Is there angular force? Tangential force? Why is Tangential F scaled twice for Torque?

[Subliminal]

I'm using a circle diagram to derive the formulae for linear quantities in terms of angular ones.

 

Radius r is used as a scaling device more than anything taken from the know formula for displacement in terms of angular displacement.

 

radius × angular displacement = |linear displacement if object continued on linear trajectory, arch length|

 

(radius × angular displacement)/time = radius×(angular velocity) = |linear velocity, tangential velocity to be precise|

 

(Radius × (angular velocity))/time = radius × (angular acceleration) = tangential acceleration.

 

radius × (angular velocity × mass) = radius × angular force = tangential force.

 

Torque = r × tangential force(mv/t) = (r^2)× angular force?? (mw/t)

 

Thus L = Iw, (I = (r^2 × mass) × angular velocity)

 

Does the direction of the torque change during the resolution? Why is the tangential torque get scaled again?

 

 

Edit: thread tittle should say why "is" tangential force scaled twice

Edited November 30, 2013 by Subliminal

[studiot]

'Scaling' is the wrong word to be using in this context.

 

Your other threads on angular motion suggest you need to understand the basics beofre moving on to more advanced material.

 

The angular motion equivalent of force is called 'moment of momentum' (or angular momentum by some).

 

In your other thread I started to help you with the basics, but you did not reply.

There I observed that the formula for linear distance is the product of two terms whereas the formula for curved distance is the product of three terms.

 

The relationship between force and moment of momentum is similarly complicated.

Edited November 30, 2013 by studiot

[Subliminal]

'Scaling' is the wrong word to be using in this context.

 

Your other threads on angular motion suggest you need to understand the basics beofre moving on to more advanced material.

 

The angular motion equivalent of force is called 'moment of momentum' (or angular momentum by some).

 

In your other thread I started to help you with the basics, but you did not reply.

There I observed that the formula for linear distance is the product of two terms whereas the formula for curved distance is the product of three terms.

 

The relationship between force and moment of momentum is similarly complicated.

Ah sorry for not replying in that thread, you did help me, though my understanding only took me this far, after some research I found that the vector system when torque and L are considered becomes three dimensional, and so I knew immediately that I had been thinking about the situation inaccurately, as I was using two dimensional systems.

 

The reason I used the word "scaling" was because of the fact that the radius was basically the quotient of the arch length and so this waa carried through to the rest of the quantities, angular velocity and angular acceleration alike.

[studiot]

 

Ah sorry for not replying in that thread, you did help me, though my understanding only took me this far, after some research I found that the vector system when torque and L are considered becomes three dimensional, and so I knew immediately that I had been thinking about the situation inaccurately, as I was using two dimensional systems

 

Well you still need to complete basic training.

 

All moments act in a plane so are two dimensional. (yes there are many 2D planes in 3D, but they are still planes.)

 

I know why you said it, but 'Scaling' is the wrong word the radius is a perfectly respectable variable, in its own right.

 

I'm glad the introduction I posted helped.

 

 

I usually pause at some suitable point for the other person to draw breath and assimilate what I have written.

That's what I did in the other thread (as well as this one).

 

The opportunity is then there to acknowledge what has been said and provide more if needed.

 

Alternatively you can carry on doing it the hard way; I wish you the best.

[imatfaal]

'Scaling' is the wrong word to be using in this context.

 

Your other threads on angular motion suggest you need to understand the basics beofre moving on to more advanced material.

 

The angular motion equivalent of force is called 'moment of momentum' (or angular momentum by some).

 

In your other thread I started to help you with the basics, but you did not reply.

There I observed that the formula for linear distance is the product of two terms whereas the formula for curved distance is the product of three terms.

 

The relationship between force and moment of momentum is similarly complicated.

 

Agree everything else - but curious here "The angular motion equivalent of force is called 'moment of momentum' (or angular momentum by some)."

 

I would view the angular analogue of force as torque. ie

velocity <-> angular velocity, m/s <-> rad/sec

momentum <-> angular momentum conserved quantity

force <-> torque --- dp/dt <-> dL/dt

work = force.distance <-> work = torque.angle

[studiot]

 

Agree everything else - but curious here "The angular motion equivalent of force is called 'moment of momentum' (or angular momentum by some)."

 

Hello imatfaal.

 

This thread (and the others) is about a body (of mass m) executing some form of curved or circular motion.

 

As such it can exert a force (which is linear) due to its linear momentum, m, given by F = dm/dt.

 

This linear momentum can also exert a turning effect, which is a true moment, and thus called the moment of momentum.

and is M = r x dm/dt.

 

The difference is that the radius is introduced as an important true variable, not a scaling factor.

Further this approach leads directly to the answer to another question by subliminal - 'where does the expression employing the radius squared come from?', which is what i guess he means by 'scaled twice'

In fact the the only quantity that could have some legitimate claim to the title scaling factor is mass itself.

 

Yes in a more general sense concerning work and energy, moment is also an angular analog of force.

 

But really we should not make comparisons since there are also differences.

 

Personally I do not like to use the term 'torque' since it has a special meaning in engineering and there is often confusion about the turning effects offered by moments, couples and torques, which are all different., which is why we need the different words. Further for consistency with that already noted, moment is a 2D effect, torque is a 3D effect.

 

1) A single force exerts a moment about any point in the plane containing both the force and the point, which may be zero if the force passes through that point.

The moment is different about any other point in that plane.

 

2) A pair of co-planar equal opposed forces exerts a couple about any point in the same plane.

This couple has the same turning effect about any point in that plane. Compare with (1) above.

 

3) If a moment or couple is acting about some point in a plane, the turning effect can be transferred to another plane parallel to the first along a line at right angles to both planes and passing through that point (called the axis of torsion). This effect is then called 'torque'. A screwdriver is a common example.

Edited December 1, 2013 by studiot

[Subliminal]

Hello imatfaal.

 

This thread (and the others) is about a body (of mass m) executing some form of curved or circular motion.

 

As such it can exert a force (which is linear) due to its linear momentum, m, given by F = dm/dt.

 

This linear momentum can also exert a turning effect, which is a true moment, and thus called the moment of momentum.

and is M = r x dm/dt.

 

The difference is that the radius is introduced as an important true variable, not a scaling factor.

Further this approach leads directly to the answer to another question by subliminal - 'where does the expression employing the radius squared come from?', which is what i guess he means by 'scaled twice'

In fact the the only quantity that could have some legitimate claim to the title scaling factor is mass itself.

 

Yes in a more general sense concerning work and energy, moment is also an angular analog of force.

 

But really we should not make comparisons since there are also differences.

 

Personally I do not like to use the term 'torque' since it has a special meaning in engineering and there is often confusion about the turning effects offered by moments, couples and torques, which are all different., which is why we need the different words. Further for consistency with that already noted, moment is a 2D effect, torque is a 3D effect.

 

1) A single force exerts a moment about any point in the plane containing both the force and the point, which may be zero if the force passes through that point.

The moment is different about any other point in that plane.

 

2) A pair of co-planar equal opposed forces exerts a couple about any point in the same plane.

This couple has the same turning effect about any point in that plane. Compare with (1) above.

 

3) If a moment or couple is acting about some point in a plane, the turning effect can be transferred to another plane parallel to the first along a line at right angles to both planes and passing through that point (called the axis of torsion). This effect is then called 'torque'. A screwdriver is a common example.

Hi, thanks again for the help so far.

 

I was thinking about what you said and also did some more reasearch, this subject of moment has really been an issue whenever I studied physics.

 

Does the moment increase with increasing radius because the angular momentum increases with a larger radius even for the same change in angular velocity?

 

That is the change in tangential velocity is greater for the same change angular velocity if the radius is larger. I still don't really know what's going though, all I know is that more KE is transferred due to the increase in v, does that mean transferring energy is easier? If the radius is doubled the KE transferred is doubled and so the angular momentum is doubled meaning the torque was doubled..l

 

Also you said that I shouldn't use "scaling" is that because cross multiplication was occuring and not dot multiplication?

Edited December 1, 2013 by Subliminal

[studiot]

You are running to far ahead of yourself again.

 

And you are under the same misunderstanding in both threads. Here is an excerpt from your other thread.

 

 

I tried to use it to explain why tangential velocity equaled angular velocity ● radius before I knew it was cross multiplication occuring, I am still unsure if what kind of vector "r" represents, that was my original reason for thinking it was just a scaling device

 

The radius r is not a vector. It is true it has some vectorlike properties, but not all of them.

 

A vector has magnitude and direction and two vectors may be added by the parallelogram rule.

 

Whilst a radius has magnitude and direction, You cannot add two radii by the parallelogram rule to get another radius, so radius is not a vector.

 

But then neither is it a scalar, since it need two quantities to define it, whilst a scalar needs only one.

 

This leads directly to your question about scaling.

 

Scaling is said to occur when a quantity eg a vector is multiplied by a fixed scalar (hence the origin of the name scalar).

 

Do you wish to continue my program of starting at the beginning and developing the theory consistently?

 

If so please tell me if you understand my comments on the three different types of turning effect, that you quoted in your last post.

 

I need to know that you are au fait with the stuff I already posted before moving on.

[Subliminal]

.

 

1) A single force exerts a moment about any point in the plane containing both the force and the point, which may be zero if the force passes through that point.

The moment is different about any other point in that plane.

 

2) A pair of co-planar equal opposed forces exerts a couple about any point in the same plane.

This couple has the same turning effect about any point in that plane. Compare with (1) above.

 

3) If a moment or couple is acting about some point in a plane, the turning effect can be transferred to another plane parallel to the first along a line at right angles to both planes and passing through that point (called the axis of torsion). This effect is then called 'torque'. A screwdriver is a common example.

I understand one and two

 

I don't understand number three, does it mean if a moment is acting about a point and something is placed on that point it too will turn?

 

Btw can the two forces which make the couple act on the same point?

Edited December 1, 2013 by Subliminal

[studiot]

OK, now we seem to be talking here are some sketches.

 

Firstly I have drawn an axis (a line in space) perpendicular to a plane and passing through the plane at point P.

 

A force F is shown at a (perp) distance d from P so this single force exerts a moment Fd about P. F exerts a different moment about another point Q in the plane.

 

A mathematician would tell you that F actually exerts a moment, Fd, about the whole line.

But an engineer would say that the moment is only about the point P.

We will return to this at the end.

 

Secondly I have replaced the source of turning with a couple, which comprises two forces.

You should work out an example for yourself to show that the turning moment is the same whatever the location of the couple, unlike the single force example above.

 

Thirdly I have shown what happens when we apply the turning effect in plane 1 (any way you like) and this effect is transmitted down a shaft to plane 2 where it acts on something in plane 2.

This is called torsion and the applied turning effect is called torque.

I have sketched in a screwdriver where we turn the handle and the blade turns a screw.

The handle has a larger radius than the blade so although the turning effect (torque) is the same at both ends, the force applied to the screw is greater than the force we need to turn the handle.

 

An engineer would tell you that there is no torque available beyond the end of the blade but a mathematician would say that the turning effect applied to the entire (infinite) length of the line axis.

 

But then mathematicians are notoriously impractical folk.

 

 

Oh and did you understand what I said about vectors and radii?

Edited December 1, 2013 by studiot

[Endercreeper01]

There is angular force, actually. The torque is τ=Fr. Torque is also the moment of inertia I times the angular acceleration α, T=Iα. The moment of inertia is the sum of mr² for each of the masses in the system, written as Σmr². Since there is a force term in torque for the first equation, we can make the angular force F=Iα/r. Since F=ma, we can also write torque as τ=mar. So yes, there is angular force.

[studiot]

 

There is angular force, actually. The torque is τ=Fr. Torque is also the moment of inertia I times the angular acceleration α, T=Iα. The moment of inertia is the sum of mr² for each of the masses in the system, written as Σmr². Since there is a force term in torque for the first equation, we can make the angular force F=Iα/r. Since F=ma, we can also write torque as τ=mar. So yes, there is angular force.

 

 

And what if there is no angular acceleration?

[Subliminal]

OK, now we seem to be talking here are some sketches.

 

Firstly I have drawn an axis (a line in space) perpendicular to a plane and passing through the plane at point P.

 

A force F is shown at a (perp) distance d from P so this single force exerts a moment Fd about P. F exerts a different moment about another point Q in the plane.

 

A mathematician would tell you that F actually exerts a moment, Fd, about the whole line.

But an engineer would say that the moment is only about the point P.

We will return to this at the end.

 

Secondly I have replaced the source of turning with a couple, which comprises two forces.

You should work out an example for yourself to show that the turning moment is the same whatever the location of the couple, unlike the single force example above.

 

Thirdly I have shown what happens when we apply the turning effect in plane 1 (any way you like) and this effect is transmitted down a shaft to plane 2 where it acts on something in plane 2.

This is called torsion and the applied turning effect is called torque.

I have sketched in a screwdriver where we turn the handle and the blade turns a screw.

The handle has a larger radius than the blade so although the turning effect (torque) is the same at both ends, the force applied to the screw is greater than the force we need to turn the handle.

 

An engineer would tell you that there is no torque available beyond the end of the blade but a mathematician would say that the turning effect applied to the entire (infinite) length of the line axis.

 

But then mathematicians are notoriously impractical folk.

turning1.jpg

 

 

Oh and did you understand what I said about vectors and radii?

So is F exerting a moment about points Q AND P in the first diagram with the moment about P equaling Fd and that about equaling F (some other perp d)?

 

Also the moment of the couple does that only depend on the distance between the two forces? What are the forces acting on?

 

Why does something rotate.

Edited December 1, 2013 by Subliminal

[studiot]

 

So is F exerting a moment about points Q AND P in the first diagram with the moment about P equaling Fd and that about equaling F (some other perp d)?

 

 

 

Yes. F exerts a moment about every point in the plane.

Just be be quite clear is exerts all these moments at the same time.

Most have no effect since there is nothing there to exert a moment on.

This moment is zero if F passes through that point. (Why?)

 

P is a particular point.

 

Q is any other point.

 

Also the moment of the couple does that only depend on the distance between the two forces? What are the forces acting on?

 

 

 

 

No it also depends upon the magnitude of the forces.

 

The two forces have zero resultant since they are equal and opposite.

That is they cancel each other out as forces.

 

But their combined effect still exerts a turning moment about any point in the plane.

 

We differentiate between (1) where there is a net or resultant force acting as well as a turning moment.

 

and (2) where there is no net or resultant force but there is still a turning moment. We call this a couple.

 

A good example is a lawn sprinkler, which spins due to the force from the water jets, but stay put on the lawn because there is not net translational force acting.

 

 

Why does something rotate.

 

Something rotates when there is a net (or resultant) turning moment applied, just like it moves when there is a net (or resultant) force applied.

 

This also shows that for true equilibrium a body has be be separately in force equilibrium and in moment equilbrium.

[Subliminal]

 

Yes. F exerts a moment about every point in the plane.

Just be be quite clear is exerts all these moments at the same time.

Most have no effect since there is nothing there to exert a moment on.

This moment is zero if F passes through that point. (Why?)

 

P is a particular point.

 

Q is any other point.

 

 

No it also depends upon the magnitude of the forces.

 

The two forces have zero resultant since they are equal and opposite.

That is they cancel each other out as forces.

 

But their combined effect still exerts a turning moment about any point in the plane.

 

We differentiate between (1) where there is a net or resultant force acting as well as a turning moment.

 

and (2) where there is no net or resultant force but there is still a turning moment. We call this a couple.

 

A good example is a lawn sprinkler, which spins due to the force from the water jets, but stay put on the lawn because there is not net translational force acting.

 

 

Something rotates when there is a net (or resultant) turning moment applied, just like it moves when there is a net (or resultant) force applied.

 

This also shows that for true equilibrium a body has be be separately in force equilibrium and in moment equilbrium.

 

So does the whole plane rotate in example one?

 

Also with the couple does the rotation occur about the point that is exactly the mid point of the two forces?

 

Is it just a law based off of observation that objects rotate when the force is applied at a distance from the COM?

 

Or can it be explained in terms of distribution of KE across the object?

 

also what determines the axis of rotation is a force is applied at a non COM point of a uniform and non uniform bodies?

 

I remember reading that the moment of a system can be calculated by the some of moments, is it something like that? I'm not sure why I find rotation so confusing.

[studiot]

It is sometimes difficult to conceptualise something that is easy to see in practice.

Or as the old saying goes, An ounce of practice is worth a ton of theory.

 

So I recommend these simple experiments.

 

Take a thin sheet of ply about 6 inches (150mm) square.

 

Score a groove across it, through the centre so that the end of a ruler fits into the groove.

 

Now tack a several nails into it somewhere off the grrove so that the nail protrudes.

 

Lightly fix the sheet to a post or other support with a nail through the centre so that it is free to move. It will probably then hang with acorner down.

 

Firstly demonstrate my condition (1) above by pushing one of the protruding nails with the end of the ruler. Try this with the other protruding nails.

 

Does the sheet rotate whichever nails you push?

 

Now explore my condition (2).

 

Insert the end of the ruler and twist it at various points along the groove. You are applying a couple formed from the two forces at the tips of the ruler pushing against the sides of the groove.

 

Does the sheet rotate in each case?

Is the centre of rotation ever the centre of the couple?

 

 

[Subliminal]

It is sometimes difficult to conceptualise something that is easy to see in practice.

Or as the old saying goes, An ounce of practice is worth a ton of theory.

 

So I recommend these simple experiments.

 

Take a thin sheet of ply about 6 inches (150mm) square.

 

Score a groove across it, through the centre so that the end of a ruler fits into the groove.

 

Now tack a several nails into it somewhere off the grrove so that the nail protrudes.

 

Lightly fix the sheet to a post or other support with a nail through the centre so that it is free to move. It will probably then hang with acorner down.

 

Firstly demonstrate my condition (1) above by pushing one of the protruding nails with the end of the ruler. Try this with the other protruding nails.

 

Does the sheet rotate whichever nails you push?

 

Now explore my condition (2).

 

Insert the end of the ruler and twist it at various points along the groove. You are applying a couple formed from the two forces at the tips of the ruler pushing against the sides of the groove.

 

Does the sheet rotate in each case?

Is the centre of rotation ever the centre of the couple?

 

 

turning2.jpg

Not sure if i did it right.

 

The sheet rotated in the second case but not in the first.

[Endercreeper01]

If there is no angular acceleration, then there will be no tangential force, but there will be a centripetal force.

The centripetal force is mv²/r, where r is radius, m is mass, and v is velocity.