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Motion in two dimensions question

  1. May 2, 2005 #1
    A car is on a circular oval-shaped track and at point F, the car is at rest. about 1/8th of the way around the oval from point F, there's a point G. Now i dont understand why the following two statements are incorrect.

    1. The acceleration at point F is zero. As point G becomes closer and closer to point F, the change in velocity vector becomes smaller and smaller, eventually it becomes zero.

    2. The acceleration at point F is perpendicular to the curve.

    2 is incorrect because acceleration at point F is zero since the car is at rest. but i dont understand why 1 is incorrect. it seems right to me..
     
  2. jcsd
  3. May 2, 2005 #2

    StatusX

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    1 is incorrect and 2 depends on the forces pushing the car around. The acceleration is the instantaneous change in velocity. Since the car is turning, its velocity vector is changing direction, and so there is an acceleration. Now, if the speed of the car is constant, there is never any acceleration parallel to its motion, and so the acceleration is perpendicular at all points, but I don't know if this is what's going on here.
     
  4. May 2, 2005 #3

    OlderDan

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    The answer to 1 makes sense if you assume the car is moving through point G with some speed. The statement of the problem appears to be incomplete.
     
  5. May 2, 2005 #4
    the car starts from rest at point F and speeds up continuously as it moves around an oval. the car is in motion when it passes the point G.
    To me, as the point G is moved closer and closer to the F, the velocity vector will indeed get smaller and eventually reach zero when it is at point G. So i dont see what's wrong with statement 1.
    for statement 2, since the car is at rest, how can it have a perpendicular acceleration?
     
  6. May 2, 2005 #5

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    I don't understand what you mean by "as G gets closer to F." I thought you said F was 1/8 of the way around the oval from G. Is the oval shrinking? Or does F move with the car?

    If the car is continuously sped up, the acceleration is never 0. How could it be?
     
  7. May 2, 2005 #6
    I think they mean that G is just a point, and it's distance from F can be varied, although it start 1/8 of the track from F. But again, that's just how I interpreted it, we can all agree they stated whatever they meant to say very poorly.
     
  8. May 2, 2005 #7
    Velocity and acceleration are two different things, since dv/dt = a. A zero instantaneous velocity tells you nothing about your instantaneous acceleration.
     
  9. May 2, 2005 #8
    its like what KingNothing said. its just a poing on the oval. if you move this point closer and closer to F, does the velocity eventually get zero??
     
  10. May 2, 2005 #9
    I actually don't understand why (2) is wrong. Is the oval perfectly circular? If not, then there will be components of the acceleration that are parallel to the curve. EDIT: actually you'll need a parallel component even in the circular case to get the car moving.

    (1) is just talking about calculating the derivative of v at point F.

    acceleration = lim (v(x+delta) - v(x))/delta, as delta approaches zero.

    Have you studied calculus before? Think about this: if I drop you out of a helicopter with no horizontal velocity, you will be at rest the moment I drop you, but your instantaneous acceleration will be g.
     
    Last edited: May 2, 2005
  11. May 2, 2005 #10

    OlderDan

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    I think you have #2 figured out. There may be an acceleration even if the car is at rest, but if so it will be in the direction of motion. There can be no perpendicular acceleration with the car at rest.

    For number 1, the key point is that the car is moving in all cases. The statement that G gets closer to F is an obscure way of asking "what if the dimensions of the track are decreasing?" If the dimesions are decreasing, then the radius of curvature of any part of the track is decreasing. If you have some velocity as before, and the radius of curvature is approaching zero, you still have the same change in velocity from one side of the track to the other or between any two points near G that have the same velocity direction as on the original track. The change is happening faster, so if you had been asked about the RATE of change of velocity (acceleration) instead of the change itself, that would approach infinity. Besides, as someone else noted, being at rest a point F does not mean there is no acceleration there.
     
  12. May 2, 2005 #11
    Why is that? It's not impossible to have a radial force component.
     
  13. May 2, 2005 #12

    StatusX

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    The component of acceleration parrallel to the velocity changes the particle's speed while the component perpendicular changes its direction. It's true, at the moment the particle begins to move, the speed can only increase, it cannot change direction because it has no direction. The speed is increasing along the direction of the track, and the acceleration will be parallel to the track. Juvenal, if there is a net radial force at the moment the particle is at rest, it will begin to move towards the center of the circle. But since the track is there, it provides an opposing force and this doesn't happen. Once the speed builds up, a radial force is required to keep the object in circular motion.
     
  14. May 2, 2005 #13
    I don't think the speed has to build up - it just has to be non-zero, which happens almost immediately. I think it would work if you started with a radial force, but also accelerated forward at time t_0, as I mentioned in my edit above.
     
  15. May 2, 2005 #14

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    Any radial force beyond the necessary centripetal force will cause the particle to accelerate towards the center of the circle. The centripetal force is mv^2/r, so even a small net radial force applied immediately after the particle begins to move (very small v) will cause this to happen. Keep in mind the centripetal force is supplied by the walls of the track, and changes as the particle moves faster and is pushed more strongly into the wall.
     
  16. May 2, 2005 #15
    Right - but I wasn't talking about an extra radial force beyond the centripetal, except exactly at time t=0. Anyhow, a small net radial force applied at ANY time will cause acceleration towards the center.
     
  17. May 2, 2005 #16

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    I'm sorry, I meant it will cause it to move towards the center, in that the radius will start to shrink. If there was any radial force when the particle is at rest, it will begin to move towards the center. I think your problem is that you know the velocity is increasing, and a radial force will be necessary any time after t=0, and so you think there must be a force at t=0. Its true the radial force will only be zero instantaneously, but if it was some finite number greater than 0, there would be a finite time for which it would be greater than mv^2/r, and then there would be some finite distance by which the radius would shrink. It's not like the force is 0 when the particle is at rest and then switches to some finite number the instant it starts to move.
     
  18. May 2, 2005 #17

    OlderDan

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    It is a car speeding up from rest on a track. Any piece of the track can be considered to have a radius of curvature perpendicular to the direction of the track. If there is no velocity at any moment in time, there is no acceleration in the direction of that radius by a = mv^2/r. If v = 0 and then a =0

    If by radial acceleration you mean toward the center of the oval, then there could be a radial component, but the problem was carefully worded to talk about acceleration perpendicular to the curve of the track, not towards the center of the oval.
     
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