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Centrifugal Force and a train

  1. Sep 14, 2006 #1
    A train is running smoothly along a curved track at the rate of 50 m/s. A passenger standing on a set of scales observes that his weight is 10% greater than when the train is at rest. The track is banked so that the force acting on the passenger is normal to the floor of the train. What is the radius of curvature of the track?

    I think i use the equation
    F= [(GMm)/r^2](r/r), and Multiply the little m by .10, and try to solve for r
     
    Last edited: Sep 14, 2006
  2. jcsd
  3. Sep 15, 2006 #2

    Andrew Mason

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    ??You do not have to use Newton's law of gravitation here (just accept that GM/r^2 = g = 9.8m/sec^2). What are the forces acting on the person?

    AM
     
  4. Sep 15, 2006 #3
    i'm interested, can someone show us how to do it?

    can someone draw a free-body diagram explaining how going around a banked track will increase weight by 10%!
     
  5. Sep 15, 2006 #4

    mrjeffy321

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    Its it not really their weight (force due to gravity) that increases by 10%, it is the normal force applied on the person from the ground which is sensed by the scales as an appearent increase in weight.

    The centrifigal force needed in order for the train and passengers to make it around the curved and banked track increases their normal force by 10%.
    If you draw the free body diagram of the object on an incline plane / banked track, the object's weight is divided up into a normal force pointing out perpendicular to the surface and a parallel force pointing a long the surface. The centripetal force is pointing in towards the center of curvature.

    parallel force = mass * gravity * sin (theta)
    normal force = mass * gravity * cos(theta)
    centripetal force = m * v^2 / r
     
    Last edited: Sep 15, 2006
  6. Sep 15, 2006 #5

    andrevdh

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    It is not really necessary to consider motion on a banked curve. Consider someone being accelerated upwards in a space shuttle while standing on a scale. He will experience two forces his weight downwards and the force from the scale upwards - the normal force. The difference between these two need to provide him with the upwards accelerating force, which means that this force need to overcome his weight and provide an additional force to accelerate him upwards. The normal force coming from the scale are therefore greater than his normal weight. According to newton's third law the scale will experience a force of similar magnitude in the opposite direction, so that the scale will register a larger mass than normal. It is as if the person is pushing down harder on it. But this only the result of the scale actually pushing harder than normal on the person in order to accelerate him upwards.

    How do the person experience this situation? Well as we saw he is pushing harder than normal down onto the scale, which means his muscles have to push much harder, which he interprets as an increase in his own weight in order to support himself.
     
    Last edited: Sep 15, 2006
  7. Sep 16, 2006 #6

    Andrew Mason

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    [correction - after reading Andrevdeh's post and thinking about it a bit more clearly! - thanks]

    Resolve the gravitational and centripetal forces into vertical and horizontal components. The vertical components sum to 0. The horizontal components sum to the centripetal force. You just have to determine what centripetal force will give this result.

    Assume the track is banked at angle [itex]\theta[/itex] to the horizontal.

    The vertical components are:

    [tex]F_{y} = mg + F_N\cos\theta = 0[/tex]

    [tex]F_{x} = F_N\sin\theta = F_c[/tex]


    [tex]F_c = F_N\sin\theta = 1.1mg\sin\theta = mv^2/R[/tex]

    [tex]F_N\cos\theta + mg = 0 \rightarrow 1.1mg\cos\theta = mg[/tex]


    Work those out and equate to find [itex]\theta[/itex]. Work out R from [itex]F_c = mv^2/R[/itex]

    AM
     
    Last edited: Sep 16, 2006
  8. Sep 16, 2006 #7

    andrevdh

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    The passanger is experiencing two forces. His weight, [itex]\vec{W}[/itex], downwards and the force from the scale [itex]\vec{S}[/itex] , which is equal to [itex]1.1W[/itex] in magnitude as Andrew remarked. This force is also perpendicular to the floor of the train. Resolving the force from the scale into its x- and y-components we note that its x-component need to provide the force that accelerates the passenger towards the centre of the curvature in the track:

    [tex]1.1W \sin(\phi) = \frac{mv^2}{r}[/tex]
     

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  9. Sep 16, 2006 #8

    Andrew Mason

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    I have corrected my earlier post. It is simpler than I thought at first. It is easier to look at the horizontal and vertical components of the normal force. The vertical component of the normal force has to balance gravity. And the horizontal component has to provide the centripetal acceleration. See corrected post above.

    AM
     
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