# Is Angular Momentum Defined Only For Spinning Objects? [duplicate]

This question already has an answer here:

I was just going through some of the lectures of angular momentum where it was defined as

Conservation of angular momentum, one of the fundamental laws of physics, observes that the angular rotation of a spinning object remains constant unless acted upon by external torque.

But I found quite a decent example that contradicts this. Consider a bullet (point mass) having a velocity that strikes a coin on its edge.The coin will definitely start spinning!! If the bullet(being a point mass) had no angular momentum then how could it transfer it to the coin? Please suggest how can I understand this..

## marked as duplicate by sammy gerbil, Chris♦, Emilio Pisanty, David Z♦Feb 21 '18 at 8:59

• of course a bullet has angular momentum, namely $\textbf{r} \times m\textbf{v}$ where $\textbf{r}$ is the position vector of the bullet at any given moment of its flight relative to some fixed point that is also the reference point relative to which the coin's momentum is measured. – hyportnex Feb 18 '18 at 15:27
• But as per definition angular momentum is defined for spinning objects and the bullet is not spinning. – Jnan Feb 18 '18 at 17:57
• a bullet flying in a straight line is rotating (ie, "spinning") relative to a point that does not lie on the line of flight. – hyportnex Feb 18 '18 at 18:08
• Any impulse acting non at the center of mass will impart angular momentum as well as linear momentum. – ja72 Feb 18 '18 at 19:56

No.

Angular momentum is a manifestation of linear momentum at a distance. For a single particle of small mass $m_i$ and velocity $\boldsymbol{v}_i$ the linear and angular momentum (about the origin) is

\begin{aligned} \boldsymbol{p}_i & = m_i \boldsymbol{v}_i \\ \boldsymbol{L}_i & = \boldsymbol{r}_i \times \boldsymbol{p}_i \end{aligned}

where $\boldsymbol{r}_i$ is the position vector of the particle from the origin.

Now in the context of rigid bodies we split the velocity of each particle into the velocity of the center of mass, and a rotation about the center of mass. Without loss of generality we can move the origin at the center of mass to simplify the equations.

The particle speed is thus $\boldsymbol{v}_i = \boldsymbol{v}_{C} + \boldsymbol{\omega} \times \boldsymbol{r}_i$.

The total linear and angular momentum are now

\require{cancel} \begin{aligned} \boldsymbol{p} = \sum_i \boldsymbol{p}_i & = \sum_i m_i (\boldsymbol{v}_{C}+ \boldsymbol{\omega} \times \boldsymbol{r}_i) = m\, \boldsymbol{v}_C + \boldsymbol{\omega} \times \cancel{\left(\sum_i m_i \boldsymbol{r}_i \right)} \\ \boldsymbol{L}_C = \sum_i \boldsymbol{L}_i & = \sum_i \boldsymbol{r}_i \times m_i (\boldsymbol{v}_{C}- \boldsymbol{r}_i \times \boldsymbol{\omega}) = \cancel{\left(\sum_i m_i \boldsymbol{r}_i \right)} \times \boldsymbol{v_C} - \sum_i m_i\, \boldsymbol{r}_i \times (\boldsymbol{r}_i \times \boldsymbol{\omega}) \\ \end{aligned}

Where $m=\sum_i m_i$ and $\sum_i m_i \boldsymbol{r}_i = 0$ from the definition of the Center of Mass.

And that is how we defined the mass moment of inertia tensor about the origin $\boldsymbol{L}_C = \mathrm{I}_C \boldsymbol{\omega}$ or

$$\mathrm{I}_C = \sum_i \left(-m_i [\boldsymbol{r}_i \times][\boldsymbol{r}_i \times] \right) = \sum_i m_i \left| \matrix{y_i^2+z_i^2 & -x_i y_i & -z_i x_i \\ -x_i y_i & x_i^2+z_i^2 & -y_i z_i \\ -z_i x_i & -y_i z_i & x_i^2+y_i^2} \right|$$

Thus by definition angular momentum as the moment of momentum (linear momentum at a distance) we derive the equations of motion at the center of mass

\left. \begin{aligned} \boldsymbol{p} & = m\, \boldsymbol{v}_C \\ \boldsymbol{L}_C & = \mathrm{I}_C \boldsymbol{\omega} \end{aligned}\, \right\} \begin{aligned} \boldsymbol{F}_{\rm net} & = \frac{\rm d}{{\rm d}t} \boldsymbol{p} = m\, \boldsymbol{a}_C \\ \boldsymbol{\tau}_{\rm net} & = \frac{\rm d}{{\rm d}t} \boldsymbol{L}_C = \mathrm{I}_C \boldsymbol{\alpha} + \boldsymbol{\omega} \times \boldsymbol{L}_C \end{aligned}

Now let's look at the relationship between linear velocity $\boldsymbol v$ and rotational velocity $\boldsymbol \omega$, as well as between angular momentum $\boldsymbol{L}$ and linear momentum $\boldsymbol p$ and finally between torque $\boldsymbol \tau$ and force $\boldsymbol F$:

$$\begin{array}{r|c|c|l} \mbox{quantity} & \mbox{direction vector} & \mbox{moment vector} & \mbox{transformation law} \\ \hline \mbox{motion} & \boldsymbol{\omega} & \boldsymbol{v} = \boldsymbol{r} \times \boldsymbol{\omega} & \boldsymbol{v}_A = \boldsymbol{v}_B + (\boldsymbol{r}_A - \boldsymbol{r}_B) \times \boldsymbol{\omega} \\ \hline \mbox{momentum} & \boldsymbol{p} & \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} & \boldsymbol{L}_A = \boldsymbol{L}_B + (\boldsymbol{r}_A - \boldsymbol{r}_B) \times \boldsymbol{p} \\ \hline \mbox{loading} & \boldsymbol{F} & \boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F} & \boldsymbol{\tau}_A = \boldsymbol{\tau}_B + (\boldsymbol{r}_A - \boldsymbol{r}_B) \times \boldsymbol{F} \\ \end{array}$$

Which leads to the following definitions:

• Linear velocity is the moment of rotation.
• Angular momentum is the moment of momentum.
• Torque is the moment of force.

• Together a direction vector and a moment vector describe a line in space.

This line is often called the axis of rotation, or line of action of force and finally the axis of percussion for momentum.

Please read this answer to get more details on how forces or impulses affect various rigid bodies.

What I think is that angular momentum and linear momentum are totally different things. You can check dimensions too. For your question, the bullet only strikes the edge of the coin. Due to transfer of momentum, the part of mass gets linear momentum and tries to move linearly in the direction of force applied. But other masses which were unaffected by the bullet stay still. So, the mass which was hit rotates and as a result the coin rotates.

The angular momentum of a point mass about some point $\vec r_0$ is $$\vec L = (\vec r-\vec r_0) \times\vec p$$

As long as the momentum $\vec p$ is nonzero, then the angular momentum is generically nonzero, though this depends on your choice of the point $\vec r_0$.

As you can see, $(\vec r - \vec r_0)\times \vec p \neq 0$, so the angular momentum of the system about the point $\vec r_0$ is nonzero.

If the bullet were rotating, then it would have an additional component of angular momentum associated with its spin. It would then follow that the total angular momentum (the sum of the spin angular momentum and the angular momentum discussed above) would be conserved.