[4th repost] What is inclination angle of the coin?

Question

After a coin was accidently dropped on the floor, it is rolling around the circle R = 0.4m. The speed of the coin v = 1.00 m/s. What is the inclination angle of the coin to the vertical?

The coin is a thin uniform disk of radius r << R.
The force of friction is sufficient to keep the coin from slipping.
Ignore all enrgy losses, air, etc..

Rufus:
in rotating frame of reference in addition
to centrifugal force you also have
Coriolis force. This Coriolis force
pushes the top of rolling coin outward,
and the bottom of the coin inward.

Answer 1

I remember struggling with that problem in classical mechanics. Here's a formulation of the problem that gives you the answer and sort of walks you through the process.

Like any problem from a textbook (Fetter & Walecka 5.3), you can probably find a complete solution online somewhere.

edit--how do you figure the coriolis force is negligible? It's going to be of similar order of magnitude as the centrifugal force.

Answer 2

Wow, it's been a long time since my physics was fresh enough to tackle this one. But here's a great resource on this general problem. Enjoy.

http://www.eulersdisk.com/rollingdisk.pdf

Personally, my suspicion is that the solution is found by equating the centripetal force pushing the coin outwards to the gravitational force pulling it downwards. The centripetal force is pretty easy - you've got a velocity and a radius.

I believe that the gravitational force acting on a leaning object is mg*sin(lean angle). It's 0 when the lean angle is 0, and mg when the lean angle is 90 degrees.

Plug and chug.

Will be curious to know if this is right.

--

Coriolos force is tiny - irrelevant.

Answer 3

Looking over the links offered by Mark P and Bekki B, I think this problem is already pretty well covered. Yes, this is a classic problem in mechanics of solid bodies. What's the point of working this one out when I can just read the explanations. I can't figure out any other considerations beyond those already included.

Answer 4

in case you think approximately the vulnerable coin comparable to an vulnerable curve, evaluate those forces Ncos teta -mg=0 ....eq a million Nsin teta= mv^2/R....eq....2 in case you divide eq 2 by using eq a million you stumble on tan teta= v^2/Rg an then teta= tan^-a million (v^2/Rg) This the only way i recognize to clean up this situation...wish it rather is nice P.A.