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I apologize if this question is dumb, but I've looked all over for a straightforward answer, and either I can't find one or either the terms are too complex for me to understand. I have only a rudimentary knowledge of Mechanics, but I do understand basic Linear Algebra.

So torque, mathematically, is the cross product of the radial distance vector and a force vector. This cross-product gives another vector that is orthogonal to both vectors and it points either outside or towards the "page" (in the context of a two-dimensional diagram).

Assuming this is correct, I do not understand what it pointing in or out means. Does it even have a physical, intuitive meaning?

The best answer I've been able to come up with is that it's just a mathematical convention with no actual physical meaning, meant to provide a framework within which operations between torque vectors, such as addition and subtraction, make sense.

Am I correct or way off the mark here?

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  • $\begingroup$ Yes, it's a convention. The 'interpretation' is that the rotation will be curling around the axis defined by the torque in the same way the RIGHT hand curls around the thumb (right-hand rule). $\endgroup$
    – Danu
    Oct 30, 2013 at 22:00
  • $\begingroup$ @Danu So, if I understand correctly, it means that if your right hand would hold the torque vector, which is the axis of rotation, with it's thumb pointing outwards from what is being rotated, and you would bend your wrist forward, it would rotate in a certain direction. And if the torque vector was facing the opposite direction, the forward motion of the right hand would produce a movement in the opposite direction due to the hand itself having changed direction to mantain the rule of the thumb pointing outwards. Would this be correct? Thank you very much. $\endgroup$
    – xxe
    Oct 30, 2013 at 22:28
  • $\begingroup$ Yes, I believe so. This right-hand rule is something that pops up quite often when dealing with cross products (for instance, the direction of the magnetic field due to a current). $\endgroup$
    – Danu
    Oct 30, 2013 at 22:32
  • $\begingroup$ To really get to the bottom of this, you must grok that torque is actually a pseudo-vector. Indeed, in the relativistic context, angular momentum is rank 2 tensor, not a vector: en.wikipedia.org/wiki/Relativistic_angular_momentum So... to get a handle on the physical meaning of this, you've got some homework to do. $\endgroup$ Oct 30, 2013 at 23:21

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As in the comments, there's certainly something of a convention at work here and it's to do with the "co-incidence" that we live in three spatial dimensions.

As in Greg's answer, torque is intimately linked with angular momentum through Euler's second law. That is, torque and angular momentum are about rotational motion. And rotations, in general, are characterized by the planes that they rotate together with the angles of rotation for each of these planes. In three dimensions, the plane of rotation can be defined by a single vector - namely the vector orthogonal to the plane. So we have the concept of the "axis" of rotation, but this is not general, its simply that a line happens to be the subspace of a three dimensional vector space that is orthogonal to the plane of rotation. In four and higher $N$ spatial dimensions, the concept of an axis is meaningless: not only does an axis not specify a plane (the space orthogonal to a plane is of dimension $N-2$), but also a general rotation rotates several planes (up to and including the biggest whole number less than or equal to $N/2$).

So the "true" information specifying a three dimensional rotation is the "bivector" $A\wedge B$, where $A, B$ are linearly independent vectors defining the plane, and a bivector is an abstract directed "plane" just like a "vector" is an abstract directed "line". Cross products in three dimensions are actually bivectors, not vectors, but we can get away with thinking of them as such in three dimensions.

Some further reading to help you out: the Wikipedia pages Plane Of Rotation, Rotation Matrix and Orthogonal Group (rotation matrices form the group $SO(N)$, the group of orthogonal matrices with unit determinant).

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While the notion of the 'meaning' is somewhat subjective so this may not provide any more meaning but, you should think about the origin of the torque. Namely as the rate of change of the angular momentum: $$\textbf{N} = \frac{d\textbf{L}}{dt},$$ Now the fact is orthogonal to the plane containing the position vector and force vector is a direct result of the definition of angular momentum $$\textbf{L} = \textbf{r} \times \textbf{p} $$ where $\textbf{p}$ is the linear momentum. So one can see the torque as the vector acting on the angular momentum.

This is where I claim that it is not a convention, but a requirement that the angular momentum and hence torque lie perpendicular to the plane. Imagine you could define it in some other way, where it didn't lie perpendicular to this plane, then for a constant angular momentum (e.g no torque) the angular momentum vector would have to rotate with the system and would not be constant! For this reason the choice of orthogonality, is not really a choice or convention but it describes the system.

This is my own personal interpretation so I don't claim it as exactly true in any sense.

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In electrical engg, itis very clear in that the current carrying conductor and the magnetic field produced by it are mutually perpendicular to each other.In mechanics it is very difficult to assign a physical meaning for directions of angular velocity,angular momentum and torque.

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When we said a direction we mean the motion is just in that direction .So torque is a vector quantity appears in the normal direction of RXF plane and not appear in any other angle than 90 degree. That is my understanding

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When I was taught about torque it seemed lifeless to me having no physical significance unlike other quantities- force, velocity etc. I developed my own intuition about torque after some analysis. I think the direction of torque is indicating the axis around which the object rotates. Also the direction in which we curl our fingers(in right hand thumb rule) indicates sense of rotation of the object around the axis. enter image description here

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The geometry of torque has nothing to do with motion and the equations of motion. You can have torques in statics and their magnitude and direction are important and insightful. You do need to consider a torque vector, as well as any force vector, applied together to extract the geometry of the situation.

Torque gives us the line of action of force.

A force vector $\boldsymbol{F}$ gives us the magnitude of the force, as well as the direction it acts upon. What it does not give us, is the location in space where the force is applied. This line of action as it is called is only available from the torque this force produces $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$.

A point on the line of action closest to the reference location where torque is measured is found by

$$ \boldsymbol{r}_{\rm action} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau}}{ \| \boldsymbol{F} \|^2} \tag{1}$$

The magnitude of torque is a measure of the perpendicular distance to the line of action, and the direction of torque is perpendicular to both the force direction and the location of the line of action. It is also the same direction velocity would be pointing of the line of action was a rotation axis. The math is identical, as shown in the linked answers below.

Just as the torque of a force $\boldsymbol{F}$ located at $\boldsymbol{r}$ is $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$, the velocity of a rigid rotating about $\boldsymbol{r}$ is also $\boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega}$. If you can understand and visualize the velocity field of a rotating body, you can understand and visualize the torque field of a force vector.


Related Answers

  1. https://physics.stackexchange.com/a/518467/392
  2. https://physics.stackexchange.com/a/516069/392

Proof of expression (1)

Transfer the torque from the reference point to the line of action and show that the torque on the line of action is zero (use vector triple product identity).

$$\require{cancel} \begin{aligned} \boldsymbol{\tau}_{\rm action} & = \boldsymbol{\tau} - \boldsymbol{r}_{\rm action} \times \boldsymbol{F} \\ & = \boldsymbol{\tau} - \frac{(\boldsymbol{F} \times \boldsymbol{\tau})}{\| \boldsymbol{F} \|^2} \times \boldsymbol{F} \\ & = \boldsymbol{\tau} - \frac{ \boldsymbol{\tau}(\boldsymbol{F}\cdot \boldsymbol{F}) - \boldsymbol{F} (\cancel{ \boldsymbol{\tau} \cdot \boldsymbol{F}}) }{\| \boldsymbol{F} \|^2} \\ & = \boldsymbol{\tau} - \frac{ \boldsymbol{\tau} \| \boldsymbol{F}\|^2}{\| \boldsymbol{F} \|^2} = \boldsymbol{0} \end{aligned} $$

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