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Simple(?) Speed bump and centripetal force problem

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A speed hump is installed. A 1800 kg car passes over it at 30km/h. The hump follows the arc of a circle with r=20.4 What force does the road exert on the car as the car passes the highest point of the hump?
    the answer is 1.15x10^4 N according to my professor and this random guy on reddit whose post I found:
    http://www.reddit.com/r/cheatatmathhomework/comments/l4aji/physics_for_engineers_1/


    2. Relevant equations

    Force Centripetal = (m*v^2)/r
    g=9.8 m/s^2


    3. The attempt at a solution
    When the car is at the top of the hump, the force of gravity (mg) is acting towards the center of the circle with radius 20.4 meters, or towards the ground. Therefore, centripetal force and gravity are acting in the same direction. The opposing force, the normal force, must be equal and opposite to keep the car in place.

    F_net= F(n)-mg-F(c)=0

    F(c)= (m*v^2)/r
    =(1800*8.33333^2)/20.4
    =6127.45
    mg=17640

    F(n)=23767 N
    But this answer is incorrect according to my professor.

    Instead, the correct answer comes when F(c) acts in the same direction as the normal force (exactly away from the center of the circle, which defies the definition of centripetal force).

    F(n)= mg - F(c)
    =17640 - 6127.45
    =11512 which is really close to 11500 which is the answer

    So my question is, why is centripetal force acting upwards with F(n) and not downwards towards the center of the circle like it should be?

    Why is centripetal force even a factor in the problem? The car is only accelerating towards the center of the circle (of which the speed hump is a part of) only when it is at the very top. I am very confused. Any help would be great. Thanks.
     
  2. jcsd
  3. Feb 15, 2012 #2

    cepheid

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    EDIT: Welcome to PF!

    There are only two actual forces acting on the car (in the vertical direction) in this problem:

    1. Gravity
    2. The Normal force

    These are real forces with real physical causes associated with them.

    We also have a requirement for circular motion:

    - if the object is moving in a circle with radius r and constant speed v, then there must be a NET force acting on that object of magnitude mv^2/r and in a direction towards the centre of the circle. We call this force the centripetal force.

    So you see, you shouldn't think of the centripetal force as a third force distinct from the other two, because then the question would be, "where does it come from?" Instead, think of the centripetal force as something that *arises from* or *is provided by* the actual real forces in the problem, and that is a requirement for circular motion. If there are no forces in the problem to *provide* or *be* the centripetal force, then there won't be any circular motion.

    As I mentioned above, the NET vertical force is what is available to act as a centripetal force in this problem. In other words, the centripetal force is provided by the combination of gravity and the normal force. Hence we have:

    F_net = F_centripetal = F_normal + F_gravitational

    (The net vertical force is the sum of the two vertical forces that are present)

    F_normal = F_centripetal - F_gravitational

    = -(mv^2)/r - (-mg)

    where the gravitational force is -mg because it is downward, and I've picked upward to be the positive vertical direction. The centripetal force is also negative, because it also acts downward (towards the centre of the circular arc)

    This becomes F_normal = mg - (mv^2)/r, which is the equation that you had, and that you were so confused about.
     
  4. Feb 15, 2012 #3
    Thanks for the reply and the welcome. Your explanation of centripetal force as a net force and net result of the normal force and mg helped a lot. I did question where centripetal force came from in certain situations like you mentioned. But I still have some questions.

    F_net = F_centripetal = F_normal + F_gravitational

    Is F_net not 0? Almost every other physics problem I've worked, if something is pushing against something and not moving in that direction, then the force and normal force have to cancel out, meaning f_net = 0

    For an object sitting there, F_normal-mg=0 correct? (I use g=+9.8 and make it directional with my signs)
     
    Last edited: Feb 15, 2012
  5. Feb 15, 2012 #4

    ehild

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    It is correct, but the car does not sit on the top of the hump but moves around it with velocity v=30km/h. So the normal force and mg has to add up to the centripetal force, mv2/r, pointing downward.


    ehild
     
  6. Feb 15, 2012 #5

    cepheid

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    What you say about gravity canceling the normal force is true for an object "just sitting there" not accelerating vertically.

    However, because this object is moving on a curved circular path, its velocity is changing direction, which means it is accelerating, which means there is a net force.
     
  7. Feb 15, 2012 #6

    cepheid

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    As an example, say you're at the very top of a moving Ferris wheel. You're "just sitting there", so the normal force from the seat will support your weight, and there will be no net force on you, right?

    WRONG.

    In this case, since you're moving on a circular path, there must be a net centripetal force acting on you. Where does it come from? Ans: the seat doesn't push up on you as hard as usual. It doesn't have to cancel/support all of your weight. It only has to support the part of the gravitational force that is not being used as a centripetal force to keep you on the circle. So you see there is a net downward force on you, since the downward gravitational pull is larger than the upward normal force from the seat. You feel lighter than usual in this instance, since the seat does not push up on you as hard as usual.

    When you reach the bottom of the Ferris wheel, there must be a net upward centripetal force. This is provided by the normal force from the seat, which not only supports your entire weight, but also pushes up even harder than that, by an amount equal to the required centripetal force. So there is a net upward force on you in this situation, since the normal force is greater than gravity. You feel heavier than usual here, since the seat pushes up on you much harder than usual.
     
  8. Feb 15, 2012 #7
    Ah, thanks you two. Ehild's way of explaining seemed to make more sense to me for some reason. So centripetal force is a force observed whenever an object follows a circular path with a constant velocity, but it is a product of other "real" forces that act upon the object, which in this case are gravity and the normal force. Therefore, the interaction between gravity and the normal force must yield a net centripetal force that must point towards the center of the path's arc.

    F(c) = -(m*v^2)/r = F(n) - mg

    Which yields approximately 1.15 E4

    Got it. Thanks again!
     
  9. Feb 15, 2012 #8

    ehild

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    You really got it! :smile: I am pleased.


    ehild
     
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