# Does the law of conservation of momentum mean both linear and angular momentum?

I was solving a problem where a bullet hit a rod hinged about one of its ends. The rod is standing vertically before the collision. We had to find the final angular velocity of the rod. The way we did it was to find the angular momentum of the bullet about the axis of rotation just at the point where it touches the rod.

I have two questions.

1. Firstly, I thought that momentum was always conserved. So when the bullet approaches the rod the total momentum of the bullet is its linear momentum, which must be equal to the total momentum of the combined system after the collision(which now becomes angular). Where am I going wrong with my logic?

2. Secondly, I am confused about the idea of having to find the angular momentum for an object that is traveling linearly. What does this mean physically?

• The question in the title doesn't make any sense. Did you type something wrong? – D. Halsey Aug 3 '19 at 1:22
• 1. Linear momentum is conserved in the collision, but right after the collision the hinge makes a force on the system (rod+bullet) so linear momentum is no longer conserved during the circular motion. 2. angular momentum is defined as L=rxp, so it is well defined for linear motion. It is mostly useful for rotating motion though. – Wolphram jonny Aug 3 '19 at 1:33
• @D.Halsey I changed it so that it makes more sense – Oishika Aug 3 '19 at 1:38

Firstly, I thought that momentum was always conserved. So when the bullet approaches the rod the total momentum of the bullet is its linear momentum, which must be equal to the total momentum of the combined system after the collision(which now becomes angular). Where am I going wrong with my logic?

Your example is a good one to differentiate between the conservation of linear momentum and the conservation of angular momentum.

Assume the velocity of the bullet just before collision is in the horizontal direction.

The first thing is to define the system that is being considered which in your case is the rod and the bullet.
Having done this external (to the system) forces can be identified.
When it is hanging at rest your rod has two forces acting on it, the force due to the pivot and the force due to the gravitational attraction of the Earth.
Ignoring air resistance the bullet has only one force acting on it, the gravitational attraction of the Earth which before the collision is ignored ie there is no vertical component of the bullet's velocity.

The bullet and the rod collide over a period which is usually taken to be short compared with other time scales related to the problem eg effectively the rod only starts moving when there is no relative movement between the bullet and the rod. The bullet imparts an impulsive on the rod and the rod imparts an equal magnitude and opposite direction impulse on the bullet - Newton's third law.
As far as the system is concerned these impulses are due to internal forces and so need not be considered when considering the two conservation laws.

During the collision the pivot exerts an external force on the rod & bullet system this immediately means that linear momentum is not conserved.
If the pivot had not been there then there would have ben no external horizontal force acting on the rod & bullet system and the initial linear momentum of the bullet would have equally the final linear momentum of the centre of mass ofthe bullet & rod system.

Now why did you use the pivot when evaluating the angular momentum?
It was because during the collision there were no external torques about the pivot acting on the system and so the conservation of angular momentum could be used.
Note that the force on the bullet due to the rod and the force on the rod due to the bullet are internal force and provide no net torque about the pivot between them.
For any other point (except those on a horizontal line through the pivot) there would have been a contribution to the external torque from the force on the rod & bullet system due to the pivot and so the angular momentum of the system would not be conserved.

Secondly, I am confused about the idea of having to find the angular momentum for an object that is travelling linearly. What does this mean physically?

When you have a body of mass $$m$$ moving along a circular path of radius $$r$$ that body has at each point in its travels an instantaneous linear velocity with speed $$v$$.
The magnitude of the angular momentum of such a body is $$mvr$$.
You could evaluate the angular momentum at any point on the line from the centre of the circle to the body as $$mvr'$$ where $$r'$$ is the distance between the point and the body.
If the radius of the circle $$r$$ is made larger and larger then the path of the body becomes closer and closer to a straight line and the angular momentum is still $$mvr'$$.
In the limit of the path becoming a straight line since there are no external forces/torques acting on it the angular momentum about the chose point $$mvr'$$ (and the linear momentum $$mv$$) must be constant.

This is all encapsulated in the vector definition of angular momentum $$\vec L = \vec r \times \vec p$$.

1) Linear and angular momentum are two different things. They do not make up a "total momentum" that is conserved, and you do not convert between linear and angular momentum (unlike energy, where we do define different types that we say are converted between each other). Linear momentum and angular momentum are conserved separately in general, and in a given system one can be conserved while the other is not.

For example, think about a planet orbiting around a star. The linear momentum of the planet is definitely not constant. This is because a net force acts on the planet. However, the angular momentum of the planet about the star is conserved because there is no net torque acting on the planet.$$^*$$

2) It is a common misconception that angular momentum means rotation about a point. And of course one could argue that linear motion is just rotation around a circle whose center is at infinity, but this isn't relevant for angular momentum. You don't need circular motion to have angular momentum about some point.

The angular momentum about some reference point of an object with mass $$m$$, velocity $$\mathbf v$$, and position $$\mathbf r$$ relative to the reference point is a pseudovector given by $$\mathbf L=m\mathbf v\times\mathbf r$$ Notice how this doesn't require circular motion. All that is required to have angular momentum about a point is a non-zero velocity that is not pointing directly towards or directly away from your reference point. Therefore, an object with linear motion that isn't heading directly towards or directly away from the reference point has an angular momentum about the reference point.

$$^*$$ Of course if you take your system to be the planet and the star then both linear and angular momentum will be conserved, but my point is that both are not conserved in all systems.

The angular momentum conservation law is independent of the linear momentum conservation law. It is two laws, that means that they have been established by experiment, together with energy conservation that are always fulfilled in the mathematical theories we at presently have for physics, both quantum and classical.( It is only in General Relativity one has to think twice about these experimentally established laws , in special cases).

Once a mass has linear momentum $$p$$ an angular momentum , $$L=r\times p$$, can always be defined for any point in space with respect to this particle, in effect assuming that there is an axis of rotation at that point ( it is as if the particle is flying off tangentially to the hypothetical axis).

Now in the specific problem, the bullet comes with a calculable angular momentum towards the axis around which the rod will rotate.

The conservation of linear momentum is taken care by the whole system ( earth + rod), since the mass of the earth is so large it can be assumed for the system to be immobile, if the rod were not well attached to the earth, the whole thing would also move to conserve linear momentum.

The angular momentum that the bullet had is conserved by the rotation of the rod, and the energy for the whole operation is given by the energy brought in by the bullet.

O can see there are lot of answer to understand it short and simple we cannot merge the two theorm because linear momentum will conserve if net external force is absent and angular momentum will conserve if net external torque is absent so yiu cannot merge two things ,now secondly the question that you have asked about bullet you can solve it by angular momentum conservation velocity of ball before collision can be written mv where v=wr r is the point about which rod and m will rotate now final momentum will be Iw where I calculate for both m and rod

The law is of conservation of the line of momentum and its magnitude.

• The direction of the line and magnitude are given by linear momentum $$\boldsymbol{p} = p \,\hat{\boldsymbol{e}}$$.

where $$p$$ is the scalar magnitude and $$\hat{\boldsymbol{e}}$$ is the unit direction vector.

• The line in space is given by the angular momentum $$\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$$.

Where $$\boldsymbol{r}$$ is the location of any point along the line.

• The location of the line is recovered with the following expression $$\boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ p^2 } \tag{1}$$

So the conservation of angular momentum means conservation of $$(\boldsymbol{r} \times \boldsymbol{p})$$ and since $$\boldsymbol{p}$$ is already conserved, it requires that $$\boldsymbol{r}$$ is conserved also.

So the location of the line is conserved also in addition to the direction and magnitude

Please note that the line of momentum is called the axis of percussion. It signifies the location in space where an impulse needs to be applied in order to completely stop the object. The impact needs magnitude (the impulse) and direction (the contact normal), in addition to a location along its line (the point of contact) to match up with the momentum of the object.

1. The total momentum is conserved before and after impact, and by total momentum where I mean a) its magnitude, b) direction, and c) location in space.

2. Angular momentum of a particle is simply a way to express where the axis of percussion is. And for a particle, the axis of percussion is equivalent to the line of motion.

There is a fourth quantity that is conserved and it is called the pitch, defined by $$h = \frac{\boldsymbol{p} \cdot \boldsymbol{L}}{p^2} \tag{2}$$ This is zero for most cases, except for a rigid body that translates and rotates about the same axis. In such a case a single impact cannot completely stop the body.

The angular momentum of a moving rigid body is completely defined by $$\boldsymbol{L} = \mathrm{I}_{\rm COM} \boldsymbol{\omega} + \boldsymbol{r}_{\rm COM} \times \boldsymbol{p} \tag{3}$$

It can also be completely defined by $$\boldsymbol{L} = \boldsymbol{r}_{\rm AOP} \times \boldsymbol{p} + h\,\boldsymbol{p} \tag{4}$$

One can go from the first to the second form with expressions (1) and (2) from above to find the location of the percussion axis $$\boldsymbol{r}_{\rm AOP}$$ and pitch $$h$$ respectively.

## Example

A vertical rod of length $$\ell$$ and mass $$m$$ is suspended from one end using a long thin inelastic string from the ceiling. Find where the rod was impacted if after the impact it is exhibiting a pure rotation about the point of suspension.

Set a coordinate frame on the point of suspension and define the following quantities

\begin{aligned} \boldsymbol{r}_{\rm COM} & = \pmatrix{0 \\ -\tfrac{\ell}{2} \\ 0} & \mathrm{I}_{\rm COM} = \left[ \matrix{ \ldots & & \\ & 0 & \\ & & \tfrac{m}{12} \ell^2} \right] \\ \boldsymbol{\omega} & = \pmatrix{0 \\ 0 \\ \dot{\theta}} & \boldsymbol{v}_{\rm COM} = \pmatrix{ \tfrac{\ell}{2} \dot{\theta} \\ 0 \\ 0} \end{aligned}

Then find the linear and angular momentum

\begin{aligned} \boldsymbol{p} & = m \boldsymbol{v}_{\rm COM} = \pmatrix{m \tfrac{\ell}{2} \dot{\theta} \\ 0 \\} \\ \boldsymbol{L} & = \mathrm{I}_{\rm COM} \boldsymbol{\omega} + \boldsymbol{r}_{\rm COM} \times (m \boldsymbol{v}_{\rm COM}) = \pmatrix{0 \\ 0 \\ \tfrac{m \ell^2}{3} \dot{\theta}} \end{aligned}

Finally, find the axis of percussion location

$$\boldsymbol{r}_{\rm AOP} = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ \| \boldsymbol{p} \|^2} = \pmatrix{0 \\ -\tfrac{2}{3} \ell \\ 0}$$

So the impact must have happened 2/3 along the rod.