Physics Q&A Game: Calculate Minimum Power for Man-Powered Helicopter

[jdavel]

Berislav,

This may be more of a guess than an "elaboration". As the bricks heat up in the sun water held in the bricks evaporates. The bricks can get much hotter than the air temperature, so in order for this vapor to leave the bricks, the air outside must have low relative humidity. So it needs to be a hot, dry day. Now there are empty spaces for water to enter, displacing the air through bubbles.

Just in case that was confusing, I'm not saying the bricks get hot enough to boil water (you said that doesn't happen) it's just that the vapor pressure of the water is very high inside the bricks.

?

Edit: Berislav, now I'm seeing that you said surface tension plays a role. I haven't figured that out yet, so I guess the game is still on.

 

[hemmul]

Edit: Berislav, now I'm seeing that you said surface tension plays a role. I haven't figured that out yet, so I guess the game is still on.

I believe the surface tension is responsible for the production of sound: when the bubble bursts the time during which the surface water will come together into single point depends on the radius of the bubble and surface tension (through the accelleration it provides).
The frequency of plop is inversely proportional to this time. And the plops from the vast number of bubbles produces hissing sound :) (like that in a hot cup of well-made coffee)

jdavel, "standing on the shoulders" and not falling down - is much more difficult than standing on the ground! YOU definitely do deserve the next turn :)

 

[Berislav]

Hemmul, you explained why we hear hissing instead of a 'plop'.

Jdavel, you are very close to the solution. If you want additional hints regarding surface tension, feel free to ask.

 

[Gokul43201]

jdavel, it is now your turn to ask a new question.

Berislav, feel free to post any additional details in the solution to your question. We shall consider the question answered for the purposes of the continuity of the game.

If jdavel does not come up with a question in 12 hrs, hemmul may go next. If neither asks a question within 24 hrs, I will.

 

[Berislav]

Very well.
The bricks have a porous structure. The pores can be viewed as interconnected by cavities. At the moment of impact when rain drops fall on a brick surface tension pulls the water down into the pores. This "pulling force" will be greater for water bubble on a smaller pore than on a larger one. Since there are more smaller pores than larger ones the air in the cavity is displaced until the large bubbles burst.

 

[jdavel]

"jdavel, it is now your turn to ask a new question."

How about this? Two 20-year-old twins live on the earth. One travels to Alpha Centauri and back at 90% the speed of light...JUST KIDDING!

Ok, here's my real question; it requires no calculation, and only high school level physics should be needed.

An astronaut is floating weightless in space when two wrist watches come drifting by. He reaches out and grabs them, sets them in front of him carefully so they don't drift away, and then observes them for a while. The first thing he notices is that they both have the same inscription on their faces, "Zurich, 1905". After watching them for a while, he notices they keep identical time, so he concludes (correctly) that these are identical watches. Then he puts one on.

Question: Do the identical watches continue to keep the same time while he is wearing one but not the other?. If so, explain. If not, explain.

 

[Crosson]

Question: Do the identical watches continue to keep the same time while he is wearing one but not the other?. If so, explain. If not, explain.

There are two effects, both of which are gravitational and minuscule, which effect the keeping of time in this situation.

Relative to the astronaut, the watch he is not wearing begins to "speed up", due to gravitational time dilation. The gravitational field is obviously stronger for the watch he is wearing.

Also, the gravitational field of the astronaut will effect the mechanical motion of the watches internal parts. Depending on the orientation of the astronaught to the watch he is wearing, this will either speed up of slow down the watch.

Despite all I have said, I conjecture that the clocks keep the same time to within one Planck time over the life of the universe (10^-52 seconds per year) under these vanishingly small gravitational differences. Therefore, the answer is yes, they continue to keep the same time.

 

[jdavel]

Crosson,

Your analysis is correct; there would be no way for the astronaut to detect the effect of his own gravitational field on how a clock on his wrist and one right in front of him keep time. I meant to convey that this is a problem in classical physics by saying that "only high school level physics should be needed". Sorry about the confusion.

That said, while your analysis was correct, it wasn't the right one and so your answer (they keep the same time) isn't necessarily correct.

 

[hemmul]

That said, while your analysis was correct, it wasn't the right one and so your answer 9they keep the same time) isn't necessarily correct.

getting back to high school physics course (luckily it is not that far!) the answer is no, however there are several points: how do we compare the time on two clocks? to be able to compare them we have to bring them to one and the same of space-time. Now, after the astronaut gets the watch on his hand his, and the other watch's world lines are different. Any watch shows its proper time.
Now, as the remained watch continues its inertial motion, let's consider the situation from it's RF. It is clear, that the astronaut is moving in the space (with some accelleration, searches the Earth, searches his spacecraft etc...). So his world line is curved, with respect to the linear line of the traveling watch. that's why when the astronaut stackles upon the traveling watch again - he will see the difference (his watch will show that less time passed)!

This is only in the case if the traveling watch is traveling at a constant velocity e.i. we can bind a single Inertial Reference Frame to it.

If no - (the other anstronaut grabs the remained watch) - then we have to integrate the proper time along the world lines for both astronauts, that is far beyond the high school level ;)

 

[jdavel]

hemmul,

You sure got further in high school physics than I did!

But again, to clarify, this is a problem in classical physics. That is, you can assume Galilean/Newtonian relativity (c is infinite) and you can assume zero uncertainty (h = 0).

 

[Gokul43201]

jdavel,

Could you clarify/describe what you mean by "then he puts one on" ? I'm imagining an astronaut in a bulky space suit. How on Earth (or in space) does he put on a wristwatch ? If this is an irrelevant detail, you may say so. But if there is any form of energy transfer between objects, this may be necessary to better estimate the sizes of effects.

Also, does it matter how the watch works, or is the the answer independent of that too ?

 

[jdavel]

Gokul,

How he gets the watch on is not important.

How the watch works IS important.

 

[hemmul]

one healthy astronaut's body-temperature is around 36.6° C. when he puts the watch on - there is certain heat transfer between the two. As the responce to the heating is different for different parts of the mechanism inside the watch - the relative size of the elements changes - so the watch becomes disbalanced :)
well, it is crazy idea, because as far as i remember the clocks produced in 1905 in Zurich were highly accurate ;)
The other possiblity is that the astronaut waves his hands, and the constant random accellerations can lead to some errors in the mechanism, but the average effect of those is zero... so getting back to the change in the conditions for the two clocks, the only parameter that is changed, AFAIK is temperature (thermal expansion)

 

[jdavel]

I'm beginning to think that my attempt to wrap a story around this problem has left the physics too obscure and wide open to interpretations other than the one I had in mind. So I'm going to give a hint (actually answer part of the question for you) to redirect and focus your thinking where I intended.

The watches don't keep the same time. The cause (at least the one I'm looking for) is purely mechanical, having to do with the way that watches made in 1905 (or in any year before about 1960) worked. Also, discovering the cause leads unambiguously to whether the watch the astronaut wears runs faster or slower.

So now the question is: Does putting the watch on cause it to speed up or to slow down, and why?

 

[hemmul]

The watches don't keep the same time. The cause (at least the one I'm looking for) is purely mechanical, having to do with the way that watches made in 1905 (or in any year before about 1960) worked. Also, discovering the cause leads unambiguously to whether the watch the astronaut wears runs faster or slower.

So now the question is: Does putting the watch on cause it to speed up or to slow down, and why?

Well, well, well :) the question turns to be more and more interesting :)

Actually, by the end of IXX's century they understood that the dust was dangerous for the watch. There were several attempts to prevent it from entering the mechanism - like making additional covers etc, but unfortunately they didn't lead to full isolation and impermeability of the watch...

in 1926 the first licenses for water-resistant cases began to appear, while in 1946 the first impermeable hand-watch - Rolex - was born!

Later, in 1969, the fully water-resistant watch and their "brothers" were very popular. and very useful! So popular, that Neil Armstrong and Edwin Eldrine had the highly-accurate Omega Speedmaster on their hands while walking on the moon :)

Well, now back in 1905 - the inner mechanism of the watch was not protected from the outer world.
While there was no danger for the watch floating in space, the one placed on the hand of an astronaut will definitely live in the environment where there is air, water dust etc... i hope it's clear why that should be the reason for different functioning ;)

 

[Gokul43201]

... i hope it's clear why that should be the reason for different functioning ;)

But jdavel makes it clear the you should be able to determine whther the watch speeds up or slows down.

 

[hemmul]

But jdavel makes it clear the you should be able to determine whther the watch speeds up or slows down.

excuse me, note taken!
IMHO it should run slower - u know friction, additional inertia due to the dust particles etc... to say the truth i was unable to find the charts/plots/diagrams of the mechanisms of hand watches from the beginning of XX'th century, to make further assumptions :) hope anyone else will succeed

 

[jdavel]

hemmul,

I bet you're kicking yourself for letting me go next!

The answer isn't messy (like particles getting in or not). It's real physicsy (if that's a word), and (at least I think) pretty cool!

In fact (and this is a hint) if you knew the mass of the astronaut and the mass and dimensions of all the watch components, you could calculate quantitatively how much the watch worn by the astronaut will run slow or fast. That's a BIG hint!

 

[Davorak]

Ok I had to do some research,
http://elginwatches.org/technical/adjustments.html
http://rustyrobin.com/ElginHistory/ElginHistory.htm

The watch will get heated up through conduction from the astronaut. The balance wheel in the watch will expand causing the moment of inertia to go up and the watch to slow down.

$$\omega = \sqrt{\frac{m g L}{ I}}=\sqrt{\frac{g}{L}}$$

Right?

After ~1920 "invar" and "elinvar" metals were introduced witch expand less when heated. Before that the balance wheel had adjustable screw to adjust it for different temperatures.

edit:
On rereading the article, there is another reason why the warm clock would run slower:
"The colder the watch gets, the stiffer the hairspring gets and the quicker the watch runs."

In fact the effect mentioned first does not make a noticable change.

 

[hemmul]

hemmul,
I bet you're kicking yourself for letting me go next!

hehe :) why so? its pleasure to think on non-standart question, and I'm happy its your turn ;)

The answer isn't messy (like particles getting in or not). It's real physicsy (if that's a word), and (at least I think) pretty cool!

In fact (and this is a hint) if you knew the mass of the astronaut and the mass and dimensions of all the watch components, you could calculate quantitatively how much the watch worn by the astronaut will run slow or fast. That's a BIG hint!

hmmm... i shall think on it...

 

[Davorak]

Am I right jdavel? I think I added in the edit after you saw my post. I would PM you but you do not seem to have that feature enabled.

 

[jdavel]

Davorak,

Well, it' not the solution I had in mind!

In my post #53 (written at almost the same time you were writing you solution) I gave a hint by saying that if you knew the dimensions and mass of the watch components and the astronaut you could get a quantitative solution to how slow the watch worn by the astronaut will run. That wouldn't be the case with your termperature change solution, since you have no info about relevant temperatures.

Hint: the solution is based purely on Newtonian mechanics.

 

[Davorak]

Well he did find them floating in space and still in working condition that places some restraint on the temperature. I guess I was still going off the original "no calculations necessary."

I gave a hint by saying that if you knew the dimensions and mass of the watch components and the astronaut you could get a quantitative solution

means of interaction:
electromagnetic "touch effects" or physical contact
electromagnetic radiation.
weak
strong
gravity

With your last post you have ruled out temperature, and therefore most the electromagnetic radiation interaction through inferred radiation.

The weak and strong forces can be ignored.

This only leaves gravity in my mind andelectromagnetic "touch effects". Your hint implies this by including mass as an important factor.

But Crossman’s answer has already been ruled out so no GR, SR. This leaves Newtonian gravity. Now Newtonian gravity with 90 kilo astronaut and 1 kilo watch at 0.5 meters would give 2.40228 × 10-08 Newton. Too small to have a large effect. If the gravitational force was much larger I could see it slowing the watch down through friction, but no friction was mentioned and you have implied that we could do a quantitative analysis just know the mass and the construction.

So effectively gravity is eliminated, unless I am missing something.

Now the Astronaut moved the watch as well witch could cause the watch to run faster or slower if it is an oddly made watch. Ok I know I am stretching here. It is probably something simple which has slipped every one as of yet.

None of the above answers are satisfying. So I will go out on a limb and say that these early clock have no anti recoil action on the hair spring. In essence as the watch ticks and the hair spring unwinds the watch will spin do to conservation of angular momentum. When the watch is attached to the astronaut the astronaut becomes part of the initial system.
Iastronaut>>Iwatch
The watch will not rotate because of the hairspring unwinding or watch hands moving. Thus the hair spring will act with more effective force on the rest of the watch making it run faster then the watch in space.
I guess this would fall under electromagnetic touch effects.

So I am stumped and no one else seems to have come up with anything I think it is time for you to set a timetable to reveal the answer even if no one gets it by then.

edit:
In addition many old clock can not run when weightless. So when put on by the asronaut the watch starts to work while before it was not working at all. This is a more extrem example of the last example above.

 

[jdavel]

Davorak,

Your analysis based on conservation of angular momentum is on the right track (or at least the track I had intended). But if I understand your conclusion, it's that the watch worn by the astronaut runs faster than the one left floating. I'm quite sure it's the other way around.

But since you've cracked through all my misdirection and obfuscation (only some of which was intentional!) maybe I should summarize the physics by simplifying my problem statement.

In the absense of any outside force, a mechanical watch will run slower when it is attached to another object (something with non-zero mass and volume) than when it is not attached to another object.

Hint: Mechanical watches have (at least) two springs. The drive spring provides the energy to move the hands while the balance spring is part of the system that regulates the average angular speed of the hands. The function of the drive spring is not affected by the watch being attached to another object.

 

[Davorak]

Mind posting some links to some pictures, it would help with understanding of the watches workings which seems necessary to solve the problem. I have found:
http://www.timezone.com/library/horologium/horologium631673198118416858
http://www.timezone.com/library/archives/archives631703030492773204

Here is how I came to the conclusion that the watch would be worn by the astronaut would run faster then the free floating watch.

To my limited understanding of mechanical watches, the hair spring, some times called the balance spring is attached to the frame of the watch. This is an assumption on my part since I could not find the specific information any where.

When the hair spring unwinds it provides a force on both ends of the spring and through fore a torque on the balance wheel and the frame of the watch.

This provides equal angular momentum to both the astronaut and the balance wheel, however a greater amount of energy is transferred to the balance wheel when compared to the astronaut since it has a greater freedom to move.

When a hair spring gets cold it will cause the clock to run faster, because it gets stiffer.
Since the frame of the watch is to be held still the hairspring can provide more force to the balance wheel, thus causing the watch to run faster.

That is how I thought about it. Ready to share why you think it would move slower?
I don’t know how to include the watches isochronic Adjustments into the picture. Before 1850 these adjustments apparently did not help much and were greatly improved after 1850. Any isochronic adjustment should simply reduce the time dilation between the watches.

 

[jdavel]

Davorak,

Nice pictures you found; ain't the www grand!?

Since there doesn't seem to be a whole lot of interest left in this question anymore, I think I'll declare you the winner, and give my answer.

My thinking goes like this. The balance wheel system is a simple harmonic oscilator, consisting of the hair spring, the balance wheel and either 1) the rest of the watch or 2) the rest of the watch plus the astronaut.

In short, the greater the mass of a SHO the lower the frequency. So with the astronaut included, the watch runs slow.

In more depth, for a SHO with two masses (connected by a spring) if m1 >> m2, then the end of the spring attached to m1 will hardly move, so the frequency will be almost exactly (k/m2)^1/2. But if m1 = m2, it's the center of the spring that doesn't move. So each mass oscillates as if it were attached to a spring that's half the length of the original spring. But when you cut a spring, k, in half, each piece has a spring constant of 2k. So when m1=m2 the frequency is (2k/m2)^1/2. So as m1 increases from m1 = m2 to m1>>m2, the frequency decreases. Attaching the astronaut doesn't make much difference in the frequency because even without him, m1>>m2. But with him m1>>>m2. So the watch slows down a little.

Whatdya think?

Ok, Davorak, your turn.

 

[Gokul43201]

Nice one Davorak. It's your turn now. If you don't post a new question here by tomorrow, I'll jump in again with a question.

 

[Gokul43201]

NEW QUESTION : A billiard ball of radius R is hit with a cue such that it starts out with a velocity $v_0$ and a backspin rate of $\omega _0$. Calculate the subsequent motion of the ball. What is required for the ball to return towards the cue ?

 

[siddharth]

Subsequent motion is sliding of the ball. That is, the ball slides on the ground with friction slowing down both the velcoity and angular velocity.

The condition fot the ball to come back must be that when v=rw, v=0.
There fore the relation between v0 and w0 is v0=(2/5)rwo.

 

[siddharth]

If v=rw when v=0, then the body won't come back at all. For any other value of w greater than this, then the sphere will undergo pure rolling as it comes back. Therefore it will then reach the inital point

Intially, a=f/m
there fore
v(t)=v0-(f/m)t
and
w(t)=w0-(5f/2mr)t

when v=0, t=(mv0/f)
therefore, t=(mv0)/f
at this time, w>0
wo-(5f/2mr)(mv0/f)>0
ie,
wo>(5v0/2r)

 

[Gokul43201]

That was quick ! Siddharth's reply is correct in all essential details.

For the sake of completeness, I shall paraphrase Siddharth's post with a few additional details thrown in.

Choose the positive direction of $\omega$, so that the velocity at the point of contact, $v_c$, is given by $v + R \omega$. Friction, $F = \mu mg$ acting backwards (only because of the backspin - this is important) at this point, is a constant. So, the motion of the center is given by

$$v = v_0 \mu gt$$ and the rotation about the center, by

$$\omega = \omega _0 -\mu Rmgt/I~~;~~~I = (2/5)mR^2$$
$$=>\omega = \omega _0 - (5 \mu g t/2R)$$

The ball is sliding forwards with a constant deceleration, and spinning backwards at a constant angular deceleration. Plugging into the first equation, the velocity of the point of contact evolves as :

$$v_c = v_0 + R\omega _0 -(7/2) \mu gt$$

When a ball is rolling without slipping, there is no relative motion between the point of contact and the floor (hence no kinetic friction on a rolling ball). So here, slipping stops and rolling begins at the instant when $v_c = 0$. Using this condition in the previous equation tells us when this happens.

$$t = 2(v_0 + R \omega_0)/7 \mu g = \tau~,~~say$$

The value of v at $t = \tau$ is given by $R\omega _0(5v_0 /R \omega_0 - 2)/7$. This gives the speed of the rolling ball after slipping has stopped. If $v_0/R\omega_0 < 2/5~$ , this value is negative and the ball will roll backwards towards the cue. If $v_0/R\omega_0 > 2/5~$ , the ball will roll forwards.

This makes sense. To get the ball to roll back it seems intuitive that you want large backspin, $\omega _0$ and small velocity $v_0$. The subsequesnt motion is that of a ball rolling at a uniform velocity (which in reality, decays due to rolling friction).

***********

Siddharth, your turn now to ask a question...

 

[siddharth]

The subsequesnt motion is that of a ball rolling at a uniform velocity (which in reality, decays due to rolling friction).

I thought the ball slows down because of the fact that when the ball is in contact with the ground, the surface of contact is deformed. As a result, there is an area of contact rather than a point of contact where the front part pushes the table more than the back part. As a result the normal does not pass through the centre and is shifted to the front. This Normal force produces a torque which slows the ball down.

Anyway, here's another question involving mechanics.

A ring of radius R is made from two semi circular rings of mass M1 and M2. The ring is then released from an inclined plane with angle of inclination of 30 degrees. If the ring rolls without slipping ,find
i) Angular Acceleration
ii) Normal Reaction
iii) Frictional force

Take M1=2kg, M2=4kg, and 2R=1m

 

[Gokul43201]

I thought the ball slows down because of the fact that when the ball is in contact with the ground, the surface of contact is deformed. As a result, there is an area of contact rather than a point of contact where the front part pushes the table more than the back part. As a result the normal does not pass through the centre and is shifted to the front. This Normal force produces a torque which slows the ball down.

This is exactly what constitutes rolling friction ! You will not have rolling friction without deformation.

I'll repeat your question (for clarity). The standing question is the following :

A ring of radius R is made from two semi circular rings of mass M1 and M2. The ring is then released from an inclined plane with angle of inclination of 30 degrees. If the ring rolls without slipping ,find
i) Angular Acceleration
ii) Normal Reaction
iii) Frictional force

Take M1=2kg, M2=4kg, and 2R=1m

 

[Stefan Udrea]

I don't know how to solve the problem, but I have a little observation :
Suppose A and B are the points where the 2 semi-circles join.
If,in the initial position,the ring is placed so that the tangential component of the gravity is ortogonal to AB,then the ring either slips or it doesn't move at all (if friction is high enough),in any case,it cannot rotate.

 

[Stefan Udrea]

With the notation above, if AB is on the horizontal axis of coordinates and the heavier semi-circle above the axis, the origin in the middle of the AB ,the center of mass of the ring has the coordinates :

X=0
$$Y= \pi \frac {R} {12} = \frac {\pi } {24} meters$$