What are the units for centrifugal force and how do you calculate it?

Click For Summary

Discussion Overview

The discussion revolves around the units and calculations related to centrifugal force, exploring its relationship with centripetal force, and the mathematical derivations involved. Participants express uncertainty about the correct application of formulas and the interpretation of forces in rotating systems.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants reference the equation F = d/dt (m*v) but express confusion about its application and the need for derivatives.
  • Others suggest that if velocity is constant, there is no net force, leading to discussions about the correct form of the force equation, F = m dv/dt.
  • One participant proposes using the equation for centripetal force, v² = F * R/M, to calculate centrifugal force, questioning the distinction between the two forces.
  • Clarifications are made regarding the MKS (meters, kilograms, seconds) system and its relation to SI units, with some noting that centrifugal force is given by F = m*v²/R.
  • Participants discuss the concept of centrifugal force as a fictitious force in rotating frames and the necessity of understanding centripetal acceleration instead.
  • There are inquiries about calculating the force exerted by an object at various points in a circular path, with some participants seeking help with the derivation of relevant equations.
  • Discussions include the conversion of weight in pounds to mass in slugs and the implications for calculating forces in different unit systems.
  • One participant attempts to calculate centrifugal force using specific values and questions the correctness of their units, leading to further clarification about the relationship between pounds and slugs.
  • There is a query about whether centrifugal force is equivalent to centripetal force, highlighting the ongoing debate about the nature of these forces.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and agreement on the definitions and calculations related to centrifugal and centripetal forces. There is no consensus on the interpretation of centrifugal force, with some viewing it as fictitious while others seek to equate it with centripetal force.

Contextual Notes

Limitations include potential misunderstandings of the terms used, the dependence on the choice of coordinate systems, and unresolved mathematical steps in deriving the equations discussed.

cgaday
Messages
57
Reaction score
0
I found this on Wikipedia for the equation relating to centrifugal force.

F = d/dt (m*v)

So I have my mass and velocity but I just multiplied the two, that can't be correct can it. You have to derive something I imagine. Also would the units just be in lbs or Newtons. I'm using feet/s and lbs.

Help I'm pretty rusty on my calculus.
 
Physics news on Phys.org
cgaday said:
I found this on Wikipedia for the equation relating to centrifugal force.

F = d/dt (m*v)

So I have my mass and velocity but I just multiplied the two, that can't be correct can it. You have to derive something I imagine. Also would the units just be in lbs or Newtons. I'm using feet/s and lbs.

Help I'm pretty rusty on my calculus.

If your velocity is constant, there is no net force (no acceleration).

A better way to write it (assuming constant mass) is F = m dv/dt

So the units in mks are Newtons = kg m / s^2
 
The odd thing I found is was this equation for centripetal force. where v^2 = F * R/M

Radius and mass.

Couldn't I just use this equation for centrifugal force since all centrifugal and centripetal are is the name for the direction which the force is going on the wheel?
 
what is mks. how do you perform that equation?
 
cgaday said:
what is mks. how do you perform that equation?

mks = meters, kilograms, seconds. It's one of the two metric systems of units. The other is cgs = centimeters, grams, seconds.

If you mean the F = m dv/dt equation, the dv/dt is a timie derivative of velocity. That is, how much the velocity chages over time, which is called the acceleration.

v = dx/dt (velocity is change in position with time, in mks units of m/s)
a = dv/dt (acceleration is change of velocity with time, in mks units of m/s^2)
 
SI is the more common name for MKS.

Centrifugal force is given by F = m*v²/R. Thats kg*(m/s)²/m = kg*m/s² = N.

So the units are exactly the same as these of F = ma.
 
Sorry I'm a little lost on figuring out the derivative. How do you set that up?
 
cgaday said:
Sorry I'm a little lost on figuring out the derivative. How do you set that up?
What exactly are you trying to calculate?
 
The centrifugal force of an object based upon the radius, weight or mass of the object.

In the long run I would like to calculate the force exerted by the object at any point in the radius of the circle while rotating around at a constant velocity.

In this case would we now be talking about acceleration?
 
  • #10
cgaday said:
The centrifugal force of an object based upon the radius, weight or mass of the object.
Realize that 'centrifugal force', at least in standard physics usage, is a fictitious force that only exists when analyzing motion from a rotating frame. You don't need centrifugal force to understand rotation.

In the long run I would like to calculate the force exerted by the object at any point in the radius of the circle while rotating around at a constant velocity.
OK. You can easily calculate the centripetal force acting on the object. And for each force acting on the object there will be a corresponding force exerted by the object.

In this case would we now be talking about acceleration?
Anything moving in a circle will have a centripetal acceleration.
 
  • #11
I understand the rotation, I would just like to know what the force of that object would be at different points from the center as it rotates. Like when you twirl a rock on a string around your finger, as you bring it in I'm sure the force changes and so does the speed. Am I making any sense?

I figured that the centripetal force would be ficticious if any, how do you even get an object to go to the inside of a rotation. Don't want to get off topic, but maybe this is a goodtime for a description of the two.
 
  • #12
It's just a choice of the coordinate system. If your coordinate system rotates, you have centrifugal force. If it doesn't, there isn't one.

If you are standing on a rotating disk, it's convenient for you to define your position relative to disk. That's your choice of coordinate system, and you then end up observing outwards pull you call Centrifugal Force. That being canceled by the friction with the ground, you remain stationary in that system.

If, however, you decide to take your bearings relative to an object that's outside the disk, you note that you aren't just standing still. You are moving. In fact, you are constantly accelerating towards the center of the disk. The force that provides that acceleration is still the same friction with the ground that was countering Centrifugal Force in the rotating frame.

So as long as you choose coordinate system without rotation, you really don't need centrifugal force.
 
  • #13
Simply put I just would like to find the force of a single object rotating in a circle relating to weight, RPM or any omega, and distance from center.

Would the equation below work for what I am trying to do.


K^2 said:
SI is the more common name for MKS.

Centrifugal force is given by F = m*v²/R. Thats kg*(m/s)²/m = kg*m/s² = N.

So the units are exactly the same as these of F = ma.

Or do I need to derive something, If this is the case, I could use some help.
 
  • #14
cgaday said:
Simply put I just would like to find the force of a single object rotating in a circle relating to weight, RPM or any omega, and distance from center.

Would the equation below work for what I am trying to do.




Or do I need to derive something, If this is the case, I could use some help.

Yes the equation F= m v^2/r will work just fine if you use SI units of kg, meters and seconds (which of course means the force is in Newtons).

Feel free to ask if you need to know how that equation is derived.
 
  • #15
cgaday said:
Simply put I just would like to find the force of a single object rotating in a circle relating to weight, RPM or any omega, and distance from center.

Would the equation below work for what I am trying to do.
For a mass m moving in a circle of radius r at constant speed v (or angular speed ω), the net force on it (also known as the centripetal force) is given by:

Fc = mv²/r = mω²r
 
  • #16
What units would that be, I would initially think lbs. but it how does the ft/s cancel.

lbs * (ft/s)^2/ft = lbs*ft/s^2 that can't be right. what am I forgetting?
 
  • #17
cgaday said:
What units would that be, I would initially think lbs. but it how does the ft/s cancel.

lbs * (ft/s)^2/ft = lbs*ft/s^2 that can't be right. what am I forgetting?
You are forgetting that a pound is a unit of force (for example, weight) not mass. In the English system the unit of mass is the slug.

slug * ft/s^2 ≡ lbs

At the Earth's surface, an object's weight in pounds is related to its mass in slugs via: Weight = mass*g = mass * (32 ft/s^2)

In the SI system:

kg * m/s^2 ≡ Newtons
 
  • #18
So for example:

you have an item which ways 5lbs which would equal 5/32.2 = .15528 slugs.

your velocity is 50ft/s and your radius is 2ft.

So according to this equation. Fc = mv²/r = mω²r

Fc = (.15528 slugs*50ft/s ^2)/2 ft = 194 (lb*f*s^2/ft)(ft^2/s^2)/(ft) = 194 lb right.

And this force would be towards the center of the cirlce?
 
  • #19
cgaday said:
So for example:

you have an item which ways 5lbs which would equal 5/32.2 = .15528 slugs.

your velocity is 50ft/s and your radius is 2ft.

So according to this equation. Fc = mv²/r = mω²r

Fc = (.15528 slugs*50ft/s ^2)/2 ft = 194 (lb*f*s^2/ft)(ft^2/s^2)/(ft) = 194 lb right.
Right.

And this force would be towards the center of the cirlce?
Yes. This is the net force towards the center required to make it go in a circle at that speed and radius.
 
  • #20
To determine the centrifugal force, would it be the same as the centripetal? equal opposing reacting force?
 
  • #21
cgaday said:
To determine the centrifugal force, would it be the same as the centripetal? equal opposing reacting force?
What do you mean by centrifugal force? In standard physics usage, centrifugal force is a fictitious force that only appears when analyzing motion from a rotating reference frame. I assume that's not what you mean, since you mentioned 'reacting force'.

Some people do use the term 'centrifugal force' for the 'reaction force' to the centripetal force, but that poses problems in complicated cases. In a simple case, such as a ball being twirled on a string, the string exerts a centripetal force on the ball and the ball, per Newton's 3rd law, exerts an equal and opposite force on the string. I assume that's what you mean?
 
  • #22
From reading other forums, I have been confused on the fictional meaning of the force. I was just wanting to know how to determine the outward force which I thought was called centrifugual force.

In your example with the string, wouldn't there be a stretching force on the string towards the outside of the circle. and what would that force be called.
 
Last edited:
  • #23
cgaday said:
In your example with the string, wouldn't there be a stretching force on the string towards the outside of the circle.
Sure (as I described). There's no need to give it a name. Some call it the reactive centrifugal force.
 
  • #24
how do you go about doing that?
 
  • #25
cgaday said:
how do you go about doing that?
Doing what?
 
  • #26
determining that force acting on the string? by the object, or determining how much force the object is putting on the string. either way.
 
  • #27
cgaday said:
determining that force acting on the string? by the object, or determining how much force the object is putting on the string. either way.
Well, you know how to calculate the force the string exerts on the ball. That's the centripetal force, given by mv²/r. The force that the ball exerts on the string is equal and opposite. (That's Newton's 3rd law.)
 
  • #28
alright. Thanks very much.
 

Similar threads

High School Centrifugal Force on Rotating 4" tire / wheel @ 100mph
  • · Replies 15 ·
Replies
15
Views
3K
Undergrad Remark on centrifugal force in Heino Falcke's black hole book
  • · Replies 22 ·
Replies
22
Views
3K
Undergrad How Do You Calculate Bank Angle to Counteract Centrifugal Force in Turns?
  • · Replies 3 ·
Replies
3
Views
3K
High School Confusion while trying to build intuition of centripetal force
  • · Replies 15 ·
Replies
15
Views
4K
Undergrad Why Doesn't Centrifugal Force Act in the Direction of Velocity?
  • · Replies 10 ·
Replies
10
Views
6K
Graduate Using a centrifuge to extract heavy water (help with the calculations please)
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad Argument on centrifugal force
  • · Replies 23 ·
Replies
23
Views
5K
Undergrad What is the meaning of the free surface in fluid mechanics?
  • · Replies 11 ·
Replies
11
Views
2K
High School What is the Net Force on a Banked Curve?
  • · Replies 3 ·
Replies
3
Views
2K
High School What is the difference between centrifugal and centripetal forces?
  • · Replies 17 ·
Replies
17
Views
14K