[Physics] Normal force of loop-the-loop at the side of the circle

centripetal-forceclassical-mechanicshomework-and-exercisesnewtonian-mechanics

In the loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has the mass of 230 kg and moves with the speed of 300 m/s. The loop-the-loop has a radius R=20 m. What would then be the magnitude of the normal force on the car when it is at the side of the circle moving upward?

I tried to solve this problem, by:

  1. gravitational force = $mg$ = 230*9.8 (downward)
  2. centripetal force = $mv^2/r$ = 300^2/20 (toward circle, which is horizontal)
    and

by vector addition/subtraction, magnitude of normal force would be $\sqrt {(mg)^2+(mv^2/r)^2}$ .

But the answer I got is wrong, so this approach must be wrong… What did I do wrong here?

Best Answer

If I'm understanding your problem correctly, then the normal force is the centripetal force.

  • $ F_N = \frac{mv^2}{r} $

In other words, the normal force from the rail causes the centripetal acceleration towards the center of the circle. There are, as I understand it, no other forces acting in the normal direction. Remember that you are only supposed to consider forces in the normal direction:

  • $ \sum F_N=ma_N $

The gravitational force is perpendicular to the normal force at this position and so has no effect in the normal direction.

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