Why do we need to consider all points in a rigid body when calculating torque?

[Tahmeed]

I wanted to know how the ,mathematical concept of Torque comes(Like force comes from Newton's law). So i went and read university physics and feynman's lecture. and here is a problem I had. I am uploading a picture of Feynman's lecture, and stating my problem below about that part.

But since in that pichttps://m.ak.fbcdn.net/photos-d.ak/hphotos-ak-xpf1/v/t1.0-0/s320x320/11836687_950467178349947_4485007967532677838_n.jpg?efg=eyJpIjoiYiJ9&oh=9cac19cabafcc34ce269c7978e08e302&oe=567BD963&__gda__=1447136990_bc04b8adf657c188d5e94fb7a0038b88 we are talking about rigid body, if a force is applied in (x,y) point and that point mass goes a displacement of del x and del y, all other point masses in that rigid body also goes by a displacement as a result of this force. then when we calculate the work, it isn't sufficient to write w=Fx.delx + Fy.dely. because all points of that rigid body moves as a result of this force.

 

[Philip Wood]

A definition of work that I've always found to give consistent results is

Work done by force = Magnitude of force x Distance moved by point of application of force x (cosine of angle between direction of force and direction of displacement of its point of application).

If the force acts on a rigid body, and the rest of the body moves as a result, this is irrelevant. The work done by the force is still as defined above.

 

[mfb]

It does not matter what moves where. This is just the definition of work as force times (infinitesimal) distance.

 

[Tahmeed]

A definition of work that I've always found to give consistent results is

Work done by force = Magnitude of force x Distance moved by point of application of force x (cosine of angle between direction of force and direction of displacement of its point of application).

If the force acts on a rigid body, and the rest of the body moves as a result, this is irrelevant. The work done by the force is still as defined above.

But, other points of rigid body moves as well! So when we consider change in energy or the total work done on the body, we have to count them right?? How is it irrelevant??

 

[Philip Wood]

But, other points of rigid body moves as well! So when we consider change in energy or the total work done on the body, we have to count them right?? How is it irrelevant??

Wrong. Not relevant.

In post 2, I consider the work done by a force. If there's more than one force you can add the quantities of work by scalar addition. That gives you the total work done on the body. There's no need to worry about how the rest of the body moves.

You may be worried about internal forces… Suppose particle A of the body exerts force \mathbf{F}_{BA} on B, so B exerts force \mathbf{F}_{AB} on A where \mathbf{F}_{BA}\ = -\mathbf{F}_{AB}.

Then total work done by these internal forces if A is displaced by \Delta \mathbf{r}_A and B by \Delta \mathbf{r}_B is
\mathbf{F}_{AB} . \Delta \mathbf{r}_A + \mathbf{F}_{BA} . \Delta \mathbf{r}_B = \mathbf{F}_{AB} .\Delta (\mathbf{r}_A - \mathbf{r}_B)
But \Delta (\mathbf{r}_A - \mathbf{r}_B) is zero for a rigid body, so the internal forces don't do any work in total.

 

[A.T.]

... we have to count them...

Go and count all the points on a rigid body. Report back how many there are.

 

[Tahmeed]

Go and count all the points on a rigid body. Report back how many there are.

All I have to do is integrate the energy in different points. I don't have to literally 'count'.

 

[Tahmeed]

Wrong. Not relevant.

In post 2, I consider the work done by a force. If there's more than one force you can add the quantities of work by scalar addition. That gives you the total work done on the body. There's no need to worry about how the rest of the body moves.

You may be worried about internal forces… Suppose particle A of the body exerts force \mathbf{F}_{BA} on B, so B exerts force \mathbf{F}_{AB} on A where \mathbf{F}_{BA}\ = -\mathbf{F}_{AB}.

Then total work done by these internal forces if A is displaced by \Delta \mathbf{r}_A and B by \Delta \mathbf{r}_B is
\mathbf{F}_{AB} . \Delta \mathbf{r}_A + \mathbf{F}_{BA} . \Delta \mathbf{r}_B = \mathbf{F}_{AB} .\Delta (\mathbf{r}_A - \mathbf{r}_B)
But \Delta (\mathbf{r}_A - \mathbf{r}_B) is zero for a rigid body, so the internal forces don't do any work in total.

This helped. Thanks :)

 

[Tahmeed]

Wrong. Not relevant.

In post 2, I consider the work done by a force. If there's more than one force you can add the quantities of work by scalar addition. That gives you the total work done on the body. There's no need to worry about how the rest of the body moves.

You may be worried about internal forces… Suppose particle A of the body exerts force \mathbf{F}_{BA} on B, so B exerts force \mathbf{F}_{AB} on A where \mathbf{F}_{BA}\ = -\mathbf{F}_{AB}.

Then total work done by these internal forces if A is displaced by \Delta \mathbf{r}_A and B by \Delta \mathbf{r}_B is
\mathbf{F}_{AB} . \Delta \mathbf{r}_A + \mathbf{F}_{BA} . \Delta \mathbf{r}_B = \mathbf{F}_{AB} .\Delta (\mathbf{r}_A - \mathbf{r}_B)
But \Delta (\mathbf{r}_A - \mathbf{r}_B) is zero for a rigid body, so the internal forces don't do any work in total.

Oops, came up with a problem. As shown in the pic taken from University Physics below, we are counting torque on all of the particles of the rigid body, if there is no work done on other particles other than the point force is applied, how there is torque working on other points?? Torque derives from work, right??

 

[Philip Wood]

Torque derives from work, right??

Not a clear enough statement to be useful, imo. Torque is the cross product of force and distance from a chosen reference point; work is the dot product of force and distance moved by point of application of force.

Feynman chose to introduce torque via work. There's no compulsion to do so.

I re-iterate that the work done on the body is the sum of the work done by individual forces on it, and these are calculated 'locally' as described. That's not to deny that these forces affect the whole body, making it move, perhaps, by both rotation and translation, but that's nothing to do with how we calculate work done on the body.

 

[Tahmeed]

Not a clear enough statement to be useful, imo. Torque is the cross product of force and distance from a chosen reference point; work is the dot product of force and distance moved by point of application of force.

Feynman chose to introduce torque via work. There's no compulsion to do so.

I re-iterate that the work done on the body is the sum of the work done by individual forces on it, and these are calculated 'locally' as described. That's not to deny that these forces affect the whole body, making it move, perhaps, by both rotation and translation, but that's nothing to do with how we calculate work done on the body.

How would you derive mathematically Torque without differentiating work??

 

[Tahmeed]

I am kinda confused. If a single force is applied on a single point of an extended body, Does this force create torques in all points of the body about rotational center??

 

[stockzahn]

Hi,

first of all I'm not sure there are confusions with terms. Torque changes the rotational speed of a body, whereas m⋅r² is the moment of inertia, which is more or less the resitance of a body against acceleration (or deceleration) in its rotational motion by torque. In your first post you wrote, that you want know the physical concept of torque like the laws of Newton for force. In fact you can compare them very easily:

F = m⋅a ⇔ T = I ⋅ α''

If you use the force F to accelerate a body, the size of the acceleration a depends on the mass m of the body - the larger the mass, the smaller the acceleration. The mass is the resistance of the body against beeing set in (linear) motion.

If you use the torque T to set a body rotating, the size of the rotational acceleration φ'' depends on the moment of inertia I of the body - the larger the moment of inertia, the smaller the acceleration. The moment of inertia is the resistance of the body against beeing set in rotation.

The parts of the formulas corrispond to each other

F ⇔ T
m ⇔ I
a ⇔ α''

The moment of inertia not only has to take in account the mass of the body which should be set rotating, but also the its distance from the rotational axis. You can think of that very easily if you compare two hollow cylinders (pipes) of the same mass and the same length, but with different diameters. You will need a higher torque to set the larger pipe rotating at the same speed in the same time, as the mass has to move faster to turn around one time when rotating. This moment of inertia is not as easy to find for a "natural" body (for simple bodies (spheres, cylinders, ...) you will find them and also can calculate them as mentioned in the text you posted - it's the some of all Δm⋅r² in a body). Of course a certain torque will set rotating the whole rigid body, but its final rotational speed depends on its moment of inertia.

Now the torque itself is the (cross!) product of a force and the normal distance l (=length of lever) between the force and the rotational axis.

T = F⋅l

To calculate the torque there are several way:

1) You can measure the rotational speed φ' of the body and the time t you needed to accelerate it (assuming constant torque/acceleration).

α'' = α' / t

Knowing you have accelerated every point of the body (due to its rigidity), you can calculate the torque, if you know the momentum of inertia.

T = I ⋅α''

With this approach you found the torque "using" all the points set in motion.

2) The Torque itself comes from a force which not necesserily acts over the whole body, but maybe just in one point (like when you open a door, you only push the door knob not the whole surface of the door). This torque must be same as calculated in 1), but now with a different approach:

T = F⋅l

Now you don't need to know the speed of all points of the body, the force and the lever are sufficient for calculating the torque.

3) A third way would be the calculation with the help of the energy consistency (now I'm referreing to your first post: "... it isn't sufficient to write w=Fx.delx + Fy.dely. because all points of that rigid body moves as a result of this force.") The energy transferred in a body, which was rotated is

T⋅α = W_rot (compare for linear motion: F⋅s = W_lin )

Due to the energy consistency, the work done by the force F, must be the same as the energy input into the body.

W_rot = W_lin
T⋅α = F⋅s
T = F⋅s / α

So it is sufficient only to use the work done by the force and it's displacement to calculate the torque, if you know the angle α, which the body has rotated. Of course you could transfer it in a formula using all the moved mass points:

T = I⋅α''
F⋅s = I⋅α''⋅α = τ⋅α''

 

[Philip Wood]

How would you derive mathematically Torque without differentiating work

Depends what you mean by 'derive torque'.

If you mean: how do you derive the concept of torque?, it's a definition, which therefore can't be derived, though it can be motivated – which would usually be done by discussing crowbars and suchlike, though you could come at it from work considerations if that appeals to you.

If you mean: how do you calculate torques?, then it would depend what other information you had. Usually it would be forces and their points of application.

 

[Tahmeed]

Hi,

first of all I'm not sure there are confusions with terms. Torque changes the rotational speed of a body, whereas m⋅r² is the moment of inertia, which is more or less the resitance of a body against acceleration (or deceleration) in its rotational motion by torque. In your first post you wrote, that you want know the physical concept of torque like the laws of Newton for force. In fact you can compare them very easily:

F = m⋅a ⇔ T = I ⋅ α''

If you use the force F to accelerate a body, the size of the acceleration a depends on the mass m of the body - the larger the mass, the smaller the acceleration. The mass is the resistance of the body against beeing set in (linear) motion.

If you use the torque T to set a body rotating, the size of the rotational acceleration φ'' depends on the moment of inertia I of the body - the larger the moment of inertia, the smaller the acceleration. The moment of inertia is the resistance of the body against beeing set in rotation.

The parts of the formulas corrispond to each other

F ⇔ T
m ⇔ I
a ⇔ α''

The moment of inertia not only has to take in account the mass of the body which should be set rotating, but also the its distance from the rotational axis. You can think of that very easily if you compare two hollow cylinders (pipes) of the same mass and the same length, but with different diameters. You will need a higher torque to set the larger pipe rotating at the same speed in the same time, as the mass has to move faster to turn around one time when rotating. This moment of inertia is not as easy to find for a "natural" body (for simple bodies (spheres, cylinders, ...) you will find them and also can calculate them as mentioned in the text you posted - it's the some of all Δm⋅r² in a body). Of course a certain torque will set rotating the whole rigid body, but its final rotational speed depends on its moment of inertia.

Now the torque itself is the (cross!) product of a force and the normal distance l (=length of lever) between the force and the rotational axis.

T = F⋅l

To calculate the torque there are several way:

1) You can measure the rotational speed φ' of the body and the time t you needed to accelerate it (assuming constant torque/acceleration).

α'' = α' / t

Knowing you have accelerated every point of the body (due to its rigidity), you can calculate the torque, if you know the momentum of inertia.

T = I ⋅α''

With this approach you found the torque "using" all the points set in motion.

2) The Torque itself comes from a force which not necesserily acts over the whole body, but maybe just in one point (like when you open a door, you only push the door knob not the whole surface of the door). This torque must be same as calculated in 1), but now with a different approach:

T = F⋅l

Now you don't need to know the speed of all points of the body, the force and the lever are sufficient for calculating the torque.

3) A third way would be the calculation with the help of the energy consistency (now I'm referreing to your first post: "... it isn't sufficient to write w=Fx.delx + Fy.dely. because all points of that rigid body moves as a result of this force.") The energy transferred in a body, which was rotated is

T⋅α = W_rot (compare for linear motion: F⋅s = W_lin )

Due to the energy consistency, the work done by the force F, must be the same as the energy input into the body.

W_rot = W_lin
T⋅α = F⋅s
T = F⋅s / α

So it is sufficient only to use the work done by the force and it's displacement to calculate the torque, if you know the angle α, which the body has rotated. Of course you could transfer it in a formula using all the moved mass points:

T = I⋅α''
F⋅s = I⋅α''⋅α = τ⋅α''

Depends what you mean by 'derive torque'.

If you mean: how do you derive the concept of torque?, it's a definition, which therefore can't be derived, though it can be motivated – which would usually be done by discussing crowbars and suchlike, though you could come at it from work considerations if that appeals to you.

If you mean: how do you calculate torques?, then it would depend what other information you had. Usually it would be forces and their points of application.

Thanks a lot Mr. Wood and Stockzahn. :)

And by derivation I actually meant how did we find out rotational analog of force is Torque= r cross f. However got my answer, thanks.

 

[A.T.]

All I have to do is integrate the energy in different points. I don't have to literally 'count'.

Since the force doesn't act on the other points, the integral over them would be zero.

 

[A.T.]

I am kinda confused. If a single force is applied on a single point of an extended body, Does this force create torques in all points of the body about rotational center??

No, the other points are accelerated by internal forces.

 

[CWatters]

Isnt this the same as asking...

When a force pushes a block up a slope how does the force act on all points within the block?