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Force Vectors and Circular Motion

  1. Oct 11, 2006 #1
    This is a case where I know the results but am fighting the math. At the top of a ferris wheel your acceleration in Y is 0. Ol' Newton says all all forces then must cancel, got that. I know my weight at this point will feel less at this point. Here is where I am troubled, vector for gravity points down so does the vector for the circle's accleration but my Normal points up and must have the same magnitude as Fg and Fc. The answer given subtracts Fc from Fg to get Fn (which feels right, but contradicts my vectors). Any conceptual help is appreciated.
     
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  3. Oct 11, 2006 #2

    OlderDan

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    Your statement in bold is false. You need to rethink everything that follows it.
     
  4. Oct 11, 2006 #3
    I realized that after I posted. I was confusing acceleration with velocity. I still have 2 force vectors working in the downward direction. I've seen the solution but they figure Fn=mv^2/r + mg, but they use a negative for the angular acceleration but a possitive for gravity. That just doesn't make since if they both point toward the center of the wheel.
     
  5. Oct 11, 2006 #4

    OlderDan

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    The centripetal force is not an additional force in the problem. It is the name we give to the combination of forces that are causing the circular motion. The only forces acting on you are gravity and the normal force, and these are in opposite directions. You know that gravity is stronger, because the acceleration is down toward the center of the circle. If you take down as positive (the direction of the centripital force) you have
    Ftotal = mg - Fn = mv^2/r , where everything represented as a letter is positive
    An alternative way of looking at the problem is to treat mv^2/r as a centrifugal force (I don't recommend it) If you do this the direction of mv^2/r is upward, as is the normal force, and gravity is downward. In this case, the net force has to be zero, so you have
    mg = Fn + mv^2/r
    Either way you do it, you get the same result
    Fn = mg - mv^2/r
    You were right to question what you wrote earlier as the given answer. Writing the mv^2/r with the + and treating the term as a negative is confusing. There is no natural way to introduce a negative into this quantity.
     
  6. Oct 12, 2006 #5
    The centrafugal way is exactly what "feels" natural in this problem but I know it's a psudo force. I still don't understand why we can ignore the centripetal acceleration because if that is so then my future trajectory would become parabolic and no longer circular. This is why I have problems because I can't "trust" any of my intuitive instincts for fear of these caviats.
     
  7. Oct 12, 2006 #6

    Doc Al

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    Both weight and centripetal acceleration point downward and are negative.

    First identify the forces:
    (1) Normal force, which points upward: Fn
    (2) Weight, which points downard: -mg

    Now add them up:
    Total force= Fn -mg

    The acceleration is centripetal and points down, so: a = -v^2/r
    Now apply Newton's 2nd law (total force = ma):
    Fn -mg = -mv^2/r

    Thus: Fn = mg -mv^2/r
     
  8. Oct 12, 2006 #7
    But why is the centripetal acceleration not accounted for in the total force? It is acting on the mass in the downward direction just as gravity is, right?
     
  9. Oct 12, 2006 #8

    OlderDan

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    No. The centripetal force is not an additional force. It is the name we give to the net force acting on an object in circular motion with constant speed. That net force comes from the real forces acting on an object, such as gravity, friction, normal force. When you write the net force in terms of the contributing forces, you know that its magnitude must be mv^2/r

    Edit: I added the "with constant speed". An object in circular motion could be accelerating tangentailly as well, but that is not part of this problem. The net force acting on such an object would have both a centrifugal force component and a tangential force component.
     
    Last edited: Oct 12, 2006
  10. Oct 12, 2006 #9

    Doc Al

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    Just to add to OlderDan's explanation: "Centripetal" just means "towards the center"--it specifies a direction, not a new kind of force. You need real forces (like the seat pushing up and gravity pulling down) to provide the net force acting towards the center.
     
  11. Oct 13, 2006 #10
    Thank you guys so much. That was exactly the point I needed clarification on.
     
  12. Oct 13, 2006 #11

    OlderDan

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    OOPs That should have said
    The net force acting on such an object would have both a centripetal force component and a tangential force component.
     
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