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A question about centripetal acceleration

  1. Feb 23, 2010 #1
    http://img517.imageshack.us/img517/4817/aholelol.png [Broken]
    http://img532.imageshack.us/img532/881/bridgelol.png [Broken]

    There are 2 images. Let ac be the centripetal acceleration vector and v the speed vector.

    Now are my drawing right or have I gotten this whole thing wrong?
    My teacher said that the centripetal acceleration on the bridge would be pointed up. And that would make the object weigh less.
    I, on the other hand, think its because of the inertia of the body.
    And the same thing in the pit/hole. Imo it should be pointed up, but my teacher claims the opposite.

    Have I understood this whole thing wrong?

    Thanks in advance,
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 23, 2010 #2
    Centripetal acceleration means a "center pointing" acceleration. With respect to an inertial frame, the object on top of the bridge is experiencing a centripetal acceleration pointing towards the center of the curve. When the object is at the bottom of the pit the centripetal acceleration is pointing up toward the center of the curve. When at the top the body will experience less weight, and when at the bottom it will experience more weight.
  4. Feb 23, 2010 #3
    So my drawings are correct and my teacher made a mistake?
  5. Feb 23, 2010 #4


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    I'm assuming gravity is involved here, although you left it out in the diagrams. In the first case, centripetal acceleration opposes gravity, increasing the force the pit applies to the object. In the second case, centripetal acceleration is in the direction of gravity (it's gravity causing the centripetal acceleration), reducing the force applied to the object by the bridge.
  6. Feb 23, 2010 #5
    Im a tad confused here now. When the centipetal acceleration opposes gravity, isnt the objects weight supposed to reduce? And vice versa with the bridge?

    Thanks in advance,
  7. Feb 23, 2010 #6


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    The force of gravity is downwards. For an object at rest on a surface, then the force exerted by the surface onto the object, and by the object exerted onto the surface equals the weight of the object. If the surface is accelerating upwards (opposing gravity), then the apparent "weight" of the object increases. If the surface is accelerating downwards (in the same direction of gravity), then the apparent "weight" of the object decreases.
  8. Feb 25, 2010 #7
    Alright. I've been thinking about this one and here is my attempt at an explanation...

    The centripetal force is not a real force, it must be supplied by a real force such as (in this case) weight, or the normal reaction force.

    If you were to draw your diagram with just the real forces then you would only have weight and the normal force. The result of these (the net force) supplies the centripetal force keeping the object on the circular path. So...

    Centripetal force = net force = weight + normal force

    At the bottom of a dip the weight is in the opposite direction to the required net force, so the normal force must be larger to supply the net force.

    At the top of the hump the weight force is in the same direction as the required net force. So some of the weight force supplies this leaving a smaller normal reaction force.

    Using the formulas...


    net force (up) = -weight (down) + normal (up) (upward forces are positive)
    so normal = net force + weight
    or normal = required centripetal force + weight


    net force (down) = weight (down) + -normal (up) (downward forces are positive)
    so normal = weight - net force
    or normal = weight - required centripetal force

    The apparent weight is given by the size of the normal force.
    I think your diagram is correct.
    I think you are on the right track. It's all a result of inertia - the object would prefer to travel in a straight line. The centripetal force must be supplied (from somewhere) to make it follow a curved path.

    A simplified summary: The centripetal force is supplied by real forces; the weight at the top of a curved vertical path and the normal at the bottom.

    Does this help?
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