Dropping a penny off a skyscraper.

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SUMMARY

The discussion centers on the physics of a penny dropped from a skyscraper, specifically calculating its terminal speed, time to reach that speed, and the force exerted upon impact. The terminal speed is calculated to be approximately 11.067 m/s using the formula V(terminal) = √((2mg)/(air density * surface area * drag coefficient)), with parameters including a mass of 0.0025 kg, air density of 1.204 kg/m³, surface area of 0.000284 m², and a drag coefficient of 1.17. The time to reach terminal speed is estimated at around 11 seconds, and the pressure exerted by the penny upon impact is determined to be insufficient to break skin, which requires a pressure of 460 psi.

PREREQUISITES
  • Understanding of terminal velocity calculations
  • Familiarity with drag force equations
  • Knowledge of pressure calculations (P=F/A)
  • Basic principles of conservation of energy
NEXT STEPS
  • Study the effects of drag coefficients on falling objects
  • Learn about the impulse-momentum relationship for impact force calculations
  • Explore the physics of terminal velocity in different mediums
  • Investigate the elasticity of human skin and its impact on pressure thresholds
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Physics students, educators, and anyone interested in the practical applications of physics principles related to falling objects and impact forces.

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Homework Statement



A penny is dropped from a skyscraper.

a) Determine the terminal speed

b) How long will it take to reach terminal speed.

c) Will it exert enough force at terminal speed to break skin (460 psi is needed).

Homework Equations



a) F(air drag)=(1/2)(air density)v^2(surface area)(drag coefficient)

F= mg

c) conservation of energy?--P=F/A→Fd=(1/2)mv^2→P=(mv^2)/(2Ad)


The Attempt at a Solution



a) V(terminal)= √((2mg)/(air density*surface area*drag coefficient))=11.067 m/s

m= 0.0025kg air density=1.204 kg/(m^3) surface area= 0.000284m^2 drag coe=1.17

b) I completed an iterative process on excel and came up with t ≈11s d ≈122m

c) When I used the energy approach I get a pressure lower than that of when the penny is just laying on your skin. I'm not sure what went wrong. Oh and I'm assuming the penny is falling flat the whole way.
 
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btpolk said:
c) conservation of energy?--P=F/A→Fd=(1/2)mv^2→P=(mv^2)/(2Ad)

How exactly did you figure out what d is here? I can't see how you'd have any way of knowing that.
 
Product of terminal velocity and time it takes to reach terminal velocity.
 
how thick is the skin?
 
460 psi is needed to break the skin.
 
btpolk said:
Product of terminal velocity and time it takes to reach terminal velocity.

This doesn't make too much sense to me. In this equation, "d" is the distance over which the work is done. Therefore, it is the distance over which the penny decelerates. In other words, it is the distance traveled between the moment of impact and the moment of coming to rest.

So what you really need to know is how far the skin is depressed. I imagine this depends on the elasticity of the skin. I really don't know how you are supposed to estimate it.
 
Of course! Simple mistake. Thank you cepheid for your help!

My professor said I would have to make a number of assumptions in this problem, and I imagine the location of the point of contact on the body is one of them.
 
You can do it this way. Take the given P (remembering to convert the units), and calculate d? What do you get?
 
If I do that I get 1.7e-4 m, but how does that help me find how much the penny is sinking in the skin?
 
  • #10
btpolk said:
If I do that I get 1.7e-4 m, but how does that help me find how much the penny is sinking in the skin?

Well, I think what Dickfore might have been getting at is that you assume the same amount of depression for the case of the falling coin. This is kind of an upper limit on the amount of depression, and hence solving at least gives you a lower limit on the amount of force (and pressure).
 
  • #11
If the falling penny hits you in the say the palm of your hand, you'd just feel a small sting, no puncture of the skin. I suppose if it hit you on your hard (bald!) head, it might cause a small skin puncture, although even that is questionable. I don't know how you would run the numbers without more data.

Now if a bullet was dropped, all bets are off, because of its high terminal velocity and small contact area.
 
  • #13
I'm not sure I understand. So can I use this to prove that the penny can't exert that much force?
 
  • #14
PhanthomJay said:
If the falling penny hits you in the say the palm of your hand, you'd just feel a small sting, no puncture of the skin. I suppose if it hit you on your hard (bald!) head, it might cause a small skin puncture, although even that is questionable. I don't know how you would run the numbers without more data.

Now if a bullet was dropped, all bets are off, because of its high terminal velocity and small contact area.

I do realize that it won't do a lot of damage, if any, but I do have to prove it mathematically.
 
  • #15
cepheid said:
You might also be interested in listening to the very last story at the bottom of the page I'm linking to below :wink:

http://www.cbc.ca/quirks/episode/2012/03/24/march-24-2012/
Thanks, that's good news. Some 50 years ago when i was a boy, barely tall, I chucked a penny, when my parents weren't looking, through the railings of that majestic building's 86th floor outdoor observation deck. I didn't know a hoot about Physics, but somehow I surmised that it would do no harm. The years came later when I became a man, and i wondered...oh how I wondered ...if I had hurt anyone that day. Now, I can rest. :biggrin:

btpolk: You say you can make assumptions, so assume the penny falls flat on a person's noggin', and it decelerates to a stop in a mere .001 second. That will allow you to calculate the average impact force, using the impulse-momentum relationship, from which you can calculate the pressure by dividing the impact force by the surface area of the penny, and compare it to the 450psi allowable. You should find that the pressure is quite small, even if he impact time is smaller than the asumed value.
 

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