Why does Torque exist?

Question

Specifically, why does force increase with the moment arm? What about making a perpendicular force farther away from the axis of rotation increases it? How does the moment arm cause that?

Answer 1

So @lemon gave a good explanation in the comments that satisfies the question for me. If you have more to contribute please do.

If you are rotating a body (applying a torque) you are doing work to rotate it over an angle. The same amount of work will be needed to cover that angle no matter what. At a greater distance from the axis the arc is larger, there is more distance to cover the same angle. So if the work required is the same but it is applied over a greater distance, you will need less force.

The moment arm affects the force in a torque because a larger moment arm, radius, or perpendicular distance means more distance to cover the same angle of rotation.

Answer 2

Wow this is actually a deep question, warranting a deep answer.

The way I interpret Newtonian mechanics is that torque, just like linear velocity is purely a result something happening at a distance. Namely, a force, or a rotation. In fact, for me forces and rotations are fundamental to the description of rigid body mechanics, and torques and velocities secondary.

Here is what you need to fully describe the loading on a rigid body:

1. Force Magnitude, $F$
2. Force Direction, $\vec{e}$
3. Force (axis) Location $\vec{r}$ or Torque at origin $\vec{\tau}$.

And here are the derived properties from this information:

1. Force Vector, $\vec{F} = F \vec{e}$
2. Torque at origin, $\vec{\tau} = \vec{r} \times \vec{F}$ or Force (axis) Location $\vec{r} = \frac{\vec{F} \times \vec{\tau}}{F^2}$
3. Force pitch (linear to angular ratio) $h = \frac{\vec{F} \cdot \vec{\tau}}{F^2}$

$$ \boldsymbol{f} = \begin{Bmatrix} \vec{F} \\ \vec{\tau} \end{Bmatrix} = F \begin{Bmatrix} \vec{e} \\ \vec{r} \times \vec{e} + h \vec{e} \end{Bmatrix} $$

Similarly for the motion of a rigid body:

1. Rotation Magnitude, $\omega$
2. Rotation Direction, $\vec{e}$
3. Rotation (axis) Location $\vec{r}$ or Speed at origin $\vec{v}$.

And here are the derived properties from this information:

1. Rotation Vector, $\vec{\omega} = \omega \vec{e}$
2. Speed at origin, $\vec{v} = \vec{r} \times \vec{\omega}$ or Rotation (axis) Location $\vec{r} = \frac{\vec{\omega} \times \vec{v}}{\omega^2}$
3. Motion pitch (linear to angular ratio) $h = \frac{\vec{\omega} \cdot \vec{v}}{\omega^2}$

$$ \boldsymbol{v} = \begin{Bmatrix} \vec{\omega} \\ \vec{v} \end{Bmatrix} = \omega \begin{Bmatrix} \vec{e} \\ \vec{r} \times \vec{e} + h \vec{e} \end{Bmatrix} $$

In addition, momentum is similarly described like forces:

1. Momentum Magnitude, $p$
2. Momentum Direction, $\vec{e}$
3. Momentum (axis) Location $\vec{r}$ or Angular momentum at origin $\vec{L}$.

And here are the derived properties from this information:

1. Momentum Vector, $\vec{p} = F \vec{e}$
2. Angular momentum at origin, $\vec{L} = \vec{r} \times \vec{p}$ or Momentum (axis) Location $\vec{r} = \frac{\vec{p} \times \vec{L}}{p^2}$
3. Momentum pitch (linear to angular ratio) $h = \frac{\vec{p} \cdot \vec{L}}{p^2}$

$$ \boldsymbol{p} = \begin{Bmatrix} \vec{p} \\ \vec{L} \end{Bmatrix} = p \begin{Bmatrix} \vec{e} \\ \vec{r} \times \vec{e} + h \vec{e} \end{Bmatrix} $$

Now the fundamental equations of mechanics are described for a rigid body as:

$$\begin{align} \boldsymbol{p} &= \mathrm{I} \boldsymbol{v}\\ \boldsymbol{f} &= \frac{{\rm d}}{{\rm d}t} \boldsymbol{p} \end{align}$$

with $\rm I$ an appropriate 6×6 spatial inertia matrix. If the center of mass is located at $\vec{r}_C$ and the 3×3 inertia matrix about the center of mass is $\mathcal{I}_C$ then

$$ {\rm I} = \begin{Bmatrix} m & -m [\vec{r}_C]\times \\ m [\vec{r}_C]\times & \mathcal{I}_C - m [\vec{r}_C]\times [\vec{r}_C]\times \end{Bmatrix} $$

So from this long way around, you see that torque is not directly needed, other than to convey the location of where forces go through. The equation $\boldsymbol{p} = \mathrm{I} \boldsymbol{v}$ has a geometrical interpretation since both $\boldsymbol{p}$ and $\boldsymbol{v}$ are lines in space. This equation is a one-to-one mapping related to the pole-polar relation in planar geometry. Finally, the moment arm of momentum $\ell$ is related by the moment arm of rotation $c$ with the expression $\ell = \frac{\kappa^2}{c}$ where $\kappa$ is the radius of gyration of the rigid body along the axis motion.

_{NOTE: $\times$ is the vector cross product and $\cdot$ is the vector dot product. Also $[\vec{c}\times]$ is a 3×3 skew symmetric matrix such that $[\vec{c}\times] \vec{a} = \vec{c} \times \vec{a}$}

_References:

1. Spatial Inertia(slides)
2. Spatial Vector Algebra(pdf)

3. Derivation of Equations of Motion(this site)

Answer 3

The answer Lemon gave is intuitive, but I think his rationale is a bit simplified for the sake of brevity.

In all generality, an object can be rotated by any angle with any amount of work done. Consider a rod is sitting in free space, the system has the same energy regardless of the orientation of the rod. When you rotate the rod, you give it some angular momentum about an axis. The work you've done on the rod is $\frac{1}{2} I \omega_\text{max}^2$ (or the tensor form for sufficiently complicated rotations) by the work-energy theorem. When the rod is stopped, it does the same work back on the object that stopped it. The work done is then either 0 or a function of the rotational velocity depending on whether you're asking about before or after the rod was stopped.

There's a couple lines of reasoning we can use to consider this issue. First consider the, perhaps deeper, question of angular momentum. Torque is to angular momentum as force is to linear momentum. The "reason torque increases with the moment arm" is that the canonical momentum for the coordinate angle $\theta$ has an $mR$ as the prefactor to $\dot{\theta}$. When we use the angle $\theta$ to consider our problem, the relevant mass for point objects is not $m$, it's $mR$.

Is there another way we can try to understand that, why should $m \to mR$ when we work in polar coordinates? Well, it's because $\theta$ is not a distance; $R \theta$ is. So the distance associated with a displacement $\delta \theta$ is $R \delta \theta$. And the virtual work associated with that displacement against a linear force $F$ is $F R \delta \theta$. That is why you can do more work with the same force at a greater distance $R$.

Your confusion is that when you are working in torques and when you are working in linear forces, you are fundamentally working in different coordinate systems. A change in $\theta$ is associated with a torque in the way a change in $x$ is associated to a force. When you want to translate from forces to torques, you are implicitly translating from linear to polar coordinates.

Answer 4

The rotational version of Newton's Second Law says,

Total net torque acting on a system is equal to the product of its moment of inertia and the angular acceleration it experiences due to that torque.

The other version of this law is:

Total net torque acting on a system is equal to the rate of change of the angular momentum of the system.

So, the mere reason behind the existence of any non-zero total net torque will be a non-conserved angular momentum in the system under consideration; the system is not isolated.

Thanks,