# The centripetal force direction on hill and valley

A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill the normal force on the driver from the car seat is 0. The driver's mass is 70.0 kg. What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley?

On the top of the hill, the centripetal force is downward. So we have $$F_N-mg-m(\frac{v^2}{R})=0$$ since we don't have motion in the vertical direction. So when $$F_N=0$$ we have $$-g=\frac{v^2}{R}$$ but this is impossible since the right-hand side is positive but the left is not. Why is that?

You got the signs wrong - the centripetal acceleration is towards the center of motion and enters the equation from the other side. This is a common mistake in analyzing free-body diagrams - you must include only the forces that act on the object. For every force you include in the free-body diagram you must be able to answer the following question:

What object is exerting this force?

In case of weight that object is Earth, and in case of normal force that object is the surface. What would be the object that exerts the centripetal force? In the second Newton's law

$$\sum_{i} \vec{F}_i = ma$$

the left-hand side is the vector sum of the forces from the free-body diagram, and the right-hand side is the consequence (motion). In your example, the centripetal force is the consequence.

If you define the positive direction for the vertical ($$\hat{\jmath}$$) axis to be upwards (away from the Earth's center), then the free-body diagram would show

$$\vec{n} + \vec{w} = \vec{F}_\text{net} \quad \rightarrow \quad N \hat{\jmath} - W \hat{\jmath} = \vec{F}_\text{net}$$

where $$N$$ is magnitude of the normal force, $$W$$ is magnitude of the weight, and $$\vec{F}_\text{net}$$ is the net force vector. We know that for the circular motion the net force points towards the center of motion (also known as the centripetal force), hence at the top of the hill the net force is

$$\vec{F}_\text{net} = -m\frac{v^2}{R} \hat{\jmath}$$

from which we could easily calculate the magnitude of the normal force

$$N = W - F_\text{rad} = mg - m\frac{v^2}{R} = m \Bigl( g - \frac{v^2}{R} \Bigr)$$

The car loses contact with the road (surface) when the normal force is zero, or when the velocity is

$$\boxed{v = \sqrt{g \cdot R}}$$

At the bottom of the hill the net force is

$$\vec{F}_\text{net} = m \frac{v^2}{R} \hat{\jmath}$$

Assuming the velocity did not change from the top to the bottom of the hill, it is trivial to calculate the normal force at the bottom.

• Isn't the weight and the centripetal force in the same direction?
– Mina
Commented Jan 18, 2022 at 15:47
• It is, but centripetal force is a consequence and not the reason. This is a common mistake - when you draw the free-body diagram, you should not draw the resultant force which is in this case the centripetal force. Commented Jan 18, 2022 at 15:48
• I get it now. Thanks a lot.
– Mina
Commented Jan 18, 2022 at 15:51
• @Mina When you draw the free-body diagram, for every force you draw you must answer one important question: what other object is exerting this force? In case of weight it is the Earth, and in case of normal force it is the surface. What would be the object exerting the centripetal force? Commented Jan 18, 2022 at 16:05
• If I could give five upvotes to the comments by @MarkoGulin I would. Centripetal force is very commonly misunderstood by novices. Be sure to understand what he is saying. Commented Jan 18, 2022 at 18:26