What is the point of torque?

Question

I am having a hard time understanding torque. What is the physical significance of torque? Why is it defined as $\mathbf{r} \times \mathbf{F}$? I know that torque helps us to calculate the angular acceleration of a rotating body, which we can use to calculate the acceleration of each point on the body. But I can do the same by considering only the external forces on the body as well. I can, in principle, draw the free body diagram of every mass element $\mathrm{d}m$ of the body, write the equations of motion for each point due to internal and external forces, and solve for all the accelerations of all points on the body. So is torque just a mathematical trick to make this easier, using $a = r \alpha$?

Answer 1

Yes, you can consider torque as a mathematical trick, to find the angular acceleration. In fact, you cannot easily find the acceleration by taking dm masses, especially if the body is rotating about an axis tilted in a weird direction. Basically the concept of torque takes advantage of the fact that the internal forces inside the body don't affect its external state (like linear momentum, and angular momentum).

The fundamental definition of torque is $\frac{dL}{dt}$ and not $r$ x $F$. (L= angular momentum vector)

$r$ x $F$ is derived from $\frac{dL}{dt}$ for a particle.

This L vector can be about any point, it doesn't have to be the center of mass. But the torque we will get from $\frac{dL}{dt}$ will also be about that point.

Also, we do not always calculate torque about the COM, we can solve questions by taking about any point. But sometimes, taking torque about COM is beneficial in case the COM is accelerating, as the torque of pseudo force will become 0 (because the psuedo force is radially outward with respect to the COM)

If you want an example of significance of torque, here's a question:

Find the distance of closest approach of a spaceship far away moving with initial speed v and at a perpendicular distance of "a" from the sun. Assume sun is stationary.

Try the question by using only free body diagram (will require high level calculus) and also try by using the concept of torque(will take less than 2 minutes to solve)

I'll share the solution after you have tried your best :D

Answer 2

There are two types of answers here.

1. Torque absent of motion (in the field called statics) and its role in balancing forces.

   $$ \vec{\tau}_{A} = \sum_i \vec{r}_{i/A} \times \vec{F}_i \tag{1.1}$$

   _{where subscript A indicates summation about some arbitrary reference point A, and $\vec{r}_{i/A}$ is relative position of some particle/point i to the reference point A.}

   Now if you take a single force, for example $\vec{F}_B$, that goes through some point B with relative position $\vec{r}_{B/A}$, torque is defined as

   $$ \vec{\tau}_{A} = \vec{r}_{B/A} \times \vec{F}_B \tag{1.2}$$

   Now note that the geometry of the situation is that the force acts through a point, and this defines like a ray or line, for which you can slide the force vector along this line and it won't change the nature of the problem. This is why forces are sometimes called sliding vectors or line vectors

   The contribution of torque to the statics is to recover where in space a force (or a combination of forces) is applied. In the study of statics, Newton's 3rd law not only requires that a support force is equal in magnitude and opposite in direction, but also must act on the same line of action.

   This means that to support an object, you not only need to provide a balance of forces

   $$ \underbrace{ \sum_i \vec{F}_i }_{\text{supports}} = \underbrace{ \sum_j \vec{F}_j }_{\text{applied}} \tag{1.3}$$

   but also a balance of torques

   $$ \underbrace{ \sum_i \vec{r}_i \times \vec{F}_i }_{\text{supports}} = \underbrace{ \sum_j \vec{r}_j \times \vec{F}_j }_{\text{applied}} \tag{1.4}$$

   And now the fun part. If you given me that a force $\vec{F}$ produces a torque $\vec{\tau}$, I can tell you where in space this force's line of action is. The point C on the line closest to the origin (or the summation point A) is

   $$\vec{r}_{C/A} = \frac{ \vec{F}_B \times \vec{\tau}_B }{ \| \vec{F}_B \|^2 } \tag{1.5}$$

   _{where $\| \vec{F}_B \|$ is the magnitude of the force vector, and $\times$ is the cross product.}

   Note that points B and C must lie on the same line because $\vec{r}_{C/A} \cdot \vec{F}_C =0$ and thus $\vec{r}_{B/A} \times \vec{F}_C = \vec{r}_{C/A} \times \vec{F}_C$.

   In reality, this is all that is encoded into torque values. When torque balance is achieved, you are just making sure the applied and support lines of action are coincident.

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2. Torque in the context of motion (in the field of rigid body dynamics) and its relationship with angular momentum.

   $$ \vec{\tau}_{C} = \tfrac{\rm d}{{\rm d}t} \vec{L}_{C} \tag{2.1} $$

   _{where subscript C indicates summation about some the center of mass point C, and $\vec{L}_C$ is angular momentum summed about point C also.}

   Of course now we need to defined the angular momentum of a body, and this is done also with a summation

   $$ \vec{L}_{C} = \sum_i \vec{r}_{i/C} \times \vec{p}_i \tag{2.2}$$

   _{where $\vec{p}_i = m_i \vec{v}_i$ is the (translational) momentum of a particle.}

   or alternatively for known geometry for a rotating body

   $$ \vec{L}_{C} = {\rm I}_C \vec{\omega} \tag{2.3}$$

   _{where ${\rm I}_C$ is the mass moment of inertia (tensor) of the body, summed about point C.}