserved, in the form of energy).
If all of the stuff were converted into photons somehow, the thing would move at the speed of light. If they were all still moving in tandem, you'd have a laser beam (but that's kind of out of point).

Come to think of it, the centre of gravity might be more convenient instead of the centroid (they're usually the same for practical purposes anyway). Whether an object is translating or rotating or both depends on the reference frame anyway.

 

[Aiss]

1. Randomly select:
What is the meaning of this sentence "X objects is randomly selected from Y objects"?
Is it "Each object have a chance of X/Y being selected"?
Is it "Each set of X objects have a chance of 1/X C Y being selected"? (X C Y is read as "X choose Y", you know what it means)
Or something else?
My maths teacher tell me it is the first one which does SOUNDS intuitive, but on the closer examination turn out to be counter-intuitive, but since my teacher have been wrong a lot before, I might as well ask for clarification from you all.

I think your teacher may be misunderstanding what you're asking them. The first statement is absolutely correct, but it doesn't define the choice in any way. If you ask "if X objects are randomly selected from Y objects, does it mean that each object has an X/Y chance of being selected?", your teacher probably confirms meaning "yes, if this is the way we choose, the chance would be X/Y". In short: the choice of the X objects from Y objects implies the probability X/Y of each object being chosen. The implication is true if 0 -> 1 - in this case, if the method of choosing objects is different, the chance may happen to be the same. The chance itself says nothing of the method of choosing, therefore the first statement doesn't define it in any way. In particular, if we limited ourselves to the first statement, and either chose or didn't choose each object with a chance of X/Y, we might end up with a different number of objects than X.

It isn't the definition of the statement, it's implied by it. I think you may be confusing shades of use of the word "meaning" here. It may either refer to a meaning as a definition (for example " 'luck' means 'a force that brings good fortune or adversity' "), or a meaning as an implication ("if X is true, it means that Y is true as well"). I'm not a native English speaker either, so I'm not sure if I'm explaining it clearly, but I think it may be where your problem is (at least based on my experience learning the language).

The second statement is just as correct. It does, however, precise that there will be exactly X objects chosen randomly, that each object won't be chosen independently of the others, each resulting subset will be equally possible (in your example there are initially mutually exclusive choices), etc. In this way, it's defining the original choice better, although I'm not sure it should be used as a definition anyway.

 

[fullerene]

Falling objects are usually not in equilibrium (acceleration due to gravity)

They can hit equilibrium in a planet with an atmosphere (e.g. this one) due to atmospheric resistance in which case there are 2 forces passing through the same point (gravity & drag; same point: for simplicity's sake), and consequently no torque.

If it is falling and there is torque around some point, the object is rotating relative to that point.

This is right, by the way. Sorry I didn't respond. When I first read it, I *facepalmed* at how stupid it was of me not to see that, but then I thought that a particle moving in a straight line (constant velocity, not free-fall) would make the point just as easily. I actually came back here to try to prove it wrong, by saying that such a particle is still rotating about that point... but in the process I convinced myself that it's perfectly right. The angular momentum is r x p (vectors), or r*p*sin(theta). That sin(theta) ensures that no matter where the particle is relative to that point, the angular momentum is always equal to p*(the shortest distance between the origin and the line of motion of the particle).

So a particle with no forces on it will still be in rotational equilibrium with respect to an off-axis point, even though it's just traveling on a straight line. The angular momentum will be different depending on where the origin is set, but it will still be conserved unless there's a torque on it due to some external force.

 

[SEPKA]

I think I still need some clarification to #6.
When radioactive decay happen, the unstable nuclei breakdown into more stable nuclei, and emit alpha/beta/gamma ray.
Alpha and beta ray are He and electron respectively, so they all have a mass, but gamma ray is electromagnetic radiation does not.
So if I put a highly radioactive into a tight box with a few cm thick of aluminum, which will make sure that product of radioactive decay will remain inside, while alpha and beta ray cannot penetrate through, only gamma ray can escape.
According to the equation E=mc^2 and some other theory, energy from gamma ray are due to the process of a nuclide with low binding energy breakdown into the same nuclide (same A and Z) but with high binding energy: this different in energy caused the mass of the product to be less than the original.
So if I throw that box into space (ie. no external force) momentum should be conserved, this means that loss mass->gain velocity. But this just sound really weird to me that the object just suddenly gain velocity by emitting a massless ray.
So I want to clarify if this really happen, or if the theory is incomplete, or what?

 

[Architectonic]

Remember that the gamma ray also has its own momentum!

http://en.wikipedia.org/wiki/Photon

If the particles that are thrown off are trapped by the box, then the momentum of the box will stay the same. If the particles escape, then the momentum of the box will decrease as some has been transferred to the escaping particles.