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Circular motion on a inclined plane

  1. Nov 18, 2005 #1
    A car turns on a inclined road at a speed of 60 km/h, the mass of the car is 3 tons and the radius of the circular motion is 20m; If there is no friction, calculate the required angle in order to keep the car turning at the same speed.
    I calculated it to be an impossible angle, but my teacher said it was 54°. Then he showed me his solution and I think it is totally bogus... He said it is the reaction of the road that gives the centripetal force, and then analysed the reaction being perpendicular to the road and with a y composant equal to the weight of the car and calculated the x component as being the required centripetal force. I disagree, as I am persuaded that the centripel force is given by the component of mg that is parrallel to the road, the reaction cancels the other component and thus the mass moves according to the parrallel component; this force is reponsable for the centripetal force. So if the angle is θ,
    mgsinθ = m(v^2/r)
    ;v = 60/3.6
    r = 20
    sinθ = 1.41723356
    Which is impossible.
    I need an awnser since i have a test on monday.
     
    Last edited: Nov 18, 2005
  2. jcsd
  3. Nov 18, 2005 #2

    Fermat

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    Homework Helper

    Your teacher is right, I'm afraid.
    That would mean that, if the road were horizontal instead of banked, there would be no centripetal force.

    We know that the car is supported by the road. And that this support acts on the car at an angle @ to the vertical.

    An alternative way of visualising this would be to imagiine that the car was moving around in a (horizontal) circle at the end of a large cable attached to the car. The other end of the cable is attached to a point that is directly above the center of the circle that the actual road runs on. The cable would be at an angle @ to the vertical. The tension in the cable would now replace completely the support reaction from the road surface. If you resolve this tension, The vertical component is equal to the weight of the car while the horizontal component is what gives the centripetal force. From these two eqns you should get the standard result:

    v² = rg.tan@

    Does it matter if the car is pushed from below (by the road) or pulled from above (by the cable)
     
  4. Nov 18, 2005 #3
    But the reaction is perpendicular to the road while the cable would be parrallel to it. For the case of the road with no inclanation, if there is no friction, the car cannot turn. Also the reaction of the road is equal to the action that is directed towards it?? Since mg acts on the mass it acts on the inclined road with a force equal to mgsinθ and the reaction is equal to that and thus this vector cancels out on the mass and the only vector left on the mass is mgcosθ.... how can the reaction possibly be greater than mg? Explain...
     
  5. Nov 18, 2005 #4
    if the car is moving centripetally, then it must also be moving around a center, and if it's on a banked curve, then the center must be at the same level as the car (not down the road), and therefore, the centripetal force must be the x component of the normal force
     
  6. Nov 18, 2005 #5

    Doc Al

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    Staff: Mentor

    Reread Fermat's post. The cable is perpendicular to the road.

    True.

    Whatever force the car exerts on the road must equal the force that the road exerts on the car. Newton's 3rd law.
    The "reaction force" of the road on the car will only equal mgcosθ if there is no acceleration normal to the surface. That's not the case here since the car is centripetally accelerated. The reaction force can easily exceed mg if the object accelerates: consider your "weight" on an elevator that accelerates upward.

    Realize that the true "reaction force" to the car's weight is the car's gravitational attraction acting on the earth.
     
  7. Nov 18, 2005 #6
    So basically the road is giving a higher reaction because of its tendency to keep its shape? But can't mgsinθ help provide the centripetal force?
     
    Last edited: Nov 18, 2005
  8. Nov 18, 2005 #7

    Doc Al

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    Staff: Mentor

    It's the road's tendency to resist deformation that allows it to provide any force at all. Just like when you walk across the room, the floor resists being crushed by your weight (when the molecular bonds are forced together, they push back strongly) thus producing the normal force that supports you. If you replace the floor with a section made of thin paper, you will find out the hard way that it cannot provide enough of a normal force to hold you up.


    Not sure what you mean. The centripetal force is the net horizontal force on the car. Since the weight acts down, it has no horizontal component.
     
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