Centripetal force, this is confusing. Why?

In summary: MOREIn summary, when an object is in uniform circular motion, there must be a net force acting on it that is equal to its mass times its centripetal acceleration. In the case of a rock tied to a string and swung in a vertical circle, the centripetal force is equal to the sum of the gravitational force and the tension in the string. At the top of the circle, it is possible for the gravitational force to provide all of the necessary centripetal force, making the tension in the string zero. This happens when the centripetal acceleration is less than or equal to the acceleration due to gravity. However, if the centripetal acceleration exceeds the acceleration due to gravity, then the tension in the string will always be
  • #1
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18

Homework Statement


Someone takes a rock and ties it to a string, and then starts swinging the rock around in a vertical circle.

The centripetal force should be...

Fc = Fg + FtIt's clear that the book sets Ft=0... but why? Can someone explain this?

Homework Equations


Fc = m * ac
ac = v^2/r

The Attempt at a Solution


But, the book says this is wrong. It implies the centripetal force is Fc = Fg at this point. This is very confusing for me. Can someone explain this? Both forces are pointing towards the middle of the circle, but why do we assume that tension isn't there?

If I use Fc = Fg, I get the same answers the book has.
If I use Fc = Fg + Ft, I get an answer that's off by the amount of tension in the string.

Fc is the net force acting in this direction, which at the top of the swing should be gravity and tension. All of the questions in my Ryerson Physics 12 book which want me to calculate something where the object is at the top of the circle set Ft = 0
 
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  • #2
One thing to observe is that the total centripetal force at the top of circle must be AT LEAST mg. Otherwise there is no circular motion to begin with. This corresponds to some minimal velocity; find it.

At the minimal velocity, the force of tension is indeed zero. At greater velocities, it becomes non-zero. Find the relationship between them.
 
  • #3
Gravity is *always* there and always points vertically downward. So it doesn't always point towards the centre of the circle, but most of the time, some component of it does, and that component contributes to the centripetal force.

Centripetal force is not a physical force that is present in addition to the other forces, gravity and tension. Rather, centripetal force is merely a *kinematic requirement*: IF the object is moving in uniform circular motion at radius r with speed v, THEN there must be a net force on the object equal in magnitude to mv^2 / r. However, this net centripetal force must be a result of some of the physical forces that are present in the problem, and if these forces cease to be available to *provide* a centripetal force, then the object will cease to be in circular motion.

In this problem, the forces summing vectorially to *provide* the centripetal force are gravity and the tension in the string. As I mentioned before, gravity always points vertically downward, which means that its "centripetal" component varies with the position of the mass in the circle. When the mass is at the top of the circle, it happens that "vertically downward" and "towards the centre of the circle" are the same thing. In other words, ALL of the gravitational force points centripetally, and gravity can contribute *maximally* to the centripetal force at this point. So, at this point in the circle, it may be the case that you don't NEED any tension to provide the necessary centripetal force for that circular motion. All of it can be provided by gravity. If that's the case, then the tension will go to zero at this instant. The condition for not needing any tension is simply that the available force from gravity is greater than the required centripetal force for the circular motion:

mg ≥ mv2/r

g ≥ v2/r = ac

So as long as ac ≤ g, gravity will be more than enough to provide the centripetal force at the top of the circle, and no additional contribution from tension will be required. But if you spin the object faster (or in a tighter circle) such that ac > g, then it will be the case that gravity won't be enough to provide the centripetal force at any point in the circle, and the string will always be under tension to provide the shortfall.

EDIT: as voko pointed out, we actually need ac >= mg, else there is no circular motion because the string goes slack. But everything else I said holds. If ac = g, the string tension goes to zero at the top of the loop.
 
  • #4
cepheid said:
Gravity is *always* there and always points vertically downward. So it doesn't always point towards the centre of the circle, but most of the time, some component of it does, and that component contributes to the centripetal force.

Centripetal force is not a physical force that is present in addition to the other forces, gravity and tension. Rather, centripetal force is merely a *kinematic requirement*: IF the object is moving in uniform circular motion at radius r with speed v, THEN there must be a net force on the object equal in magnitude to mv^2 / r. However, this net centripetal force must be a result of some of the physical forces that are present in the problem, and if these forces cease to be available to *provide* a centripetal force, then the object will cease to be in circular motion.

In this problem, the forces summing vectorially to *provide* the centripetal force are gravity and the tension in the string. As I mentioned before, gravity always points vertically downward, which means that its "centripetal" component varies with the position of the mass in the circle. When the mass is at the top of the circle, it happens that "vertically downward" and "towards the centre of the circle" are the same thing. In other words, ALL of the gravitational force points centripetally, and gravity can contribute *maximally* to the centripetal force at this point. So, at this point in the circle, it may be the case that you don't NEED any tension to provide the necessary centripetal force for that circular motion. All of it can be provided by gravity. If that's the case, then the tension will go to zero at this instant. The condition for not needing any tension is simply that the available force from gravity is greater than the required centripetal force for the circular motion:

mg ≥ mv2/r

g ≥ v2/r = ac

So as long as ac ≤ g, gravity will be more than enough to provide the centripetal force at the top of the circle, and no additional contribution from tension will be required. But if you spin the object faster (or in a tighter circle) such that ac > g, then it will be the case that gravity won't be enough to provide the centripetal force at any point in the circle, and the string will always be under tension to provide the shortfall.

EDIT: as voko pointed out, we actually need ac >= mg, else there is no circular motion because the string goes slack. But everything else I said holds. If ac = mg, the string tension goes to zero at the top of the loop.

Thank you, this clears everything up very well.

voko said:
One thing to observe is that the total centripetal force at the top of circle must be AT LEAST mg. Otherwise there is no circular motion to begin with. This corresponds to some minimal velocity; find it.

At the minimal velocity, the force of tension is indeed zero. At greater velocities, it becomes non-zero. Find the relationship between them.

Okay, thank you. This makes sense.
 
  • #5
.

The reason for setting Ft = 0 at the top of the circle is because at that point, the tension in the string is not actually contributing to the centripetal force. The centripetal force is the force that keeps an object moving in a circular path, and it always acts towards the center of the circle. In this case, the centripetal force is provided by the force of gravity (Fg) which pulls the rock towards the center of the circle.

At the top of the circle, the tension in the string is pulling the rock upwards, but it is not contributing to the centripetal force. This is because the direction of the tension force is not towards the center of the circle, but rather it is perpendicular to the centripetal force. Therefore, we can ignore the tension force when calculating the centripetal force.

In other words, at the top of the circle, the rock is moving in a horizontal direction and there is no force acting in that direction to keep it moving in a circular path. The only force acting in the horizontal direction is the force of gravity, which is the centripetal force.

I hope this explanation helps to clarify the concept of centripetal force and why we set Ft = 0 at the top of the circle. Keep in mind that the tension force is still present and is necessary to keep the rock moving in a circular path, but it is not contributing to the centripetal force.
 

Related to Centripetal force, this is confusing. Why?

1. What is centripetal force?

Centripetal force is a force that acts towards the center of a circular path, keeping an object moving in a curved path. It is responsible for keeping objects in a circular motion and preventing them from flying off in a straight line.

2. How does centripetal force work?

Centripetal force works by constantly changing the direction of an object's velocity, keeping it moving in a circular path. This change in direction is caused by a force acting perpendicular to the object's velocity, towards the center of the circle.

3. Why is centripetal force considered confusing?

Centripetal force can be confusing because it is often misunderstood as a separate type of force, when in reality it is simply the net force acting on an object in circular motion. Additionally, the concept can be difficult to grasp because it is not a tangible force that can be directly observed.

4. What are some common examples of centripetal force?

Some common examples of centripetal force include the motion of planets around the sun, the rotation of a ball on a string, and the circular motion of a car around a curve. Any object moving in a circular path is experiencing centripetal force.

5. How is centripetal force related to centripetal acceleration?

Centripetal force and centripetal acceleration are directly related. Centripetal acceleration is the acceleration towards the center of a circular path, and it is caused by the centripetal force acting on an object. The greater the centripetal force, the greater the centripetal acceleration will be.

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