# Normal force of loop-the-loop at the side of the circle

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?

normal force when the car is at the side of the loop is equal to centripetal force only.

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.

From physics we know that the net force on a mass moving in a circular path at constant speed always points towards the center and has magnitude $$|F_{net}| = m*a_{c}=\frac{mv^2}{r}$$

For an object moving in a vertical circle, when the object reaches the side the net force must be pointing towards the center (west). This implies that the force of weight is counteracted by some other force directed upwards. The reason why is, if there was no counteracting force for the weight vector, then the resultant net force would not point towards the center. To see this, from the point on side draw the normal force supplied by track towards the west, and weight vector towards the south at right angles. $$|F_{net}| = \sqrt{N^2 + (mg)^2}$$

The direction of $|F_{net}|$ will point towards the southwest, which contradicts our assumption of uniform circular motion that net force must point towards center (here west).

Now a good question is how is this counter-acting force supplied while maintaining a constant speed. I guess you could increase gas as you are going up.

So that leaves us only with the normal force $N$ supplied by track. $$|F_{net}|=N$$ and using our theorem above $$|F_{net}| = \frac{mv^2}{r}$$ by substitution we get $$N = \frac{mv^2}{r}$$