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What does the centripetal force equal? (generally)

  1. Jan 10, 2015 #1
    Hey guys, new to the forum here and I have a general question regarding the centripetal force. In the example of a ferris wheel where there is a normal force pushing up against the person and the gravitational force pulling the person down, which force is centripetal? I know that the centripetal force counters the linear velocity, tangent to the circle of motion, which allows the object or person to stay in circular motion but which force is actually pulling it towards the center, the gravitational force or the normal force? Also, would it be correct to say that the net force equals zero (since the person is neither moving towards or away from the center) in this example or does the net force equal the centripetal force (since the centripetal force has to counter the linear velocity --- if this is correct, how would I compare the two since linear velocity is not a force)?

    I know that if a car is moving around a banked curve, the horizontal normal force will be centripetal but what about in other examples such as the ferris wheel? Also would the net force of a car moving around a bank curved be zero since it is neither moving towards or away the center?

    tl;dr - is the net force in a centripetal force example zero or is the net force equal to the centripetal force? Also, how would I relate this to the linear velocity that cancels it out?

    Thanks for the help!
     
  2. jcsd
  3. Jan 10, 2015 #2

    jbriggs444

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    The word centripetal literally means "toward the center".

    Normally when one talks about uniform circular motion, the "centripetal" force that is talked about is the net force -- the sum of all of the forces on the object. This net force must be inward, toward the center (hence the name) if the motion is to be circular.

    At the top of the ride, gravity points inward (downward) toward the center of the wheel and the supporting force from the wheel points outward. At the bottom of the ride, gravity points outward (still downward) away from the center of the wheel and the supporting force from the wheel points inward. The net force is inward in both cases.

    No. The net force is decidedly non-zero. The person's speed is unchanging. But the direction of their velocity is changing. That is an acceleration and requires a force.
     
  4. Jan 10, 2015 #3

    Andrew Mason

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    Welcome to PF Ghost Koi!

    In the case of the person on the ferris wheel moving at constant angular speed, the only acceleration that he experiences is centripetal acceleration. So the net force is the centripetal force. The forces acting are on him are gravity and the normal force. So the sum of those forces has to constantly change in order to provide the constantly changing centripetal acceleration. Since gravity is constant (ignoring the slight decrease in gravity at the top compared to the bottom), the normal force has to keep changing magnitude and direction. That constant change is part of what makes the ferris wheel fun to ride in.

    AM
     
  5. Jan 10, 2015 #4
    So I would always look at the Net Force to find the centripetal force if I have more than one force acting on an object? (I.e. could setup my equation like this)...

    m(ν2/r) = FN + Fg
     
  6. Jan 10, 2015 #5

    Andrew Mason

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    The net force on a body (which is the vector sum of all the forces) is always equal to its mass x its acceleration:

    [tex]\sum_i \vec{F}_i = \vec{F}_{net} = ma[/tex]

    If the body is experiencing constant angular velocity circular motion, the acceleration is entirely centripetal and is constant in magnitude. So [itex]\sum_i \vec{F}_i = -\frac{mv^2}{r}\hat{r}[/itex]

    AM
     
    Last edited: Jan 12, 2015
  7. Jan 10, 2015 #6
    Thanks... I think I understand but just to confirm, is my previous statement, and equation, correct?
     
  8. Jan 11, 2015 #7

    Andrew Mason

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    As I said, your equation, [itex]\sum_{i}\vec{F}_i = -\frac{mv^2}{r}\hat{r}[/itex] is correct ONLY IF the acceleration is ENTIRELY centripetal (i.e. constant circular motion). ie. there is no other acceleration.

    AM
     
    Last edited: Jan 12, 2015
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