Why is torque defined as $\vec{r} \times F$?

Question

Here I cannot convince myself myself that it is units because the torque is defined to be in units of Newton meter is a reiteration of the law stated above. Why was it not $r^2 \times F$ or $r^3 \times F$ or $r^2 \times F^2$ etc. The argument "in our experience how much something rotates depends on the lever length and the force applied" is really insufficient. Can someone outline a more rigorous proof or motivation?

Answer 1

It's not like someone said "Ah ha, torque! What should the definition of torque be?" That doesn't make sense; you don't think up a term and then try to assign a definition.

Instead, it was found that this thing $\mathbf r\times\mathbf F$ turned out to be really useful in explaining physical phenomena. Particularly rotational dynamics of systems. So it got its own special definition.

If you want to define $r^2F$ or $r^2F^2$ as something, you can go right ahead. No one will stop you. The goal would then be to show that your new definition has physical significance.

P.S. I think the answers that have "Define torque as the time-derivative of angular momentum" or something like it is showing the physical significance of $\mathbf r\times\mathbf F$ by linking it to other things with physical significance (assuming you think angular momentum and its rate of change is significant). But I think they miss the point of showing that you can't derive or prove definitions and that questions like "why is insert physics term here not equal to insert modified definition here instead?" misunderstand why something would be defined in physics in the first place.

What you are doing here is not actually asking why is torque defined as $\mathbf r\times\mathbf F$, but rather (possibly unaware to yourself) you have some other notion of what torque is/should be, and you want to know why we can go from what you are thinking of to $\mathbf r\times\mathbf F$. Of course, we would need to know what your actual starting point is that you desire in order to get there. Other answers assume rate of change of angular momentum, or use virtual work and equilibrium, but that isn't necessarily what you might be thinking of. The fact that there are a variety of other answers here shows how flawed the question (not the post though) is.

Answer 2

Usually, the motivation behind introducing true is to determine the equilibrium position of a rigid body allowed to rotate about a point. A fundamental theorem of statics says that such a body is at equilibrium iff the resulting torques of all the forces applied on it with respect to the rotation point is zero. Contrary to most derivations, angular momentum is hardly relevant for statics. It is certainly a relevant quantity to consider when a system is invariant by rotation as a consequence of Noether's theorem, but does not shed much insight for statics.

In order to characterise equilibrium, the usual approach for constrained systems is to use the principle of virtual work. Take a set of points indexed by $i$ at position $\vec r_i$, on which are applied respectively the forces $\vec F_i$. This gives first formula of vitual work for a general displacement: $$ \delta W = \sum \vec F_i \cdot \delta\vec r_i $$ Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $\delta\vec r_i =\delta\vec \phi \times \vec r_i$ where $\vec \phi$ is the differential angular displacement. Injecting in the work you get: $$ \begin{align} \delta W &= \sum \vec F_i \cdot \left(\delta\vec \phi \times \vec r_i\right)\\ &= \left(\sum \vec r_i \times\vec F_i \right) \cdot \delta\vec \phi \end{align} $$

The principle of virtual work says that static equilibrium is equivalent to a vanishing virtual work for any possible virtual displacement (obeying the constraints). Therefore, equilibrium is equivalent to: $$ \sum \vec r_i \times\vec F_i = 0 $$ and you naturally introduce torque as the LHS. This proves the aforementioned theorem. It also explains also the useful power formula for rotation (with angular velocity $\vec \omega$): $$ P = \left(\sum \vec r_i \times\vec F_i \right) \cdot \vec \omega $$

Note that the theorem can be extended to any solid body, and you just need to add the prescription that the resulting force is also zero. This guarantees that if torque vanishes about a certain point, it also vanishes about any point.

Answer 3

Torque is change of angular momentum: $$ \vec{\tau} = \frac{d\vec{L}}{dt}$$

Angular momentum is defined as $$ \vec{L} = \vec{r} \times\vec{p} $$

Using the chain rule: $$ \vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times\vec{p})$$

$$\frac{d}{dt}(\vec{r} \times\vec{p}) = \frac{d \vec{r}}{dt}\times \vec{p}+ \vec{r} \times\frac{d \vec{p}}{dt} = \vec{0} + \vec{r} \times\frac{d \vec{p}}{dt} \tag{1}$$

Remembering that $|\vec{a} \times \vec{b} |= |\vec{a}||\vec{b}| \sin \theta$ and that the velocity vector is parallel to the linear momentum vector we get:$$ \frac{d \vec{r}}{dt}\times \vec{p} = \vec{v} \times \vec{p} = \vec{v} \times m\vec{v} = m|\vec{v}| |\vec{v}| \sin \theta \ \hat r=m|\vec{v}| |\vec{v}| \sin 0 \ \hat r = \vec{0}$$

$\theta$ is the angle between two vectors, which is $0$ for any vector with itself.

So we get for torque, using the result in $\text{(1)}$:

$$ \vec{\tau} = \frac{d\vec{L}}{dt} = \vec{r} \times\frac{d \vec{p}}{dt} = \boxed{\vec{r} \times\vec{F}}, \text{ since } \vec{F}=\frac{d \vec{p}}{dt} $$

But the question then reduces to why angular momentum is defined as $ \vec{L} = \vec{r} \times\vec{p} $. I think this has to do with Noether's theorem, that this quantity is conserved when a system stays the same under a change of angle.

Answer 4

Consider a point particle of mass $m$ with velocity $\vec v$. The particle is located at some position $\vec r$ with respect to the origin $O$.

I will start with the angular momentum calculated about $O$. The angular momentum is $\vec L = I\vec \omega$, where $I=mr^2$.

Since we know $v=\omega r$, you can work out that $\vec \omega = \dfrac{\vec r\times \vec v}{r^2}$.

Next, we can substitute that into our angular momentum to get $$\vec L = m \vec r\times \vec v.$$

Define torque as the time-derivative of angular momentum, and we have that $$\vec \tau \equiv \dfrac {d\vec L}{dt}=m\vec r\times \dfrac{d\vec v}{dt}.$$

Since $\vec F=m\vec a$, we then have $$\boxed{\vec \tau =\vec r \times \vec F}$$

Answer 5

If we consider a small arc of length dl (almost 0) in a circle of radius R which subtends a small angle du As arc is almost 0 thus angle subtended will also be approximately 0

As dl is very small thus dl length can be assumed as straight line (as we can assume circle is made of very small straight line kept at some angle) So it forms a triangle with 2 sides as radius and one as dl And considering one angle as 0 the other 2 angle will be 90°

Using sin rule: [sin(du)/dl] = [sin(90)/R]

As we know sin(y)=y if y is very small So sin(du) =du

Thus dl = R(du)

Now differentiating with respect to time we get (angular-velocity)=(linear-velocity)(Radius)

Differentiating with respect to time again (angular-acceleration)=(linear-acceleration)(Radius)

Multiplying by mass We get (angular acc)(mass)=torque And (linear acc)(mass)=force

Torque=(Force)(Radius)

Now the cross product is due to force being vector and in circular motion we take resultant as vector obtained by right hand thumb rule due to convention.