Is the force for centripetal acceleration inversely proportional to the radius?

In summary, the conversation discusses two formulas for centripetal acceleration and how they show the relationship between force, velocity, and radius. The first formula, F=mv2/r, shows that as the radius increases, the speed also increases, resulting in a constant value for the term v2/r. The second formula, F=mω2r, shows that as the radius increases, the centripetal force required also increases. The conversation also addresses the misconception that the force decreases with an increase in radius, and explains that it depends on which variable is held constant.
  • #1
avito009
184
4
Since F=mv2/r. Does that mean the force required for centripetal acceleration is inversely proportional to the radius? If radius is more the lesser centripetal force is required, is this correct?
 
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  • #2
avito009 said:
F=mv2/r
What's "v2?"
 
  • #3
avito009 said:
radius is more the lesser centripetal force is required, is this correct?
That's if v is constant,
In most situations, ##\omega##(angular velocity) is constant, so try expressing centripetal force in terms of ##\omega##
 
  • #4
Now there are 2 formulas for centripetal acceleration. 1. F= mv2/r. So in this formula we know that as radius increases the speed or velocity increases. So the term v2/r remains constant. So how can we prove from this formula that the Centripetal force is more if the radius is more.

But if we take the formula 2. F=mω2r. we can easily prove that as the radius increases the Centripetal force required increases. But this inference can't be derived from the first formula.
 
  • #5
avito009 said:
v2/r remains constant
Think again. What is "v?"
 
  • #6
avito009 said:
F= mv2/r. So in this formula we know that as radius increases the speed or velocity increases
That would be if F is a constant.
Can you specify a situation!
 
  • #7
Bystander said:
Think again. What is "v?"
V is the velocity. But I didnt understand your point.
 
  • #8
What is "v" in terms of "r?"
 
  • #9
Bystander said:
What is "v" in terms of "r?"
V is velocity which is distance/time. So you mean to say that the distance or the radius in this case is more so the velocity is more at the edge. So the Centripetal Force is more. Is this what you meant to tell me?
 
  • #10
In your original question, you've ignored the fact that v equals omega times r and when looking at the dependence of centripetal acceleration on r.
avito009 said:
mv2/r
avito009 said:
inversely proportional to the radius?
 
  • #11
avito009 said:
Now there are 2 formulas for centripetal acceleration. 1. F= mv2/r. So in this formula we know that as radius increases the speed or velocity increases. So the term v2/r remains constant. So how can we prove from this formula that the Centripetal force is more if the radius is more.

But if we take the formula 2. F=mω2r. we can easily prove that as the radius increases the Centripetal force required increases. But this inference can't be derived from the first formula.

I'm tryng to interpretate the law: in the first equation the force, that deviate the particle along the orbit, increases together with the speed if we fix the radius, and it decreases together with the radius if we fix the speed.
In the second equation the force increases togheter with the radius, because fixing the angolar speed and increasing the radius, the speed (not explicit here) must increase and then also the force.
 
  • #12
I said badly "it decreases together with the radius" but I would say "the force decreases when the radius increases in the first equation" (if we fix the speed).

I hope that I can convince you there isn't a paradox about the force vs radius that sometime it increases, and sometime it decreases: it's enough that you fix one of the indipendent variabiles and then you'll se how the force changes when at the same time also the other variabile changes.

In other words, if we fix the angular speed, we know that the periferic speed increases when the radius increases and so the force. But if we fix the periferic speed, the angular speed decreases if the radius increases and then the force decreases.

I'm italian and I don't use to write in english but if I see anything that it's named physics it may be that I could try to speak japanese too.
 
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1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is directed towards the center of the circular path and is responsible for keeping the object moving in a circular motion.

2. How is centripetal acceleration calculated?

The formula for calculating centripetal acceleration is a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

3. What is the difference between centripetal acceleration and centrifugal acceleration?

Centripetal acceleration is the acceleration that keeps an object moving in a circular path, while centrifugal acceleration is the fictitious force that appears to push the object away from the center of the circular path. Centrifugal acceleration is the result of inertia, while centripetal acceleration is caused by a force acting on the object.

4. What are some real-life examples of centripetal acceleration?

Some common examples of centripetal acceleration include the motion of a car around a curved road, the rotation of planets around the sun, and the movement of a roller coaster around a loop.

5. How does centripetal acceleration affect the motion of an object?

Centripetal acceleration causes an object to continuously change its direction, even if its speed remains constant. It also creates a force that is perpendicular to the direction of motion, which is why objects moving in a circular path experience a centrifugal force that pulls them away from the center of the circle.

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