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I What is the effect of centripetal force on velocity?

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  1. Sep 13, 2016 #1
    If a car is being driven with a constant force (the angle of the pedal is kept constant) and it goes around a curve, will the velocity change? If so, is this due to centripetal force?
     
  2. jcsd
  3. Sep 13, 2016 #2
    Of course, it will change. If it did not change then you will go in a straight line and not around a curve.
     
  4. Sep 13, 2016 #3
    Don't you mean that the velocity changes because you are going around a curve?
    Could you clarify why it is changing, given that the forward force is the same? Why does going around a curve exert a counter force?
     
  5. Sep 13, 2016 #4

    jbriggs444

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    Ummm, because there is a sideways force?
     
  6. Sep 13, 2016 #5
    What is the source of this sideways force?
    People, it might be obvious to you, but it is not to me, which is why I am asking the question, so a more in depth answer would be much appreciated.
     
  7. Sep 13, 2016 #6

    CWatters

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    Velocity has components speed and direction. Going around a corner means the direction is changing so the velocity is changing. The speed need not be changing.

    Forum rules discourage us just giving people the answers because they learn more when they figure things out for themselves. Have a think about the forces that act on a car. Have you tried to push one sideways? Why would that be hard? Why do you have to slow down in wintery conditions?
     
  8. Sep 14, 2016 #7
    Ok, I thought velocity and speed were the same thing. So, if a car is being driven with a constant force and it goes around a curve, does that mean the speed also remains constant?
    Because it is heavy?
    Because the road is slippery?
     
  9. Sep 14, 2016 #8

    jbriggs444

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    The idea that "velocity" and "speed" are different is drilled into one's head constantly in first year physics. As others have said, "speed" is how fast you are going without regard to direction. By contrast, "velocity" takes direction into account. 50 miles per hour forward is a different "velocity" than 40 miles per hour forward and 30 miles per hour leftward even though both are exactly the same "speed".

    In an ideal car -- perfect wheels with no rolling resistance and no sideways slip, and no air resistance, there would be no reason to put your foot on the accelerator at all. You could coast at constant speed forever. Since your car requires force to maintain a constant speed, we need to consider what makes your car less than ideal.

    So.. how about you tell us? Why does your car slow down if you do not keep your foot on the gas?
     
  10. Sep 14, 2016 #9

    CWatters

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    So what is the car relying on to turn a corner?
     
  11. Sep 14, 2016 #10

    A.T.

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    What does this mean in terms of forces?
     
  12. Sep 14, 2016 #11
    Well, I'm not a physicist, and I'm hoping this forum is open to those physicsally challenged!

    So, if you add a leftward force (wind, say) of 30 mph to a car that was going 40 mph it would then be going 50 mph? (computed by Pythagoras I guess)

    I feel you want me to say friction. But is that a force? Air resistance? Both?
     
  13. Sep 14, 2016 #12
    Turning the wheel? Is that exerting a sideways force? Does that effect the speed?
     
  14. Sep 14, 2016 #13

    jbriggs444

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    Yes, that's Pythagoras in action.

    But you you cannot directly equate a leftward force with a leftward addition to velocity. There is a relationship via quantities known as momentum or impulse. A leftward force applied for a specified time interval will produce an impulse which is the product of the two. The leftward impulse divided by the mass of the car will tell you how much its leftward velocity changes as a result.
     
  15. Sep 14, 2016 #14
    Just to clarify, are you answering yes to the question as I'd thought the answer would be no. Are you saying that if the forward speed of a car is 40 mph, which is then exposed to a leftward force (wind, say) of 30 mph, the car would then speed up (over time) to go at a forward speed of 50 mph?

    I would have thought that the speed would still be 40 mph, but it would be pushed to the right to change direction, if the steering wheel was not turned.

    If the steering wheel was turned, into the wind, the speed and direction of the car would remain the same. Yes?
     
  16. Sep 14, 2016 #15

    jbriggs444

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    No, as I said, the relationship between force and speed is not direct. The relationship between wind speed, then force, then car speed is even less direct. And if you do not turn the steering wheel, a lateral force will (ideally) have zero effect on the vehicle.

    If you turn the steering wheel, the car turns. When you normalize the wheel, the car retains its current heading. The wind does not enter in.

    Edit: If you want to imagine an airplane, then wind velocity adds as a vector to airplane velocity to get ground velocity, but that's not the subject matter here.
     
    Last edited: Sep 14, 2016
  17. Sep 14, 2016 #16

    A.T.

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  18. Sep 14, 2016 #17
    Let's come back to speed later, but first can we clarify this point.

    This doesn't seem right to me. If there is a lateral force on an object surely that object will move laterally. If it is wind then the car will be moved laterally (maybe only slightly), so it would be necessary to turn the steering wheel into the lateral force to maintain the heading. Generally we don't notice the effects of wind except when we go across a windy bridge and we have to make an abrupt correction with the steering wheel to maintain our heading.

    Similarly, if a car is being driven along a straight road but one which slopes from left to right the force of gravity would draw the car to the right. However, by turning the steering wheel to the left the straight heading of the car is maintained. In essence we would be applying an upward force to compensate for the downward force of gravity. Do you not agree?
     
  19. Sep 14, 2016 #18
  20. Sep 14, 2016 #19

    A.T.

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    The fundamental force behind friction is electromagnetic.
     
  21. Sep 14, 2016 #20

    jbriggs444

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    No, I do not agree. The rightward turning of the steering wheel creates a rightward angle of the front wheels on the car which tends to cause the car to progressively turn farther and farther right. This is desirable if it maintains the desired course while the vehicle "crabs" left under the influence of a leftward wind. However, if one maintains the steering in this position (ignoring caster effects on the steering mechanism) the car will continue to turn further right than is desired. Ideally you want a one-time rightward steering input to counter a continuing leftward force.

    In addition, I used the word "ideally" in my earlier posting. Ringing in non-ideal effects is not appropriate.
     
    Last edited: Sep 14, 2016
  22. Sep 17, 2016 #21
    I am not quite sure what you are disagreeing with as your first statement seems to agree and the second doesn't. Could you clarify? My query regards your statement that a leftward force has no effect on an object (ideally or otherwise), which doesn't seem right, or I am misunderstanding what you mean?

    So, in this,
    are you saying that it is necessary to turn the wheel to the right to compensate for a force coming from the right? If so, that makes sense and agrees with what I said.

    In this,
    you appear to be saying that you need to turn the wheel to the right and than back again to "normal" to compensate, i.e. an impulse turn. Is that what you are saying? I don't think that could be correct because after the impulse (one-time rightward steering input) you would be back to square one with the lateral force once again moving the car laterally. I'd say you have to maintain a turned wheel, all the time the lateral force exists, in order to maintain the straight-ahead heading. In other words, although the wheel is turned the car does not turn, due to the opposing lateral force. Yes?
     
  23. Sep 17, 2016 #22

    olivermsun

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    I think the confusion/disagreement arises because in one case what seems to be described is an impulsive force leading to a small permanent yaw which would continue to compensate for the slippage caused by the side force. In the second case what is being described is a continuing force exerted through the steering mechanism and tires which exerts the compensating force.
     
  24. Sep 17, 2016 #23

    A.T.

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    You don't necessarily need steering to compensate for a lateral force. If you have sufficient traction, the static friction at all the wheels can have a lateral component, which balances the lateral force.
     
  25. Sep 17, 2016 #24

    olivermsun

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    Perhaps in an idealized case, but in the real world case, the car/wheels/tires will probably want to move sideways at some rate that the driver will simply compensate for.
     
  26. Sep 17, 2016 #25

    sophiecentaur

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    Instead of using a car on a road as your 'simple' model, there is a simpler one of a locomotive on a Perfect, frictionless) railway line. The force on the loco, when it follows a curve is not friction; it is a lateral force resulting from the fact that the track in front is angled slightly. to one side If the track is frictionless (a further simplification) then no energy can be lost so there will just be a change of direction and no change in speed. There are practical issues on a real track and one of them will be that the resistive forces from the track (rolling resistance) may actually increase when going round a curve, so a freewheeling loco would slow down - or a driven loco would need more fuel to maintain its speed. But that is a step too far if you want to get the basics first.
    If you read and absorb the basics of speed, velocity and a few other quantities for moving bodies then there will be no conflict in your mind. The Science of basic stuff like this works on consistent rules and people can easily get confused when they are told something that follows those rules but their 'experience' is of situations involves other factors that the simple rules don't cover. So that's the next step . . . . . . I'm afraid that the dreaded Maths comes into this stuff very early on.
     
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