Problem statementSolution video |

**DISCUSSION THREAD**

Any questions??

As P moves around on the circular track, two things occur:

- The normal force
*N*on P due to the circular guide is proportional to the centripetal acceleration of P:*N = mv*.^{2}/R - A friction force opposes its motion, where the sliding friction force is proportional to the normal force between the circular guide and P:
*f = μ*_{k}N = mμ_{k}v^{2}/R.

From this, we see that the friction force goes to zero as the speed goes to zero. What does this imply about P coming to rest? Can you see this in the animation of the motion below?

*HINTS*:

You will need to use the chain rule of differentiation to set up this problem: *dv/dt = (dv/ds)(ds/dt) = v (dv/ds)*.

I got an asymptomatic solution for C, and the animation seems to concur. So technically speaking, would the ball ever come to rest?

Let's hear from some more people on this.

I got that as well. I decided that since using e^-5 is 0.0067, that is a close enough approximation for zero. However, I wasn't sure if just declaring a time closest to zero and then solving is correct since it should technically go to infinity.

I got the same problem. What I did is used e^-3, which is 0.0498, and that can be considered as no speed. You can also use -4, -5, and down, but that would not significantly change the distance traveled.

i am also getting an exponential decay function, which never technically touches zero. as s goes to infinity, i would never touch the x axis signifying that the velocity is equal to zero.