Why is the force of the spring equal to the centripetal force?

AI Thread Summary
The discussion centers on the relationship between the force of a spring and centripetal force when an object is rotated in a circular path. It highlights that while both forces point inward, the centripetal force is not an independent force but rather a result of other forces acting on the object. The tension in the string or spring must equal the mass multiplied by the centripetal acceleration to maintain circular motion. Additionally, the concept of centripetal force is clarified as a directional term rather than a distinct force itself. Understanding these dynamics is crucial for calculating potential energy stored in the spring during rotation.
Perseverence
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Homework Statement


A string is rotated around a point with a radius of 4 meters. Calculate the potential energy stored in the spring

Homework Equations


F = KX

The Attempt at a Solution


The solution to the problem involves making the force of a spring equal to the centripetal force. I don't really understand why that would be. Both forces Point towards the center of the circle. How do we know that they are equal? They are not counterbalancing each other if they are both going in the same direction.
 
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But you know that's not the case by experiment as the string stretches and the object moves around in a circular path. Hence while the centripetal force is directed inward, you must look more carefully at the direction of the string tensile force.

http://en.citizendium.org/wiki/Centripetal_force

The figure shows the radially inward centripetal force FC provided by the string that is necessary to force the object to travel the circular path of radius R at a constant speed v. (The "speed" of the body is the magnitude of its velocity, without regard to direction.) The body in turn provides an equal but oppositely directed tensile force FT on the string as shown to the left (a consequence of Newton's law that to every force there is an equal and opposite reaction force), and the string also is subject to the same tensile force at its other end, in opposite direction, as provided by the fixed centerpost.
 
It's a common confusion regarding the centripetal force.
It's not a real force - it's just a value some other force has to have in order to make a mass go around in circles.
So, if you have a mass on a string (or spring) going in circles, then you know that the net radial force is precisely equal to the centripetal force.
 
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Perseverence said:

Homework Statement


A spring is rotated around a point with a radius of 4 meters. Calculate the potential energy stored in the spring.
Just to check you have the right picture in mind: how long is the rotating spring ?
 
Bandersnatch said:
It's a common confusion regarding the centripetal force.
It's not a real force - it's just a value some other force has to have in order to make a mass go around in circles.
So, if you have a mass on a string (or spring) going in circles, then you know that the net radial force is precisely equal to the centripetal force.

Some other force or forces. Sometimes several forces combine to provide the centripetal force needed to make the mass move in a circle.
 
Bandersnatch said:
It's a common confusion regarding the centripetal force.
It's not a real force - it's just a value some other force has to have in order to make a mass go around in circles.
It's probably better to say the centripetal force is not yet another force; rather, it's a resultant force that points radially inward. I wouldn't say it's not a real force, as that might cause confusion with the concept of a fictitious force, like the centrifugal force.

I usually avoid the idea of a centripetal force altogether. When an object follows a circular path, it experiences a centripetal acceleration, ##a_r = \frac{v^2}{r}.## It shows up on the ##ma## side of the second law, not on the ##\sum F## side.
 
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vela said:
I usually avoid the idea of a centripetal force altogether. When an object follows a circular path, it experiences a centripetal acceleration, ##a_r = \frac{v^2}{r}.## It shows up on the ##ma## side of the second law, not on the ##\sum F## side.
I like that. So the tension in the string/rope/etc. is equal to the mass times the centripetal acceleration.
 
I mostly agree with @vela in that "centripetal" is an adjective that indicates the direction of a force relative to some coordinate system, but I wouldn't necessarily call it the resultant. "Centripetal" is used the same way as the adjectives "normal", "horizontal", vertical", etc. to label the direction of forces. For example, the Earth exerts a vertical force on a book that is at rest on a table top; the Earth exerts a centripetal force on the book when it is in a circular orbit around the Earth. Same Earth, same book, same interaction between the two but different reference frames.
 
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