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Suppose we "Keep" a particle on a circular, rotating turn-table, rotating with constant angular velocity $\dot{\theta}$, (i.e, the at $t=0$,when the particle makes contact with the table, the turntable is already rotating,but the particle is at rest). We need to find the subsequent trajectory. Now, intuitively, The particle should simply be at the same distance from the axis of rotation at all times, since it has absolutely no interaction with the table. If we approach this mathematically:

Using polar coordinates,and using $\ddot{\theta}=0$, we get $\vec{a}= \left(\ddot{r} - r \dot{\theta}^2\right)\hat{r} + \left( 2\dot{r}\dot{\theta}\right)\hat{\theta}$. Where $r$ is the distance of the particle from the axis of rotation of the turn table. Since The force acting on the particle is $0$, The acceleration should be the the zero vector, i.e, $0\hat{r}+0\hat{\theta}$.

Solving $2\dot{r}\dot{\theta}=0$, we get $r$=constant. Now, solving $\ddot{r} - r\dot{\theta}^2=0$, since $r$ is constant, $\ddot{r}$=$0$ ,which then implies $r\dot{\theta}^2=0$. Hence , we see the only solution is $r=0$.

If we try to make physical sense of this solution, this would mean that the particle immediately get gets sucked in to the center of the table, Which really doesnt seem to make any sense. What is the reason behind this anomaly using polar coordinates? And how exactly then, Can we find out the trajectory of the particle,mathematically?

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If there is no interaction between table and particle, and the particle is at rest initially, it shouldn't move, so both $\dot{r}, \dot{\theta}$ should be zero, shouldn't it?

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  • $\begingroup$ Oh gosh. $\dot{\theta}$ in the equation (polar cordinates) is the rate at which the PARTICLE changes the angle, I put in the value of that of the table it seems.... $\endgroup$ – satan 29 May 1 at 11:59
  • $\begingroup$ so the trajectory: The particle will simply do nothing right? It simply "stays" on the table motionless, am i correct? $\endgroup$ – satan 29 May 1 at 12:01

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