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A car is driving on a frictionless banked curve of radius 45 metres and 25 degrees. Is it possible to find centripetal force? If not possible, explain why.

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  • $\begingroup$ Yes, in fact you have been given more information than you need. $\endgroup$
    – garyp
    Commented Feb 13, 2015 at 21:01
  • $\begingroup$ So how would one find it? $\endgroup$
    – user40096
    Commented Feb 13, 2015 at 21:02
  • $\begingroup$ Well, what forces are present on the car, and which one is the centripetal force? In somewhat different words, how will the car stay on the frictionless banked curve? $\endgroup$
    – Jon Custer
    Commented Feb 13, 2015 at 21:26
  • $\begingroup$ Normal force ^2 = Centripetal Force^2 + Weight ^2. $\endgroup$
    – user40096
    Commented Feb 13, 2015 at 21:31
  • $\begingroup$ I can find velocity and acceleration, but I don't think it's possible to find an actual value for centripetal force. $\endgroup$
    – user40096
    Commented Feb 13, 2015 at 21:31

2 Answers 2

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If the track is circular and the radius of the path is constant, then the net force must be purely horizontal, toward the center of the track. That net force is the force providing the centripetal acceleration (which some people call the centripetal force). Viewed from a reference frame in which the track is at rest, the forces which sum to be parallel to the ground (horizontal) are the normal force of the track on the car and the weight of the car.

The vertical components of these must add to zero. The horizontal components add to give the centripetal force. Break the forces into vertical and horizontal components and sum them. You need the mass of the car.

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The centripetal force is responsible for keeping the car in a circular path and its equal to $$F = m \frac{v^2}{r}$$, since the mass is not given you cannot calculate a force. You can calculate the speed and hence centripetal acceleration since $$ g \tan \theta = \frac{v^2}{r} $$

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