[7th 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..

It is 7th repost. No luck so far.

I still have trouble understanding why everybody tries to apply "4th law of Newton of universal inclination"

One answer mentioned that "this is the way they bunk the highway ramps".

Bekki, I tried to use your reference, but
screwed up my calculations again, as I
always do. My teacher is gradually getting frustrated with my inability to
solve this simple problem.

In addition to usual 10 points, I will also
thumb up 20 of answers of anyone
who posts the correct answer for 20
more points!

Answer 1

I gave you a good reference about 4 posts back that tells you the answer just about (you just have to do some algebra). Lemme dig it up, since you may be ready to give it a shot again.

There.

Okay, I'm assuming you can follow the solution given to equation (5), which relates d(phi)/dt to that yucky fuction of theta. Theta is what you want. d(phi)/dt is just v/r. Their (b) is just your (R)--the circle's radius. Their (a) is your (r)--the coin's radius. Now you said that the circle is much bigger than the coin. So just drop the 5a sin (theta) term. It's small. Now do some algebra.

(theta) = tan-1 (6R/4g (v/r)^2)

Check my algebra, plug in the numbers, and you're done.

Answer 2

You need to equate the force due to the coin leaning in toward the center as a function of the angle and equate that to the centrifugal force pushing outward and solve for the angle.

Answer 3

atan(g*R/v^2)

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