Hold on
A particle is in an upright circular path of radius r. At what speed, u, must it be injected (at the bottom) in order that it can go right round the track without falling off?
Neglect friction, drag and variation of g with height. Assume that the mass slides (it doesn't roll).
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The time it takes to go round the loop involves non-analytic elliptic integrals. I can do it but only thanks to wolframalpha.com
Neglect friction, drag and variation of g with height. Assume that the mass slides (it doesn't roll).
-
The time it takes to go round the loop involves non-analytic elliptic integrals. I can do it but only thanks to wolframalpha.com
Labels: funphysics
posted by Chris at
4:44 PM
10 Comments:
I assume that the mass has to be on some sort of rail or track on the inside of the vertical circle, a bit like those fairground rides that hang you from the top. If the rail was on the outside of the circle the mass would simply fall off three quarters of the way around.
So, to avoid falling off at the top the centrifugal force at that point would have to at least equal g. Would that be enough for it to stay on track for the rest of the trip to the bottom?
I've done four of these puzzles this morning (for three firsts and a second). I have a life to live so I'll leave it to someone else to finish!
- Wizard of Oz
Hi, I did have "Wizard of Oz" in my mind when I said "Oz". It's only a tiny bit harder sign in with a nickname, rather than been Anonymous. (woz, zardoz....)
I haven't been clear enough. The mass is on the inside of the track, so the only thing keeping it on the upper portions of the loop is the centrifugal force. Any track is to stop it slipping "out of the page". You have understood the problem correctly. Greetz from pongo land.
Hey Chris, i was real surprised when i saw this question!!
It had come in a competitive examination i had taken on Sunday!!!!
I disnt know how to solve it then either :P
There are some clues in the "beetle on a sphere" that was posted 05 Aug 09.
I'll give a shot: u=Sqrt(g*r)
Hi t:b:H. over twice that.
Looks like I've posted one that takes more than 5 minutes - yahoo! :)
Hi Chris. Particle will not fall down at the top point if centrifugal force is not less than its weight. The first is equal to mv^2/r, where v is the velosity on the top. If u is bottom velosity, energy conservation gives an eq.
v^2=u^2-2gh, where h=2r. Force equlibrium gives an eq.
v^2/r=g, thus u=(5gr)^0.5. It's possible more than this too.
Hi quantense, you got it. Well done.
I probably won't publish my long-winded version as your explanation is clear enough. Greetz.
woot i guessed it right!!!!(in my exam)
Brag time. I first came across that in one of my wife's old textbooks. Never seen it before, did it in my head and a couple of variations on the theme.
I had to do it on a piece of paper this time :-(
Anonymous, that was a good guess :) I hope you did well.
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