1
$\begingroup$

If a ball collides with and sticks to an unhinged uniform rod on a smooth table (the path of the ball does not necessarily pass through the center of the uniform rod), is angular momentum conserved about the center of mass of the rod and ball or about the center of the rod? Why? Is angular momentum about both conserved, and, if so, how can we calculate this?

$\endgroup$
  • 1
    $\begingroup$ Hi 333 and welcome to Physics.SE! I added the homework tag because as you can see from the definition of the tag that the homework tag “applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright. $\endgroup$ – Gonenc May 14 '15 at 19:02
1
$\begingroup$

Angular momentum should be conserved in any inertial frame of reference - if you move with the center of mass, the motion you see will be rotation about the center of mass; if you move with a different frame of reference, you will see rotation about a different axis.

So the short answer is "it doesn't matter".

Whether they "give the same answer" depends on whether you do the calculations properly...

$\endgroup$
  • $\begingroup$ There will be some point on the rod which will be purely in translation after that collision and and I think it'll be combined center of mass of the stuck particle and the rod. So if I conserve angular momentum about any other axis( name it AB) and not combined center of mass axis ( name it PQ), the final ang. velocity that I get upon solving the equation about AB, will it be omega relative and not absolute omega. If i have to match answers, should I consider that, as axis AB wasn't in pure rotation and itself rotating about PQ, I have to add or subtract some value to get omega w.r.t. ground? $\endgroup$ – 333 May 15 '15 at 18:32
  • $\begingroup$ Are you struggling with the calculation of omega, or worrying about conservation of angular momentum? Or you want to use one to calculate the other? There should not be any point that is not rotating after the impact - although there is a point called the "center of percussion" that is stationary right after the collision. Note than an inertial frame of reference does not in fact rotate. It just translates. $\endgroup$ – Floris May 15 '15 at 19:11
1
$\begingroup$

Angular momentum depends on the axis about which you calculate it. That means you'll get a different number depending on the axis, but no matter the choice, it's always conserved (assuming no external forces act on the system).

For example, imagine a ball of mass $m$ and velocity $v$ hits and sticks to the rod a distance $d$ from its center:

  • If you choose the rotation axis to be the center of the stationary rod, you would calculate the angular momentum before the collision to be $mvd$. After the collision the system ang. mom. is still $mvd$ about the fixed point where the rod was initially.
  • Now say you choose an axis at the point where the ball hits the rod. The initial ang. mom. is zero, so the final ang. mom. is also zero around that point. After the collision, the system is rotating, but its linear momentum is offset from the chosen axis, so there are two contributions to ang. mom. that exactly cancel.
$\endgroup$
  • $\begingroup$ Very nicely put. $\endgroup$ – Floris May 15 '15 at 11:24

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.