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

[cgaday]

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.

 

[berkeman]

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

 

[cgaday]

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?

 

[cgaday]

what is mks. how do you perform that equation?

 

[berkeman]

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)

 

[K^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.

 

[cgaday]

Sorry I'm a little lost on figuring out the derivative. How do you set that up?

 

[Doc Al]

Sorry I'm a little lost on figuring out the derivative. How do you set that up?

What exactly are you trying to calculate?

 

[cgaday]

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?

 

[Doc Al]

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.

 

[cgaday]

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.

 

[K^2]

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.

 

[cgaday]

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.

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.

 

[uart]

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.

 

[Doc Al]

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:

F_c = mv²/r = mω²r

 

[cgaday]

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?

 

[Doc Al]

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

 

[cgaday]

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?

 

[Doc Al]

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.

 

[cgaday]

To determine the centrifugal force, would it be the same as the centripetal? equal opposing reacting force?

 

[Doc Al]

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?

 

[cgaday]

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.

 

[Doc Al]

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.

 

[cgaday]

how do you go about doing that?

 

[Doc Al]

how do you go about doing that?

Doing what?

 

[cgaday]

determining that force acting on the string? by the object, or determining how much force the object is putting on the string. either way.

 

[Doc Al]

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.)

 

[cgaday]

alright. Thanks very much.