0
$\begingroup$

Assume there is a round plate whose moment of inertia is $I$ with regard to the axis cross its center, supported by a stick at its center. On the plate there is an insect sleeping at some point on the plate. At first, the plate and insect are static. Then the insect starts to move. In the end, the insect moves back to where it slept. Will the plate rotate back to its original position? We ignore

By conservation of angular momentum, the initial angular momentum is $0$, so we have $$ \vec{r}(t) \times m\vec{v}(t)-I\vec{\omega}(t) = 0 $$ $\vec{r}$ denotes the position of insect while taking the center as the origin. $m$ denotes the mass of insect, and $\vec{\omega}$ denotes the angular velocity of the plate. And we know $$ \int_{0}^{t} \vec{v}(t)\ dt = \vec{0} $$

Intuitively I know that $r \times v$ represents the speed of the area swept by $r$. So the plate should go back to its original position when the insect does not go around the center. If the insect go around the center by $n$ rounds, the plate should also rotate by $2n\pi$(Edit: Now I think this seems not right). But I don't know any rigorous proof. Thank you for any help!

$\endgroup$
  • $\begingroup$ Which reference frame are you observing the insect from? In other words, does the "original position" correspond to the same spot on the plate, or the same spot in space for an observer sitting to the side of the plate? $\endgroup$ – David White Sep 11 '18 at 19:06
  • $\begingroup$ @DavidWhite Just the inertial frame of reference. We can assume a person sits besides. $\endgroup$ – Edward Wang Sep 11 '18 at 19:08
  • $\begingroup$ So does the bug return to its original spot on the plate or to the same spot in space? $\endgroup$ – David White Sep 11 '18 at 19:11
  • $\begingroup$ @DavidWhite The bug return to the same spot in space. $\endgroup$ – Edward Wang Sep 11 '18 at 19:12
0
$\begingroup$

In the end, the insect moves back to where it slept. Will the plate rotate back to its original position?

Here we assume that there is no friction between the plate and the stick, i.e., there is no external torque acting on the plate-insect system, i.e., its net angular momentum should always be zero.

Let's, for simplicity, assume that the insect stays at the same distance from the center of the plate, at radius r.

As the insect starts moving along a circular path, its angular acceleration, angular speed and angular distance will be matched by the opposite angular acceleration, angular speed and angular distance of the plate at the ratio of $\frac I {mr^2}$, where $mr^2$ is angular momentum of the insect.

So, if the insect keeps moving around the circle, it'll arrive at its original spot after traveling by the angular distance of $2\pi \frac I {I+mr^2}$. By that time, the plate will turn by the angle of $2\pi \frac {mr^2} {I+mr^2}$, so it won't be in its original position.

If the insect turn around and starts moving back, the plate will start turning back as well, with the matching angular acceleration and speed (at the same ratio as before) and, therefore, by the time the insect reaches its original spot, the plate will reach its original spot as well, as you've anticipated.

$\endgroup$
0
$\begingroup$

To construct a counterexample where returning to his resting spot does not return the disc to its initial position, make the (unreasonable) assumption that the plate and the insect have the same moment of inertia. Then conservation of angular momentum demands that the insect and the plate have equal and opposite angular velocities. If the insect starts out at rest at the top of the wheel, he'll find that resting spot again at the bottom of the wheel.

If the ratio of the moments of inertia of the insect and the plate are any irrational number, then every time the insect loops back around to the top, the disc will have rotated by a noninteger number of turns.

If the insect turns around, however, then time-reversal symmetry holds.

$\endgroup$
0
$\begingroup$

Say that the plate is not perfect. It has a bulge at some location. Center of mass of the plate is not at the center of the plate.

No external torque is applied on the combined system (plate + insect). So the combined system is static. This implies that the angle to the combined center of mass is fixed.

If insect returns to the same place in space, so does the bulge, else center-of-mass of the system is rotated from its original location.

The bulge may be as small as we like with the same results. I would use continuity arguments to conclude the same for the no-bulge-case.

$\endgroup$

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.