Physics Q&A Game

  • #1
Gokul43201
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A Q&A game is simple: One person asks a relevant question (it can be research, calculation, off-the-top-of-the-head, anything as long as it is a physics question) and other people try to answer. The person who posts the first correct answer (as recognized by s/he who asked the question) gets to ask the next question, and so on.

I'll get this rolling with a simple back-of-the-envelope calculation question.

It is proposed to make a man powered helicopter, with a rotor 10 m in diameter. Assuming that the rotor blows a cylindrical column of air uniformly downwards, the cylinder diameter being the same as the rotor diameter, and the weight of the man plus machine is 200 kg, calculate the minimum mechanical power (in watts) that the man must generate, if he is to remain airborne. (take density of air ~ 1.2 kg/m3) Is the system feasible ?

(show, in a couple of lines, the essential steps in the calculation)

If a sufficient time passes with no correct answer, then the best attempt determines who goes next.
 
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Answers and Replies

  • #2
Galileo
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Great initiative Gokul!

Here's my attempt. Call the radius of the rotor r.
The power generated is the kinetic energy transferred to the air per second.
The mass of the air column is the density of the air times it's volume, so

[tex]K=\frac{1}{2}Mv^2=\frac{1}{2}\rho \pi r^2 vt v^2[/tex]
[tex]P=\frac{dK}{dt}=\frac{r^2\pi}{2}\rho v^3[/tex]

The momentum imparted to the air is:

[tex]P_m=\rho \pi r^2 v^2t[/tex]
so the force that keeps the copter floating and should equal mg is:
[tex]\rho \pi r^2 v^2=mg[/tex]
which we can solve for the speed of the air particles.
We find for the power:

[tex]P=\frac{r^2\pi}{2}\rho \left(\frac{mg}{r^2\rho \pi}\right)^{3/2}[/tex]
Plugging in the numbers, we find about 4,5 kW.
I had to look up some power consumption data. If a human runs at 24 km/h, the power consumption is about 1.7 kW. So it's not a feasible construction (not for me at least).
 
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  • #3
Gokul43201
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Nice one Galileo !!

World class sprinters and trained athletes can deliver close to 5 kW for no more than a minute. I certainly can not sustain more than about a kW for any reasonable length of time.

Your turn...
 
  • #4
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Gokul and Galileo,

Notice though that the required power goes as 1/r. This makes sense because you're getting your momentum with more mass and less speed, and that reduces the energy which goes as speed squared.

So maybe with 50m rotors (made out something with a very high strength/weight ratio) you could do it. Of course your ideal model of the air column moving down monolithically will probably fail at low speed. Oh, well!

PS Gokul, be aware that your Q&A game could be abused by students wanting a quick solution to a homework without doing any work themselves! Come to think of it, maybe that's what you were doing. :wink:
 
  • #5
Gokul43201
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jdavel said:
Gokul and Galileo,

Notice though that the required power goes as 1/r. This makes sense because you're getting your momentum with more mass and less speed, and that reduces the energy which goes as speed squared.

So maybe with 50m rotors (made out something with a very high strength/weight ratio) you could do it. Of course your ideal model of the air column moving down monolithically will probably fail at low speed. Oh, well!

PS Gokul, be aware that your Q&A game could be abused by students wanting a quick solution to a homework without doing any work themselves! Come to think of it, maybe that's what you were doing. :wink:
Point taken, on the abuse possibility (I'll not comment on the snide insinuation just yet). I'll leave it to the mentors to make that judgement.

As for the 50 m blades : you find me the material, and we're in business ! :biggrin:
 
  • #6
Galileo
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I think homework questions posted here are easily detected. Such are usually not the kind of questions you want to post here.

As a result, it took me a little time to think up a nice question (that I have to solve myself), so I actually should have taken that into consideration before posting a solution :redface: , but anyhoo:

There's an astronaut going for a space walk and enjoying the awesome sight (who wouldn't).
Just when he's ready to return to his spaceship he notices the line that connected him to the spaceship has broken!! This line was supposed to bring him back aboard the ship. The astronaut has only 5 minutes of oxygen left in his tank and he fears for his life. If something can be done, it has to be done quickly.
The astronaut realizes he has a small fire extinguisher on his spacesuit which he can use as a thrustrocket. He immediately grabs the extinguisher and wants to start spraying.

The astronaut is now 120 m away from his spaceship and has a neglegible velocity with respect to the ship. His mass (with suit, but without extinguisher) is 94.5 kg. De mass of the contents of the extinguisher is 5.0 kg and the mass of the empty container is 5.5 kg. The contents are sprayed outwards with a speed of 10 m/s, regardless of the remaining content. After 2 minutes of spraying, the container will be empty.

Will the astronaut arrive at the spaceship before his oxygen supply runs out?
Can he save his hide from this perilous predicament? Stay tuned!
 
  • #7
Gokul43201
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I think it may be important to specify that the extinguisher is attached to the spacesuit, and can not be removed. Else, there's an easy cheat.
 
  • #8
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[tex]v_f = v_i + v_{ex}*ln(m_i/m_f)[/tex]

[tex]v_f = 10ln(\frac{105}{100}) = 0.487m/s[/tex]

At t = 120s he will have travelled 120x0.487= 58.55m

At t = 600s when his gas is empty he will have travelled 600x0.487 = 292m.

He makes it with time to stop for donuts.
 
  • #9
Galileo
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whozum said:
[tex]v_f = v_i + v_{ex}*ln(m_i/m_f)[/tex]

[tex]v_f = 10ln(\frac{105}{100}) = 0.487m/s[/tex]

At t = 120s he will have travelled 120x0.487= 58.55m

At t = 600s when his gas is empty he will have travelled 600x0.487 = 292m.

He makes it with time to stop for donuts.
I`m sorry, your [itex]v_f[/itex] is roughly equal to the final velocity of the astronaut, so you cannot use the distance formula for constant velocity.
Also, there is 5 minutes of oxygen left, not 10.
 
  • #10
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I have an idea for the solution:

From the law of conservation of momentum

[tex]m_{e}v_e=(m_a+m_c)v [/tex]

Deriving with respect to t we find:

[tex] \frac{d(m_e v_e)}{d t}=m_aa+\frac{dm_c}{d t}v+m_ca [/tex]

We get an ODE which is easy to solve, since we know that

[tex]\frac{dm_e}{d t}=-\frac{dm_c}{d t}=5 kg/120s[/tex]
[tex] v_e=10 m/s[/tex]

The total time required to transverse distance s is

[tex]t=\frac{s-\int_{0}^{2min} v(t)dt}{v(120s)}+120s [/tex]

P.S.
Sorry I didn't do any numerical calculations. My MS Calculator crashes everytime I try to start it, and I don't know where my pocket calculator is. :frown:

P.P.S.
I read your question again and made the necessary correction.
 
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  • #11
Galileo
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Good try Berislav.

I can make the calculations for you if you tell me what numbers to plug in. Since you haven't got a formula for v(t) I can't solve your equation for t yet.
 
  • #12
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For [itex] v(t) [/itex] I get:

[tex]v(t)=\frac{5/120v_et}{-5/120t+10.5+m_a}[/tex],

since [tex]m_c(t)=10.5 kg-5/120t [/tex]
 
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  • #13
Galileo
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Berislav said:
For [itex] v(t) [/itex] I get:

[tex]v(t)=\frac{5/120v_et}{-5/120t+10.5+m_a}[/tex],

since [tex]m_c(t)=10.5 kg-5/120t [/tex]
Ok, well you'll have to integrate it yourself. There are some internet tools to help you with it if your require it: http://integrals.wolfram.com/
Online calculators also exist, just goooogle it up.
 
  • #14
Galileo
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Anyone?

Too hard? Too easy? I`ll start posting mild hints tonight (on my clock) if no one answers.
 
  • #15
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Wouldn't average velocity for the first two minutes work?

[tex] v_avg = 0+0.487 / 2 = 0.2435/s [/tex]

[tex]0.2435m/s * 120s = 29.22m plus[/tex]

[tex]180 * 0.487 = 87.66m[/tex]

= 116.88m. I'm not sure though, my algebra based physics skills are no match for you guys.
 
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  • #16
Gokul43201
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Galileo said:
Anyone?

Too hard? Too easy? I`ll start posting mild hints tonight (on my clock) if no one answers.
Can I suggest that you give it about 24 hrs before releasing hints ?
 
  • #17
Chronos
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Try using seconds as the time unit and see what you get. All the relevant constants defining work are so based.
 
  • #18
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Integrator gives me:

[tex]s(t)=\frac{5}{120}v_et \log(10.5+m_a-0,0416667t)[/tex]

EDIT:
No, scratch that! I meant at t=120s this is 100m. So,

[tex]t=\frac{s-\int_{0}^{2min} v(t)dt}{v(120s)}+120s [/tex]

is 160 s. So, he makes it. It seems that mulitplication and division elude me. :cry:

P.S.

Sorry I posted this so late. :frown:
 
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  • #19
Galileo
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Whozum. Taking the average velocity gives a good approximation, but from your anwer you cannot answer with certainty whether the astronaut makes it or not.

Berislav. You are close (i.e. on the right track), but there are some errors in your calculations.
(160s is off, it would mean he'd make it easily and I can guarantee the outcome will be a close call)

I wanted to post some hints at this time, since it's about two days after posting the problem, but at Gokul's request I'll wait a bit and post them in the morning (about 8:00 GMT) if no correct answer is in yet.
 
  • #20
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Can I waive my ability to solve this problem and request you message me the math needed to solve it?
 
  • #21
Galileo
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whozum said:
Can I waive my ability to solve this problem and request you message me the math needed to solve it?
That'd be a hint, wouldn't it? :biggrin:
You'll have to wait a little longer for that. :wink:

Don't forfeit your chance yet.
 
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  • #22
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I'm dying here. Atleast let me know why average velocity wouldn't work? As far as I can think the acceleration is pretty much constant, no?
 
  • #23
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Berislav. You are close (i.e. on the right track), but there are some errors in your calculations.
Yes, there are. And I think I found all of them. Apperantly, in Integrator c*x is not the same as cx.

I couldn't find any errors in my solution to the differential equation.

When I integrate v(t) from 0 to 120 s, I get 29.5121 m.

So t=300.9758 s. Close, indeed.
 
  • #24
Galileo
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Okay Berislav, you have basically solved the problem so you get to go next, although I would like you to give explicit answers to the actual question(s):
Does he make it before the oxygen supply runs out?
And: Can he (still) save himself (i.e. survive)?

Anyway, I wanted a question that was elementary on the surface, but not very easy. Also, I wanted one that could be solved by anyone, not just the 'elite'. You can answer this question using physical insight and very basic (Grade-K12) physics.
Notice the difficulty comes from the fact that the mass of the astronaut changes (the force acting on him is constant, but his acceleration isn't), but how big IS this effect? The guy starts out with 105 kg and ends with 100 kg, that's a small enough difference to consider the following: Can we solve it if we assume his mass is constant?
Ofcourse! It will be a simple linear acceleration problem. Moreover, if we assume his mass is and stays 100 kg, we can argue that he will arrive at the ship faster than in actuality. If he doesn't make it in this approximation, he surely won't make it in the real case. Similarly, if his mass stays 105 kg, you can use similar arguments.
Work it out yourself and you can easily estimate how long it will take the astronaut to reach the ship after his supply runs out. Then you can answer whether he can still save himself somehow (Gokul already hinted at an somewhat obvious idea).

In any case good job, Berislav. Your turn.
 
  • #25
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Does he make it before the oxygen supply runs out?
No, he spends about a second is outerspace without oxygen.

Can he (still) save himself (i.e. survive)?
Since he is a astronut he's supposed to be at least in fair physical condition. So, he can go without oxygen for at least 30 s before falling unconcious. This is enough time repressurize a chamber in any modern spacecraft. All in all, he survives, although not a very pleasant experience.

Very good question.

In any case good job, Berislav. Your turn.
Thank you. :smile:

In spirit of your idea, my question can be answered with physical insight alone (no mathematics needed). Here it is:

In summer after a hot and dry period raindrops falling on wall made of red brick make a hissing sound. Explain the hissing sound.

NOTE: I am not the author of this question. I will post the reference per request.
 

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