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I have first been introduced to the notion of torque in regards to a rod (or any rigid body) rotating about a fixed axle. Then I kind of got the intuition of torque as if it is the equivalent of force from linear motion but in rotational motion. I have seen how torque can be thought of as what is needed to cause angular acceleration.

Then I have been introduced to angular momentum as to being defined as $\vec{L}=\vec{r}\times\vec{p}$ however this time I was introduced to this quantity not necessarily in the context of rotation. I have seen that if you do $\dfrac{d\vec{L}}{dt}=\vec{\tau_{net}}$ now this is where the confusion comes from. Lets say that a particle is moving and I apply a force in a arbitrary direction and I change its momentum, then this would mean that I change its angular momentum and thus my applied force should produce a torque. However all this time I thought torque has to do with rotation and with rigid bodies, what does torque on a particle really mean, since I am not making it rotate about any axis?

I am just confused as to what does torque mean in the sense of a particle. The same question can be posed about angular momentum, what is the connection between angular momentum of a single particle, and the angular momentum of a rigid body rotating about an axis eg $\vec{L}=I\vec{\omega}$ how are these two same ideas the same even though they seem so different?

Thank You

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  • $\begingroup$ May I suggest you read this post as it might help you understand the nature of torque. $\endgroup$ – ja72 Nov 5 at 4:50
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These are the fundamental definitions you must know and the rest follow from these. Let's say you are in reference frame $S$ with origin $O$. Assume that there's a particle of mass $m$ located at position $\vec{r}$ (called the position vector) with respect to $O$ and its velocity is $\vec{v}$. Then, $$\vec{L}_{\text{of the point particle with respect to a reference point $P$}} := (\vec{r}-\vec{r}_P) \times m\vec{v} \tag{1}$$ $$\vec{\tau}_{\text{acting on the point particle with respect to a reference point $P$}} := (\vec{r}-\vec{r}_P) \times \vec{F}_{\text{net acting on the particle}} \tag{2}$$

For a point particle observed in an inertial frame, the following equation is true (the equation you've seen). But keep in mind that both angular momentum and torque are defined with respect to some reference point $P$ (unlike momentum and force).

$$\frac{d\vec{L}_P}{dt}=\vec{\tau}_P$$

This can now be extended to a collection of interacting point particles. And further, to rigid bodies. Classical Mechanics by Goldstein is an excellent reference.

$$\vec{L}_{\text{of a rigid body with respect to $P$}} = \int dm ((\vec{r}-\vec{r}_P) \times \vec{v})$$ $$\vec{\tau}_{\text{of the rigid body with respect to point $P$}}=\sum_{\text{$i$ : running over all external forces only}} (\vec{r}_i-\vec{r}_P) \times {\vec{F}_i}^\text{external force}$$ $$\frac{d}{dt}(\vec{L}_{\text{of a rigid body with respect to $P$}}) = \vec{\tau}_{\text{of the rigid body with respect to point $P$}}$$

Anyway, I instead will focus on the confusion that might be on your mind. "If angular momentum is defined with respect to a point, why do I regularly hear the phrase 'angular momentum about the axis of rotation'?"

Consider the pure rotation of a 3-dimensional rigid body with respect to an arbitrary axis that might not pass through the center of mass. If this is the case, then every infinitesimal piece of the rigid body is executing a circular motion whose center lies somewhere in this axis.

Diagram

Consider a point $O$ on the axis of rotation, with respect to which we are going to find the angular momentum. It's a figure I got from the internet, so focus on only the labels $O$, $C$, $P$, and the circular orbit. Forget the rest of the inscriptions.

$$d\vec{L}_O = \vec{OP} \times dm \vec{v}=(\vec{OC} + \vec{CP}) \times m \vec{v} = \vec{OC} \times dm \vec{v} + \vec{CP} \times dm \vec{v}$$

Note that both $\vec{CP}$ and $\vec{OC}$ are perpendicular to $\vec{v}$. The component of $d\vec{L}_O$ that is parallel to the axis of rotation comes from the cross product of $\vec{CP}$ with $\vec{v}$. The other one provides the component perpendicular to the axis of rotation. Also $v= CP \omega$ since the rigid body is under pure rotation about the axis.

Let's find the component net angular momentum of the rigid body with respect to point $O$ that is parallel to the axis of rotation. Let's call $\vec{CP}=\vec{r}'$. $$L_z = \int dm CP^2 \omega = \int dm r'^2 \omega = I \omega$$

This component is the same irrespective of the location of the reference point $O$.

This is useful because if we know the component of the net external torque on the rigid body with respect to any point on the axis of rotation parallel to the axis of rotation, we can equate the components to get

$$ I \omega = \tau^{\text{ext}}_{z}$$

The equation $\vec{L} = I \vec{\omega}$ is not true always. And you need to be careful.

Let me leave you with one more interesting result that you must be familiar with. The net torque due to the gravitational force on a rigid body with respect to point $P$ is given by $$\vec{\tau}=\int dm (\vec{r}-\vec{r}_P) \times \vec{g} = (\int dm (\vec{r}-\vec{r}_P))\times \vec{g} = M (\vec{r}_{\text{COM of the rigid body}} - \vec{r}_P) \times \vec{g}$$

It all comes from the two fundamental definitions.

EDIT (Response to comments are made here) :

Angular momentum is fundamentally defined with respect to a point (not an axis). Angular momentum about an axis is a derived notion that's not universally useful to talk about. You can show that the angular momentum of a rigid body of mass $M$ with respect to point $P$ under pure translation obeys

$$\vec{L}_P = M (\vec{R}_{COM}-\vec{r}_P) \times \vec{v}_{COM}$$

Now consider the rigid body to be a disc that lies in the $x$-$y$ plane and the center of the disc moves in a straight line along the path $y=a$ with uniform velocity $\vec{v}=v \hat{i}$. We're considering pure translation here. Let's say the center of the disc (which is also the position of the COM of the disc) is at $(x=x_0+vt,y=b)$ at time $t$. Let's calculate the angular momentum of this disc at $t$ with respect to the origin $O$,

$$\vec{L}_O (t) = M ((x_0+vt) \hat{i} + b \hat{j}) \times v \hat{i} = -Mbv \hat{k}$$

Now you see that the disc's angular momentum is independent of the position of the disc : it's independent of $t$. Since the entire angular momentum is along the $\hat{k}$ direction for this particular example, one can say that the angular momentum of the disc about the $z$ axis is $-Mbv$. But does it really add anything new to your knowledge of the angular momentum of the disc that was not previously already in $\vec{L}_O$?


$$\mathbf{\text{EDIT II}}$$ $$\mathbf{\text{Definition :}}\text{ A collection of $N$ particles is called a rigid collection if and only if}$$ $$|\mathbf{r}_i-\mathbf{r}_j|=c_{ij} \text{ (some constant that doesn't vary with time) } \forall i,j$$

It means that the distance between any two particles belonging to a rigid collection is fixed and unchanging. A rigid body is a solid body that is a natural generalization of this rigid collection of $N$ particles where $N \to \infty$. In a rigid body, the distance between any two points is fixed and immutable. In general, different points on the rigid body move with different velocities ($\mathbf{v}=\mathbf{v}(\mathbf{r})$). Consider a point labelled by its position vector $\mathbf{r}$ on the rigid body. Now, also consider an infinitesimal volume $dV$ centered at the point which houses a mass $dm=\rho(\mathbf{r})dV$ (where $\rho(\mathbf{r})$ gives the density of the rigid body at $\mathbf{r}$). And this infinitesimal mass, moving with velocity $\mathbf{v}(\mathbf{r})$, behaves just like a particle.

The angular momentum of any collection (can be rigid or unrigid) of $N$ particles with respect to $P$ is given by

$$ \mathbf{L}_P = \sum_{i=1}^N m_i((\mathbf{r}_i-\mathbf{r}_P) \times \mathbf{v}_i)$$

Can we extend this to a rigid body? Let's say we have a rigid body occupying volume $V$. Instead of a particle with mass $m_i$ located at position $\mathbf{r}_i$ having velocity $\mathbf{v}_i$, we have an infinitesimal mass $dm=\rho(\mathbf{r})dV$ located at $\mathbf{r}$ moving with velocity $\mathbf{v}=\mathbf{v}(\mathbf{r})$. So, instead of summation, we will have to integrate over the volume $V$

$$\mathbf{L}_P = \int_{V} dm ((\mathbf{r}-\mathbf{r}_P) \times \mathbf{v})$$

Goldstein introduces rigid bodies in $Pg. 11$ of chapter $1$. I'd strongly urge you to read the chapter from the beginning as it explains how certain concepts that you may already be familiar with (like center of mass of a collection of particles) emerges.


$$\mathbf{\text{EDIT III}}$$ Let's say we have a collection of $N$ particles. What is the angular momentum of this collection of particles with respect to a general point $P$ that moves with velocity $\mathbf{v}_P$? Well, there are two ways to define that.

$$\text{Absolute angular momentum : } \mathbf{L}_P = \sum_{i=1}^Nm_i(\mathbf{r}_i-\mathbf{r}_P) \times \mathbf{v}_i$$ $$\text{Relative angular momentum : } \mathbf{L}'_P=\sum_{i=1}^N m_i (\mathbf{r}-\mathbf{r}_P)\times (\mathbf{v}_i-\mathbf{v}_P)$$

$$\text{In general, }\mathbf{L}_P \neq \mathbf{L}'_P$$

However, what happens when our moving point $P$ is the center of mass ($CM$) of the particle-collection? $$\mathbf{L}'_{CM}=\sum_i m_i (\mathbf{r}_i-\mathbf{r}_{CM})\times(\mathbf{v}_i-\mathbf{v}_{CM})=\sum_i m_i (\mathbf{r}_i-\mathbf{r}_{CM})\times \mathbf{v}_i - \sum_i m_i (\mathbf{r}_i-\mathbf{r}_{CM}) \times \mathbf{v}_{CM}$$ $$\Rightarrow \mathbf{L}'_{CM}= \mathbf{L}_{CM} - \left((\sum_i m_i \mathbf{r}_i) - (\sum_i m_i \mathbf{r}_{CM})\right)\times \mathbf{v}_{CM}=\mathbf{L}_{CM} - (M\mathbf{r}_{CM} - M\mathbf{r}_{CM})\times \mathbf{v}_{CM}$$ $$ \Rightarrow \mathbf{L}'_{CM} = \mathbf{L}_{CM}$$

Since we normally calculate angular momentum either with respect to a stationary point or with respect to the $CM$, this distinction between absolute and relative angular momentum is never encountered. But now you must be onto many more questions such as, $$ \text{Which is/are correct : } \frac{d\mathbf{L} _P}{dt}=\mathbf{\tau}_P \;\;\text{ | }\;\; \frac{d\mathbf{L}'_P}{dt}=\mathbf{\tau}_P$$ $$\text{where }\mathbf{\tau}_P = \sum_{i=1}^N (\mathbf{r}_i-\mathbf{r}_P) \times {\mathbf{F}_i}^{\text{ext}} \text{ irrespective of whether $P$ is in motion or not}$$

It turns out that neither is correct in general. I'd highly recommend you consult this source (link) to know more about this.

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  • $\begingroup$ but if a rigid body undergoes only translational motion, surely it must also have angular momentum about an axis ? $\endgroup$ – Luca Ion Nov 1 at 9:30
  • $\begingroup$ Hey @LucaIon, check my edit. $\endgroup$ – Ajay Mohan Nov 1 at 10:05
  • $\begingroup$ ye I meant point not axis, my bad $\endgroup$ – Luca Ion Nov 1 at 15:21
  • $\begingroup$ @LucaIon Oh, that's okay. But you are right, angular momentum is just another physical quantity like momentum and it's certainly not something that should only be invoked in problems involving rotational motion. $\endgroup$ – Ajay Mohan Nov 1 at 16:02
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    $\begingroup$ $\mathbf{L}_P = \sum_{i=1}^N m_i((\mathbf{r}_i-\mathbf{r}_P) \times \mathbf{v}_i)$ do we have this because we are assuming that our coordinate system is not located exactly at P, and $r_i$ is the vector pointing to ith particle from our coordinate system? Also if we wouldn't assume that P is stationary with respect to us would the equation become $\mathbf{L}_P = \sum_{i=1}^N m_i((\mathbf{r}_i-\mathbf{r}_P) \times (\mathbf{v}_i-\mathbf{v}_P))$? $\endgroup$ – Luca Ion Nov 3 at 15:02
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Imagine we have a ball rotating around a pole due to a light string (sort of like a tetherball setup).

We can calculate the angular momentum of the ball about the pole. If we imagine drag is minimal, then the ball spins quite freely. In this case, since the only force on the ball is from the string, and that force goes through the axis we are considering for our angular momentum, then angular momentum of the ball is constant.

Now, cut the string. Since there are no forces on the ball, it moves away from the axis at a constant speed, and it must still maintain a constant angular velocity. So point particles in motion do have an angular velocity. It's just that angular velocity must be considered about an axis (at least at this level), so the choice of axis will modify the value of momentum.

The angular momentum of a particle is equal to the product of the mass and velocity of the particle along with the perpendicular distance from the axis. $$L = mvd$$ http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html

In your example of applying a force to an arbitrary particle, you might change its speed/velocity, or you might change its trajectory. Either of those would change the angular momentum, so the idea of it being a torque is correct.

At the same time, your axis might be chosen in such a way that the applied force always points in a line to it. Those forces will not change the calculated angular momentum and the torque is zero.

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As mentioned in this answer torque is just the manifestation of a force at a distance, just as angular momentum is the manifestation of momentum at a distance.

As measured at the origin, both are similar in definition $$ \begin{aligned} \boldsymbol{\tau} & = \boldsymbol{r} \times \boldsymbol{F} & \boldsymbol{L} & = \boldsymbol{r} \times \boldsymbol{p} \end{aligned} \tag{1,2}$$

Here $\boldsymbol{r}$ designates the location of the line of action of the force or the location of momentum. Also, $\boldsymbol{p} = m \boldsymbol{v}$ is momentum and $\boldsymbol{F}$ is force. $\boldsymbol{\tau}$ is torque, and $\boldsymbol{L}$ is angular momentum.

But also consider the velocity of a rotating rigid body as it is passing over the origin if the axis of rotation is at $\boldsymbol{r}$. The definition of velocity is $$ \boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega} \tag{3}$$

Does it look familiar? All of the above quantities are moments: a) force moment, b) moment of momentum and c) moment of rotation. All of them designate something happening away from the origin.

You can use Newton's 2nd law of $\boldsymbol{F} = \frac{{\rm d}}{{\rm d}t} \boldsymbol{p} $ and (1,2) to derive the law of rotational motion

$$\require{cancel} \tfrac{\rm d}{{\rm d}t} \boldsymbol{L} = (\tfrac{\rm d}{{\rm d}t} \boldsymbol{r}) \times \boldsymbol{p} + \boldsymbol{r} \times ( \tfrac{\rm d}{{\rm d}t} \boldsymbol{p}) =\cancel{ \boldsymbol{v} \times m \boldsymbol{v}} + \boldsymbol{r} \times \boldsymbol{F} = \boldsymbol{\tau} $$

So the equivalent law of rotational motion (called Euler's law) is

$$ \boxed{ \boldsymbol{\tau} = \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}} $$

And again, all the torque means is that there is a force applied at a distance. A special case exists when two equal and opposite forces act but at different locations causing a torque, but no net force. This is called a pure torque.

Sometimes when a rigid body is purely translating it is said that it is rotating about a point at infinity with zero rotational speed. Equivalently, a pure torque is a force applied at infinity with zero magnitude.

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