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Centripetal Force questions

  1. Jul 28, 2011 #1
    Hi, I got 2 questions concerning Centripetal force. I am thankful for every answer I get !

    1) Let's say a car is driving very fast in a loop with radius r on Earth where air resistance and friction doesn't exist.

    Let's say the car's engine stops and when the car gets to the top of the loop with unacceptable velocity v so that g>(v^2)/r. This will mean that the normal force stops existing and that gravity will make the car fall. But.. why? As long as v is big enough the gravity and normal force will provide the car with centripetal force..

    I realize due to common sense that the car will fall but I simply can't wrap my head around and answer why it'll fall.. other than simply saying the centripetal acceleration is smaller than g......

    2) Let's say the Earth's spinning velocity v increases to a speed so that (v^2)/r=0.5g

    This means that the measured gravity on earth would decrease by half... Again I can't really prove exactly why this happens.. I try to picture it in my head that in this instance half of the gravitational force G is "used up" trying to keep people from flying away from Earth..

    Is this correct? Can you tell me a more correct way of picturing this if not?
     
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  3. Jul 28, 2011 #2

    Doc Al

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    Well, is the speed big enough or what?
    It requires a minimum speed to maintain contact. Why? Maintaining contact means that there is some normal force between the road and the car. Go fast enough and gravity alone is not enough of a force to turn you in a circle, the road must push down; Go too slow and gravity is too strong a force--it pulls you away from the loop surface.

    Imagine you were standing on a scale. What would it read? It reads the normal force between you and the scale. If the earth's not rotating, that normal force equals the gravitational force of the earth since your acceleration is zero and the forces must balance. But if the earth is rotating quickly (and you're on the equator) you are centripetally accelerating. Thus there's a net force on you. The normal force (your apparent weight as measured by a scale) is reduced because of that acceleration. (Apply Newton's 2nd law.)
     
  4. Jul 28, 2011 #3
    To help wrap your head around it imagine that the car is going around the loop very slow. Hell, imagine it is stopped at the top of the loop. What will happen? Of course it will fall.

    Now make it go veeeeeeeeeery slowly and you will still see that it will fall. It takes a decent velocity to make it "stick" to the top.

    If that doesn't work, get a yo-yo and play with it. Spin it around fast, then spin it around slow. You'll clearly see that if you don't spin it fast enough, the yo-yo doesn't keep the string tight and/ or it simply falls.
     
  5. Jul 28, 2011 #4

    sophiecentaur

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    Here's another way of thinking about it - and there are many ways. This doesn't need to involve talking about an 'outwards' or centrifugal force. The wheels are providing an inwards force to make the car turn in a circle. This inward force acts at ground level, towards the centre of the curve (i.e. centripetally). There will be an equal and opposite force which acts through the centre of mass of the vehicle (forming a couple). This couple has a turning effect on the car which is about a horizontal axis, through the line of contact with the outer wheels (i.e it will tend to roll the car away from the direction of turn). There is also a moment, in the opposite sense, due to the weight of the car acting downwards and the upwards force from the outer wheels, which is tending to keep the car horizontal. When the moment of the centripetal force is greater than the restoring moment due to the weight, the car will roll around the axis between the outer wheels. The two forces we're talking about are actually at right angles to each other (one vertical and one sideways) but this doesn't matter because they are both tending to rotate the car about the outer wheel line. Levers don't need to be in the same plane.

    Thinking in terms of these two forces and taking moments this way shows how a low CM and a wide track will allow faster cornering without rolling.
     
  6. Jul 28, 2011 #5
    Sophiecentaur
    It seems that you are describing a car going around a turn of a horizontal road, correct?
    I thought that the OP was talking about a vertical loop. At least in part 1.
     
  7. Jul 28, 2011 #6

    sophiecentaur

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    Oh Jeez yes. it was the key word 'car' that got me going in the wrong direction.

    The middle sentence of the OP is spot on and doesn't need any more added, really-

    "I realize due to common sense that the car will fall but I simply can't wrap my head around and answer why it'll fall.. other than simply saying the centripetal acceleration is smaller than g......"

    Once you accept that forces are vectors and that they add up, it's not difficult.
     
  8. Jul 28, 2011 #7
    hi so thanks for the help everyone I think I got 1.. Except this: what if the car's velocity at the top of the loop is so that its centripetal acceleration equals exactly g? the car would fall, right?

    but concerning 2: can i get a longer explanation?

    I think cap AI is saying that centripetal force doesn't accelerate the person downwards into the ground and thus doesn't increase the normal force from the ground..

    I realize this is right but there is something fundamental I am misunderstanding about centripetal acceleration and force.. I thought centripetal force would accelerate the person in question towards the centre of the planet, to keep the person rotating along with the planet.

    But this acceleration is not in the same way as gravity?
     
    Last edited: Jul 28, 2011
  9. Jul 28, 2011 #8

    rcgldr

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    A persons momentum tends to keep the person moving in a straigh line, but gravity pulls the person inwards towards the center of the earth, and it can be considered to be a centripetal force or acceleration, since it's directed towards the center of a circular path and perpendicular to the path a person travels. There's an equal and opposite reaction force that pulls the earth towards the person, but since the earth's mass is so much larger than the person, almost all of the acceleration occurs with the person. At the slow rotation speed of the earth, attractive force from gravity is opposed by a compressive force between the person and the ground. The person exerts a downwards force onto the ground due to gravity, with the ground exerting an equal and upwards force due to compression, equal in magnitude to m g (where m is the mass of the person, and g is the rate of acceleration at the earth surface, about 9.8 m / s2.

    If the earth were spinning faster, so that the acceleration perpendicular to the circular path, v2/r = .5 g, then the accelerations and forces related to gravity remain the same, but the component of force between the person and the earth's surface are now 1/2 m g (where m is mass of the person).

    If v2/r = 1.0 g, then the force between person and earth surface would be zero.
     
  10. Jul 29, 2011 #9
    Well...no. It (and everything in it) would essentially be weightless. All the forces would balance out and be 0. Any slower than that and indeed it would begin to fall, but since the forces are 0 then it just stays in place, which is the top of the loop.
     
  11. Jul 29, 2011 #10
    Centripetal force is not a physical force, it's a condition for circular motion, so it doesn't go on force diagrams or contribute to any net force. Once you add up all your physical forces like gravity, drag, reaction from the tarmac, etc, you get a net force, and if that net force is equal to the centripetal force, the body will move in a circle.

    Then it will move in a circle. Whether it falls depends on your definition of "fall". I'd personally say it is (free) falling, in the sense that it is moving solely due to gravity.
     
  12. Jul 29, 2011 #11

    sophiecentaur

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    How can it not be a "physical" force, if it causes a radial acceleration? If you have no string, track, or gravity to provide that force then things will not go round in curves or orbits. I think what you have said is likely to be very confusing for anyone who is not quite happy with this topic. Centripetal just means towards the centre - and it needs to be provided in order to get circular motion.
     
  13. Jul 29, 2011 #12

    Doc Al

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    I think he just meant that it's not a separate force unto itself. It's just the name we give to the net force towards the center when we have centripetal acceleration.

    When students are asked to identify all the forces involved with centripetal acceleration they often list 'centripetal force' itself and show it on their diagrams in addition to all the actual forces acting. That's not good.
     
  14. Jul 29, 2011 #13
    allright guys, thanks 4 the help. i now understand this completely i think
     
  15. Jul 29, 2011 #14

    sophiecentaur

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    In many ways it's a pity we're not allowed to use the term 'centrifugal force' to describe the outwards directed radial force. All because they don't want people to say that things are 'thrown outwards'. After all, it's there because its one of a third law pair.
     
  16. Jul 29, 2011 #15

    Doc Al

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    Uh... what force? :wink:
    As long as one realizes that it's just an artifact of viewing things from a rotating frame, there's nothing wrong with using 'centrifugal force'. But not for elementary problems that can easily be dealt with from an inertial frame.
    :eek: Say what?
     
  17. Jul 29, 2011 #16

    sophiecentaur

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    When you're on a fairground ride (which is the experience that most / many people need an explanation for) your body is constrained to move in a circle but. of course, you make a dent in the seat cushion. That dent has to be because of a force that is equal and opposite to the centripetal force which is pushing you into the curve.
    So you need to be able to explain even a simple situation like a fairground ride which is not an inertial frame. Conkers on strings is one thing - the Waltzer is another, much more interesting thing and people actually FEEL that centrifugal force as they try to lift their head against it. It can't be ignored and needs to be explained.
     
  18. Jul 29, 2011 #17

    Doc Al

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    The force that you exert on the cushion is not what we usually call the centrifugal force. That's just the 'reaction' to the cushion pushing you inwards. (Centrifugal force, on the other hand, acts on you, not the cushion.)

    Centrifugal force is not a third law pair, since it's not a 'real' force. (Here I'm using the standard meaning of centrifugal force.)

    I agree that the concept of centrifugal force can be very useful. But it requires careful handling.
     
  19. Jul 29, 2011 #18

    sophiecentaur

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    I think you are over complicating this and the reason is that people can be sloppy and you feel the need to be purist about it..

    The force that is acting on you is being resisted by the cushion and it is squashing the cushion outwards. How is that any less of a real force than the force that a spade applies to the ground when you push it with your foot?
    The centrifugal force acting on you is also acting on the seat cushion. How can it not be? It's the same value and in the same direction.

    But, OK, how do you explain what happens on a fairground ride? This would be someone who believes that they can feel their weight pushing them down on the ground when they're standing up when all they can actually feel is the reaction of the ground upwards on their feet. (That is your typical Joe Public and I've met dozens of them)
     
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