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Monday, 06 October 2008

Physics Challenge

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Written by Michael /Nurta   
Tuesday, 12 February 2008

I couldn't sleep last night so my brain naturally went to physics.  I came up with a problem I couldn't get the answer to right away.  Over the course of the day I was able to come up with a solution.  Can you figure out this challenge?

 

Click "Read More" to see the problem...

You should all be familiar with the equation for centripetal acceleration:

a=V2/r

this turns into

dv/dt=v2/r

∫(r/V2)dv=∫dt

-r/V=t+C

V=-r/(t+C)

This of course when you consider, is nonsensical.  Velocity does not change over time in such a way.  In circular motion it changes direction, but not magnitude.  So where is the error?

 

I'll post the answer tomorrow!  Put your solution in the comments or take it to the forums!  To post in the forums you need an account, but if you make an account, you get full access to the site!

 

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Charles - Math Issues   | 69.243.80.234 | 2008-02-23 03:10:50
[color=white][/ color][size=medium][
/size]The problem I see is that you have (-r/V=(t+C))=(V=-r/(t+C)). This cannot be true, as you have divided both sides by -r, meaning you should have V=(t+C)/(-r). Outside of mathematical mistakes, I'm sure someone else will catch the fundamental physical mistakes before I do.
Daniel   | 68.55.74.66 | 2008-02-23 12:33:54
Charles, the math is correct. The last step involves multiplying both sides by V/(t+C).
The physical problem is that centripetal acceleration and tangential velocity are vectors and, ***uming constant tangential speed, are always orthogonal. For the purposes of calculating magnitudes, this equation is useful, but performing calculus with it is kind of an abuse, since this isn't actually a linear system.
moose007   | Author | 2008-04-02 06:57:14
I now hate centripital force. Congrates.
Nurta   | Super Administrator | 2008-04-02 17:38:35
Heh, I've hated centripetal force for some time now...
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