How far will the puck move?

Question

Rapidly rotating puck (f = 25 revolutions per second) moves
along level carpet floor. Intial speed of the puck is V = 50 cm/s,
and its radius r = 4cm, friction is μ = 0.1.

*Estimate* how far will the puck move before it stops.

The puck spins around vertical axis
and slides along level floor.

It's a good idea to try and conduct
some experiments, as usaually.

odu83:

you did notice the *Estimate* thing,
did not you?

Answer 1

Alexander, I'm not sure why I am answering your question here since this particular question solves with some very nasty eliptical ingtegrals, at least according to some work done by Z´en´o Farkas, et, al, published in
Frictional coupling between sliding and spinning motion
(September 27, 2002)
They determined through experimentation that the puck stops moving and spinning at the same time. They also found right hand and left hand convergence of the equations of motion using limits. Solving the integrals, however, required some powerful mathematics software that I don't have access to.
I gues I'm just curious about physics, so I did some research and found that there are interesting solutions out there, but that the math is very complicated. I invite you to follow the link for the solution.

Here are some highlights:
They determined that the equations of motion depend on an initial condition (also proved emperically and theoretically)

e = v/(R*w)
with v = |v| and w = |w|,

In the case you posted, e>1 (4.77462, no dimension as you probably already noted)

so the force and torque on the disk can be calculated through an integration, which is shown in their paper
http://arxiv.org/PS_cache/physics/pdf/0210/0210024v1.pdf

Since you asked about how far the puck will move before stopping, that is found through
m*dv/dt=-μm*g*F(e)
where F(e) for e>1 is
(4/(3*pi))*
((e^2+1)*E(1/e)-
(e^2-1)*K(1/e))

K(1/e) and E1/e) are the complete elliptic integral
functions of the first and the second kind, respectively.

j

Answer 2

In general, equate the initial energy KE to the energy dissipated (W = FD) by the friction force F = kN. This follows because the only energy the puck received is that goose to get it going. From then on, energy is lost due to friction. So, in the end, all that initial energy will be lost to the work of overcoming friction forces. In other words, KE - W = 0 energy left when the puck stops.

Thus, KE = 1/2 mv^2 = kND= k mg D = FD = W; where D is the distance the puck travels with the friction force (F = kmg) acting on it. k = your mu, N = normal weight of the puck, and g = 9.81 ~ 10 m/sec^2 at Earth's surface.

Thus, D = KE/F = 1/2 mv^2/kmg = 1/2 (v^2/kg); you can do the math. Don't forget to change your cm units into m units to be consistent with the g units.

I remain unclear where the rotation (not revolution) of the puck enters into your problem. If you meant the puck to be lying on its face and spinning while moving along the carpet, there would be frictional energy lost there as well. But given the size of the puck, compared to D, I would think that energy lost would be relatively small. It would have little effect on how far D went.

Answer 3

i imagine there will be a distinction on condition that friction develop into non-linear. operating example, as Alexander has stated with a floor cleanser, might want to friction be a function below linear, then spinning the %. would effectively cut back the linear element contained in the speed vector field of the %.'s bottom. on the different hand, might want to friction be a function more desirable than linear, e.g. parabolic, the opposite is real. All circumstances assume an similar linear speed, the purely distinction being 2) no matter if a spin is imparted, and three) the nature of the friction at play. A difficulty i have not totally seen yet is not any matter if such spinning disks would even holiday in a striaght line. keep that one for yet another question. apart from, as stated with the help of countless links presented with the help of persons speaking about this problem, the rotational and linear-action friction factors received't be autonomous, which fairly complicates issues, so it really is genuine not undemanding to furnish any pithy answer to this one. Addendum: If friction is a continuing, then each and every thing I reported is out the window, because then we do not inevitably have a case the position all velocities are decreased in share, yet fairly are decreased linearly. Then we ought to guage the vector sum of all those velocities, and that would want to describe the fairly complicated mathematics in the back of this. per chance contained in the APPROXIMATING case we may be able to take care of it as if all of them did cut back linearly. supply Alexander his 10 factors for this one, as he has of route spent extra time in this problem earlier.

Answer 4

Another trick question? Why do you mention rotation along with a fixed coefficient of friction?
_____
Odu: It is interesting that you say the puck will stop spinning and moving at the same time. I suspect the coefficient of friction is not static and varies with velocity. We already know it changes dramatically between static and dynamic.