Suppose a 1-kg ball A is fixed to a spoke 0.2 m long, which is attached to an axle so that the ball can rotate (v=10m/s, KE=50J, $\omega$=50 rps, L=2, p=0)

Now, there is a second ball B (m=1kg), attached to a 0.1 m stick that can likewise rotate, but at rest. An elastic collision takes place at the point where the two circular paths intersect, ball A stops dead and ball B starts moving at 10 m/s (KE = 50J v, KE is always conserved),

case a) - if ball B is not attached to a stick it acquires linear momentum (v=p = 10 m/s), but ball A had no linear momentum (p = 0),

case b) - if it is, it acquires angular momentum L = 1, ($\omega$= 100 rps), but ball A had L = 2. In this second case is angular momentum conserved?

@JohnRennie, your answer covers the angular momentum of B in case a), what about its linear momentum (from 0 to 10)?

can you also address the angular momentum of B in case b) ?

  • $\begingroup$ Since the KE is equal, it would be easy to determine if angular momentum is conserved. $\endgroup$
    – LDC3
    Aug 8, 2014 at 4:45

2 Answers 2


Consider the following diagram:


This shows a mass $m$ moving past a point $P$ in a straight line. Note that the mass isn't connected to $P$ in any way - it's just moving past in a straight line.

The angular momentum of $m$ about $P$ is given by:

$$ \vec{L} = \vec{r} \times m\vec{v} $$

So the direction of $\vec{L}$ is normal to the screen and the magnitude is:

$$ L = rmv \sin\theta \tag{1} $$

and since:

$$ \sin\theta = \frac{d}{r} $$

Substituting this into equation (1) gives:

$$ L = mvd \tag{2} $$

Note that everything in equation (2) is a constant, so that tells us that $L$ is a constant as well, so angular momentum is conserved even though at first glance this isn't a system that is rotating.

And this provides the answer to your question. Even though in your second example ball $B$ isn't attached to anything it still has an angular momentum and this angular momentum is still conserved.

Response to comment:

Your distinction between angular and linear momentum is an artificial one. If you look at my working above the particle obviously has a linear momentum, but also an angular momentum. What's more the value of the angular momentum depends on where you fix the point $P$ so there is no unique value of angular momentum.

When you have a ball on a rotating rod the direction of linear momentum is not conserved because there is a force acting (through the rod) and Newton's laws tells that force is the rate of change of momentum:

$$ \vec{F} = \frac{d\vec{p}}{dt} $$

However there is a theorem (Noether's theorem) thats tell us angular momentum is conserved if the force is independant of angle. So if we calculate the angular momentum about the pivot point we'll find that this quantity is a constant. That's why it's useful for calculating trajectories.

But it is incorrect to say that the linear momentum of the ball attached to the rod is zero. At any time the linear momentum of the ball $A$ is $\vec{p} = m\vec{v}$, but the direction of $\vec{p}$ (though not its magnitude) is changing continuously with time due to the force applied by the connecting rod.

Back to your problem: before the collision the linear momentum of $A$ is mv, but it's direction is continually changing with time. However the angular momentum is constant because the force on $A$ is central.

During the collision a force acts between $A$ and $B$. This force acts normally to the connecting rods, i.e. it's not a central force, so the angular momenta will not be constant. If we consider the collision to take an instant then the only force acting is the one between the two balls so the total linear momentum will be conserved. Before the collision $p_A = mv$ and $L_B = 0$. After the collision $p_A = 0$ and $p_B = mv$, so the total linear momentum $p_A + p_B$ is conserved.

Immediately after the collision $B$ starts rotating about its pivot due to the force applied by its connecting rod. The force applied by its rod means the direction of $p_B$ now changes with time, though its magnitude doesn't. The angular momentum $L_B$ is constant because the force applied by the rod is central around $B$.

Note that it still makes perfect physical sense to calculate $L_A$ - it's just:

$$ L_A = \vec{r}_A \times m\vec{v_B} $$

but the force applied to $B$ by its rod isn't centrally symmetric about $A$, so $L_A$ is not constant (and therefore not terribly useful).

If $B$ isn't connected to a pivot, so it moves off in a straight line, then no forces act on $B$. That means both $\vec{p}_B$ and $\vec{L}_B$ calculated about any point are constant.

  • $\begingroup$ @bobie: I've extended my answer to respond to your comment $\endgroup$ Aug 8, 2014 at 9:34

Ball A has linear momentum pointing in the direction of its flight path. But it is not conserved, because you have a force acting on it (centripetal force), that as no counter part (unless you specify where the axle is mounted and allow for the mount to move as well).

Observe, that you can ascribe an angular momentum to B as well even when it does not move on a circular trajectory (As explained by John Rennie).


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