2
$\begingroup$

I am considering a problem where there is a rod at rest on a smooth table, and a particle is incident on the end of the rod with some given speed. The particle collides with the rod and sticks to the end of the rod. I am thinking about the subsequent motion of the system. So far, I think:

  • The centre of mass of the combined system will now continue at a constant velocity in the same direction as the incident particle, and with a speed such as to conserve the linear momentum.
  • Since the centre of mass must continue on at a constant velocity, any rotation, if it occurs, must be about the centre of mass of the new/combined system.
  • There must be some rotation because the speed of the particle is reduced. Therefore the incident particle has received an impulse, and must have exerted an impulse on the end of the rod. This would provide an angular impulse about the centre of mass of the system, and therfore there must be rotation about the centre of mass.

However how can this be? The initial angular momentum of the system is zero as there is no rotation, and yet afterwards there is some angular momnetum as the system has rotation. What am I missing here?

Improve this question
$\endgroup$
  • $\begingroup$ Just because there isn't any 'rotation' (as in circular motion) that doesn't mean the angular momentum is zero. Note that angular momentum is reference frame dependent. What is the angular momentum relative to the C.O.M. of the rod? $\endgroup$ – DilithiumMatrix Mar 1 '17 at 20:50
  • $\begingroup$ @DilithiumMatrix Thank you for your reply. Can I not say that I am considering this from the reference frame of the table? This is how I am assessing the linear momentum after all: initially the rod is at rest with the table, and the particle is moving. In the table frame of reference, initially there is no rotation and afterwards there is a rotation... $\endgroup$ – Meep Mar 1 '17 at 20:58
  • 2
    $\begingroup$ Work in the rest frame of the table. Angular momentum is defined relative to a point, and here an obvious choice of a point is the centre of the rod. The particle then initially has angular momentum, given by $\vec{r} \times \vec{p}$ where $\vec{r}$ and $\vec{p}$ are the particle's radius vector (measured from the point) and momentum vector respectively. The magnitude of this angular momentum is $mbv$ where $m$, $v$ are the particle's mass and speed, and $b$ is half the length of the rod. $\endgroup$ – diracula Mar 1 '17 at 21:46
  • $\begingroup$ @diracula Ah I see so linear velocity/momentum are considered with regards to a reference frame with x and y axis etc whereas angular momentum and velocity are considered with respect to a point. $\endgroup$ – Meep Mar 1 '17 at 22:09
  • 1
    $\begingroup$ Yes, angular momentum (or angular velocity) is defined relative to a point because it depends on the radius vector (which is itself defined relative to a point). $\endgroup$ – diracula Mar 2 '17 at 9:16
1
$\begingroup$

The angular momentum will be conserved. It will be different depending on which frame of reference you are in, but it will be preserved.

Assume that the particle is traveling parallel to the edge of the table, with speed V.

Assume that the end of the rod that the particle strikes is a distance H away from the edge of the table.

Now take a coordinate system with the x axis aligned on the edge of the table.

Before striking the rod, the particle, mass M, has angular velocity HMV.

After the collision, the combined system will still have angular momentum HMV.

The angular momentum of the system will have two contributions. One contribution will be the center of mass of the system moving parallel to the path of the original particle. The other contribution will be the rotation of the combined system about the center of mass.

Note that you could choose a coordinate system where the x axis is in line with the initial path of the particle. In this system, the initial angular momentum is zero. After the collision, the two contributions to the angular momentum will be in opposite directions so that the angular momentum is still zero.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.