What does the centripetal force equal? (generally)

AI Thread Summary
Centripetal force is defined as the net force acting on an object moving in a circular path, directed toward the center of the circle. In the case of a ferris wheel, the gravitational force acts downward while the normal force varies to provide the necessary centripetal force. The net force is not zero; it equals the centripetal force required to maintain circular motion, as the object's direction changes even if its speed remains constant. The normal force must adjust in magnitude and direction to account for the constant centripetal acceleration. Understanding these forces is crucial for analyzing circular motion dynamics.
Ghost Koi
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Hey guys, new to the forum here and I have a general question regarding the centripetal force. In the example of a ferris wheel where there is a normal force pushing up against the person and the gravitational force pulling the person down, which force is centripetal? I know that the centripetal force counters the linear velocity, tangent to the circle of motion, which allows the object or person to stay in circular motion but which force is actually pulling it towards the center, the gravitational force or the normal force? Also, would it be correct to say that the net force equals zero (since the person is neither moving towards or away from the center) in this example or does the net force equal the centripetal force (since the centripetal force has to counter the linear velocity --- if this is correct, how would I compare the two since linear velocity is not a force)?

I know that if a car is moving around a banked curve, the horizontal normal force will be centripetal but what about in other examples such as the ferris wheel? Also would the net force of a car moving around a bank curved be zero since it is neither moving towards or away the center?

tl;dr - is the net force in a centripetal force example zero or is the net force equal to the centripetal force? Also, how would I relate this to the linear velocity that cancels it out?

Thanks for the help!
 
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Ghost Koi said:
Hey guys, new to the forum here and I have a general question regarding the centripetal force. In the example of a ferris wheel where there is a normal force pushing up against the person and the gravitational force pulling the person down, which force is centripetal?

The word centripetal literally means "toward the center".

Normally when one talks about uniform circular motion, the "centripetal" force that is talked about is the net force -- the sum of all of the forces on the object. This net force must be inward, toward the center (hence the name) if the motion is to be circular.

I know that the centripetal force counters the linear velocity, tangent to the circle of motion, which allows the object or person to stay in circular motion but which force is actually pulling it towards the center, the gravitational force or the normal force?
At the top of the ride, gravity points inward (downward) toward the center of the wheel and the supporting force from the wheel points outward. At the bottom of the ride, gravity points outward (still downward) away from the center of the wheel and the supporting force from the wheel points inward. The net force is inward in both cases.

Also, would it be correct to say that the net force equals zero (since the person is neither moving towards or away from the center)
No. The net force is decidedly non-zero. The person's speed is unchanging. But the direction of their velocity is changing. That is an acceleration and requires a force.
 
Ghost Koi said:
In the example of a ferris wheel where there is a normal force pushing up against the person and the gravitational force pulling the person down, which force is centripetal?
Welcome to PF Ghost Koi!

In the case of the person on the ferris wheel moving at constant angular speed, the only acceleration that he experiences is centripetal acceleration. So the net force is the centripetal force. The forces acting are on him are gravity and the normal force. So the sum of those forces has to constantly change in order to provide the constantly changing centripetal acceleration. Since gravity is constant (ignoring the slight decrease in gravity at the top compared to the bottom), the normal force has to keep changing magnitude and direction. That constant change is part of what makes the ferris wheel fun to ride in.

AM
 
Andrew Mason said:
Welcome to PF Ghost Koi!

In the case of the person on the ferris wheel moving at constant angular speed, the only acceleration that he experiences is centripetal acceleration. So the net force is the centripetal force. The forces acting are on him are gravity and the normal force. So the sum of those forces has to constantly change in order to provide the constantly changing centripetal acceleration. Since gravity is constant (ignoring the slight decrease in gravity at the top compared to the bottom), the normal force has to keep changing magnitude and direction. That constant change is part of what makes the ferris wheel fun to ride in.

AM
So I would always look at the Net Force to find the centripetal force if I have more than one force acting on an object? (I.e. could setup my equation like this)...

m(ν2/r) = FN + Fg
 
Ghost Koi said:
So I would always look at the Net Force to find the centripetal force if I have more than one force acting on an object? (I.e. could setup my equation like this)...

m(ν2/r) = FN + Fg
The net force on a body (which is the vector sum of all the forces) is always equal to its mass x its acceleration:

\sum_i \vec{F}_i = \vec{F}_{net} = ma

If the body is experiencing constant angular velocity circular motion, the acceleration is entirely centripetal and is constant in magnitude. So \sum_i \vec{F}_i = -\frac{mv^2}{r}\hat{r}

AM
 
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Thanks... I think I understand but just to confirm, is my previous statement, and equation, correct?
 
Ghost Koi said:
Thanks... I think I understand but just to confirm, is my previous statement, and equation, correct?

As I said, your equation, \sum_{i}\vec{F}_i = -\frac{mv^2}{r}\hat{r} is correct ONLY IF the acceleration is ENTIRELY centripetal (i.e. constant circular motion). ie. there is no other acceleration.

AM
 
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