| Topic: |
Science > Physics |
| User: |
"Fallingeagle" |
| Date: |
26 Sep 2006 01:38:14 PM |
| Object: |
motion in a loop |
Hi. Is there a smart person out there who can help me?
during motion at the inside of a vertical cirle, you will be able to
"hang on" to the sircle, even if you are up-side down. (if
a(centripetal) is big enough)
I don't get this...
of cource I get that: N+G = F (total) , and that, you will have N if F
(total) > G
What I don't get is why the Normal force (N) keep you "attatched" to
the sircle, since it act down, not up....
this is probably the answer to another question: why do you get pulled
out of the sircle when you are travelling along a sircular road. since
the forces act inwards,
thank you
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| User: "CWatters" |
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| Title: Re: motion in a loop |
27 Sep 2006 05:48:24 AM |
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"Fallingeagle" <falling_eagle1@hotmail.com> wrote in message
news:1159295894.082953.254120@e3g2000cwe.googlegroups.com...
this is probably the answer to another question: why do you get pulled
out of the sircle when you are travelling along a sircular road. since
the forces act inwards,
The wheels on your car accelerate it towards the center of the curve. The
force required to do that produces an equal and opposite reaction pushing
the car away from the center.
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| User: "PD" |
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| Title: Re: motion in a loop |
26 Sep 2006 05:06:57 PM |
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Fallingeagle wrote:
Hi. Is there a smart person out there who can help me?
during motion at the inside of a vertical cirle, you will be able to
"hang on" to the sircle, even if you are up-side down. (if
a(centripetal) is big enough)
I don't get this...
of cource I get that: N+G = F (total) , and that, you will have N if F
(total) > G
What I don't get is why the Normal force (N) keep you "attatched" to
the sircle, since it act down, not up....
this is probably the answer to another question: why do you get pulled
out of the sircle when you are travelling along a sircular road. since
the forces act inwards,
Look at it this way. The normal force the surface of the circle can
only PUSH you downward, it can't pull up at all. Now when you're at the
top, there are *two* forces acting down on you: the normal force and
gravity. No matter how the normal force changes, gravity isn't going to
change. That force is *always* going to be there, and as a result of
this force, you're going to accelerate downwards AT LEAST 9.8 m/s^2. If
there is a nonzero normal force also, then you will accelerate downward
faster than that.
Now, in order for you to stay on the circle, you must have *exactly*
the following acceleration: a = v^2/r.
If v is so small that the acceleration from this equation is less than
9.8 m/s^2, then there is no hope of staying on the circle, because that
gravity is going to force the acceleration to be at least that big, and
there is nothing the normal force can do to counter it.
However, if v is large enough that v^2/r is *bigger* than 9.8 m/s^2,
then the normal force can help, because it can add to the force of
gravity, and the two forces together can be big enough to produce that
acceleration.
PD
thank you
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| User: "Timo A. Nieminen" |
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| Title: Re: motion in a loop |
26 Sep 2006 04:00:16 PM |
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On Wed, 26 Sep 2006, Fallingeagle wrote:
during motion at the inside of a vertical cirle, you will be able to
"hang on" to the sircle, even if you are up-side down. (if
a(centripetal) is big enough)
I don't get this...
of cource I get that: N+G = F (total) , and that, you will have N if F
(total) > G
What I don't get is why the Normal force (N) keep you "attatched" to
the sircle, since it act down, not up....
Take gravity out of the picture. Consider motion on the inside of a
circular strip out in deep space. If the car isn't moving, there is no
force acting on either the car or the circle. When the car is moving, it's
moving in a circle, and therefore is accelerating at a=v^2/r toward the
centre of the circle. So, the circle is pushing the car towards the centre
of the circle with F=mv^2/r.
Since the push is towards the centre, why does the car "stick" to the
circular loop? Looks like a quite sensible question!
Simplify more. Don't worry about circular motion. Consider an astronaut in
space next to a flat rocket-powered platform. If they're just floating
there in space, there is no force acting on either the astronaut or the
platform. If the platform accelerates towards the astronaut, with the
astronaut standing on it, it will push the astronaut along with it. Draw a
diagram of the forces - the platform pushing on the astronaut and the
astronaut pushing on the platform - and it will look the same as the
diagram you would draw for the same astronaut standing on the surface of
the Earth. It feels the same to the astronaut, and has the same effect -
the astronaut is "pulled" towards the surface.
Think about what happens when you sit in a car and it accelerates - you
get pushed back into the seat, even though you are accelerating forwards.
this is probably the answer to another question: why do you get pulled
out of the sircle when you are travelling along a sircular road. since
the forces act inwards,
Without any inwards force, the car would leave the road, just going in a
straight line. Without any force acting on you, you would do likewise. The
required force is exerted on you by the seat, which doesn't exert this on
the upper part of your body. So your upper body tries to keep going in a
straight line, but is pulled along by the rest.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Fallingeagle" |
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| Title: Re: motion in a loop |
26 Sep 2006 03:41:17 PM |
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this is easy if you know it, and someone must know it since this is a
physics message board...?? please help.
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