Intuition behind torque, rotational inertia and angular momentum

Question

I'm reading about conservation of linear momentum and angular momentum. I understand the idea that angular momentum should be thought of as the "rotational analogue" of linear momentum, just as torque is the analogue of force and the rotational inertia $mr^2$ is the analogue of mass. What I don't understand is the intuition of why these specifically are the "correct" analogues (in a way deeper than "because it works out nicely").

For simplicity let's focus on a rotating point mass in two dimensions. If I was asked to intuitively formulate rotational analogues to the "linear" terms, without reading a physics textbook beforehand, I would probably do so as following:

• The analogue of velocity $v=\dot{x}$ is the angular velocity $\omega=\dot{\theta}$.
• Linear force is an interaction that changes the velocity (assuming constant mass). So "rotational force" should be an interaction that changes the angular velocity. The natural choice is the tangential component $F_{\rm{tangential}}$ of a force $F$, which I'll denote $\tau'$ (my "wrong version" of the torque $\tau$).
• At radius $r$, the change in angular velocity caused by $\tau'$ is $\dot{\omega}=\frac{\tau'}{mr}$ (since $\tau'=m\dot{v}$ and $v=r\omega$). So the "rotational inertia" is $I'=mr$, and the analogue of Newton's second law is $\tau'=I'\dot{\omega}$.

Why is my intuitive approach more "wrong" than the accepted concepts? Why would the "rotational force" (torque) be $rF$ instead of $F$ – What natural rotational quantity is affected stronger when the same force is applied further from the center of rotation? Because $\omega$ isn't – on the contrary, $\dot{\omega}=\frac{F_{\rm{tangential}}}{r}$ so it is less affected when $r$ is increased. Sure, the answer is $mr^2\omega$, but is there an intuitive explanation for why this is the natural way to quantify "amount of rotation"? The only answer I have is "because it's conserved", but that comes from mere mathematical manipulations, and I find it hard to gain deep insight or intuition about them.

Answer 1

The starting point should not be trying to connect the physics of rotational motion to translational, rather you should try to find some conserved quantity for rotational motion and derive the laws from there.

Let's take a step back, when looking at Newtons Laws let's think about what are are actually saying, and lets try and reduce it to the purest physical principle; and that is the Conservation of Momentum. Newton's First Law says when a system is isolated its momentum is conserved, and the combination of the Second and Third Law is that when many systems are interacting the sum of their momentums are conserved.

We find that when analyzing rotational motion the quantity which is conserved is angular momentum; this is a physical fact. Angular Momentum is measured with reference to a pivot point and can be defined for a point mass and then extended to all other shapes and mass distributions through linearity (i.e. integration). This conserved quantity of Angular Momentum is: $L=mvr=mr^2\omega$.

It is from this point that we draw a connection between rotational and angular motion. $p=mv$ and $L=mr^2 \omega$, $F=\frac{d}{dt}p=m\frac{d}{dt}v$ and $\tau=\frac{d}{dt}L=mr^2 \frac{d}{dt}\omega$. The analogue of Force is this new thing we have defined $\tau$ which is Torque, and the analogue of mass is this new thing $mr^2$ which is Moment of Inertia.

I should note that the existence of these conserved quantities is not random or completely unexplained, they are consequences of symmetries or invariances in "our world". This is whats known as Noether's Theorem should you be interested in further reading.

Answer 2

Actually, the change is velocity is greater when the torque is applied further away from the centre of rotation.

You have considered your rotational inertia (I') to be varying according to r but it is constant for a given body as it is analogous to mass.

Since the moment of inertia (I) is a constant and angular velocity is dependent on the point of application of the force, there is a necessity to take torque as rF

Answer 3

You ask for something deeper than "it works out nicely"

Angular momentum, as per the standard definition, satisfies a consistency criterium. That consistency is about the following: whenever an object is in angular motion you can define an instantaneous tangent line, and motion along that line is linear motion.

Take for example kinetic energy. You want consistency between rotational kinetic energy and linear kinetic energy.

For an object in circumnavigating motion with angular velocity $\omega$ the corresponding linear velocity is $\omega r$

The relation is $v=\omega r$ is a geometric statement, not a physics statement. If you accept Euclidean geometry you must accept $v=\omega r$

In order to be self-consistent:

The circumnavigating object has a particular kinetic energy (this is kinetic energy relative to the inertial coordinate system that is stationary with respect to the center of circumnavigation.) If the constraint that maintained the circumnavigating motion is instantaneously removed the motion of the object changes instantaneously from circumnavigating motion to linear motion. For consistency we should still be attributing the same kinetic energy to that motion (kinetic energy relative to the inertial coordinate system that is stationary with respect to what an instant ago was the center of circumnavigation.)

linear kinetic energy: $\tfrac{1}{2}mv^2$

rotational kinetic energy $\tfrac{1}{2}m(r \omega)^2 $

If the radial distance is constant we can move the radial distance $r$ out of the round brackets

rotational kinetic energy $\tfrac{1}{2}(m r^2) \omega^2 $

The constraint is self-consistency.
The idea is to set up circumstances where there is an instantaneous transition from angular motion to linear motion; the kinetic energy that you attribute to the motion must remain the same.

Answer 4

You kind of have to reject the analogy to understand things deeper.

Linear and angular momentum are not two separate things but exist together to describe the overall momentum state of a rigid body. Just as linear and angular velocity exist together describing two qualities of the motion of a body. In fact, force and torque also exist together to describe the loading of a body.

1. You can think of how an offset force $\boldsymbol{F}$ causes a torque about the reference point $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$.

2. Or the velocity of a body due to a rotation $\boldsymbol{\omega}$ through an offset point is $\boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega}$.

3. Finally, the angular momentum of a particle with momentum $\boldsymbol{p}$ in an offset trajectory is $\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$

The key insight here is that with the above were given the offset $\boldsymbol{r}$ and some quantity $\boldsymbol{A}$ and we compute the moment-of this quantity $\boldsymbol{M} = \boldsymbol{r} \times \boldsymbol{A}$. So torque is moment-of force, angular momentum is moment-of momentum and velocity is moment-of rotation.

But this can work in reverse also. The moment-of quantity $\boldsymbol{M}$ can be used and tell us where in space the vector $\boldsymbol{A}$ is applied though. This is done with $$\boldsymbol{r} = \frac{\boldsymbol{A} \times \boldsymbol{M}}{\| \boldsymbol{A} \|^2}$$

So given the pairs of quantities below this is how we find where they act

1. The torque $\boldsymbol{\tau}$ represents the offset $\boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{\tau}}{F^2}$ where the force $\boldsymbol{F}$ acts through.

2. The velocity $\boldsymbol{v}$ represents the offset $\boldsymbol{r} = \frac{ \boldsymbol{\omega} \times \boldsymbol{v}}{\omega^2}$ where the rotation $\boldsymbol{\omega}$ acts through.

3. The angular momentum $\boldsymbol{L}$ represents the offset $\boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L}}{p^2}$ where momentum $\boldsymbol{p}$ acts through.

The moment-of quantities describe the geometry of the situation. The direction and magnitude of a quantity is described by its vector, but where in space this vector exists is solely described by the moment-of quantity.

Let me qualify what the axis where momentum acts though means. For a particle, this axis goes through the center of mass, but for a rigid body, it does not. Sometimes this is called the axis of percussion, and it represents the exact point and direction in space where if you hit an object you can remove all its motion (translation and rotation).

Example

Think of a tumbling rod in space that is coming towards you and you want to stop it. If you punch it on its center of mass, you will stop its translation in space, but not its rotation. To stop both the translation and rotation you need to punch it at a specific location given by the formula above.

If at some instant the combined motion causes the rotation axis to be a distance $c$ from the center of mass, and the body has a radius of gyration $\kappa$, then the percussion point $x$ is located on the other side of the center of mass from the rotation axis at a distance

$$ x= \frac{\kappa^2}{c} $$

This calculation is purely geometrical as it involves distances only, and can be derived easily using the vector expression above.

Answer 5

In the classical sense, a force is an external agency that tends to change the state of rest or uniform motion of a body.

Torque is the rotational analogue to force that tends to rotate the body. Now, why isn't the tangential force taken into consideration only? It's because the tangential force will only impart a tangential acceleration to a body and won't change its direction, which is necessary for angular motion. It needs a centripetal acceleration as well for the change in direction.

In a body undergoing linear motion, the mass of the body is the measure of its inertia. More massive the body, the more inertia. But for a rotating body, more the mass is concentrated near the rotation axis, less is the inertia.

The torque increases with the increase in $r$ because you require a smaller force per unit distance to rotate the body, and hence a smaller power is required to turn the object.

Answer 6

Consider a thin mass-less rod with a friction-less axle at one end. The rod supports a point mass (m) at a distance (r) from the axle. A force (F) (perpendicular to the rod and axle) pushes on the rod at a distance of (R) from the axle, causing an angular acceleration (α) and angular displacement (θ). The work done by the force (acting along an arc), F(Rθ), is transmitted by the rod to the point mass, (ma)(rθ) = (mrα)(rθ). Equating these and dividing by (θ) yields: FR = (m$r^2$)α or τ = Iα. This logic can be extended to any number of forces and masses, and yield the best definition: “Torque is the work per unit angle of rotation (in Joules/radian) that can be done by a force acting in a manner which tends to cause a rotation.” Note that this requires a tangential component of force (and a distance).