Dropping a penny off a skyscraper.

  • Thread starter Thread starter btpolk
  • Start date Start date
AI Thread Summary
A penny dropped from a skyscraper reaches a terminal speed of approximately 11.067 m/s, taking about 11 seconds to do so. The pressure exerted by the penny upon impact is calculated to be lower than the 460 psi required to break skin, indicating minimal risk of injury. Discussions highlight the importance of assumptions regarding the penny's orientation and the skin's elasticity in determining impact effects. The penny's impact is unlikely to cause significant harm, as it would only produce a small sting upon hitting soft skin. Overall, the mathematical analysis supports the conclusion that a falling penny poses little danger to a person.
btpolk
Messages
38
Reaction score
0

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.
 
Last edited:
Physics news on Phys.org
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.
 
Back
Top