0
$\begingroup$

I wonder if there is a meaning of rotation for a point particle. Does a point particle have angular momentum and does he reply to torque?

$\endgroup$
  • $\begingroup$ Hi David. The term circular motion means moving in a circle and point particles can of course move in a circle. Do you mean rotation i.e. can a point particle rotate? $\endgroup$ – John Rennie May 17 at 10:58
  • $\begingroup$ yes that's what I mean.. sorry for that $\endgroup$ – david May 17 at 11:02
  • 1
    $\begingroup$ Hi david: Welcome to Phys.SE. Are you talking about a purely classical point particle, or are you also allowing internal quantum spin of the point particle? $\endgroup$ – Qmechanic May 17 at 11:55
  • $\begingroup$ hey, I'm talking about purely classical point particle $\endgroup$ – david May 17 at 12:05
0
$\begingroup$

The one-word answer is "no".

Consider a cube with side $a$ made of material with density $\rho$. The mass of the cube is $\rho a^3$. The moment of inertia about the center is $\rho a^5/6$.

As $a \to 0$, the moment of inertia gets smaller much faster than the mass, because of the $a^5$ factor compared with $a^3$.

Another way to see this is to think about the rotational inertia of a uniform rod of length $l$. You can find the moment of inertia about the center by the "standard" formula, or you can cut the rod into $n$ equal pieces and find the moment of inertia using the parallel axis theorem. If you do that, you find that as $n$ increases, the contribution from the moment of inertia of each piece about its own center is negligible compared with the "$\text{mass}\times\text{length}^2$" contributions from the parallel axis theorem. So as $n \to \infty$ the moment of inertia of each piece (i.e. each "point particle") about its own center is $0$.

$\endgroup$
  • $\begingroup$ so a point particle can't rotate itself, and therefore can't have angular momentum in this specific situation. but a point particle can rotate around another point particle and therefore he can have angular momentum in this case? $\endgroup$ – david May 17 at 15:04

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.