[5th 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's 5th repost.
I already have a dozen of solutions,
all of which invoke "the fourth Newton's law of inclination: tan(a) = F/N". There are only three Newton's laws, sorry.

Answer 1

the force exerted from floor to the coin is parallel to the coins face so the angle the force inclined to the vertical is equal to the inclination angle of the coin to the vertical we may call the angle α. and the magnitude of the force is R
if the coins mass is m
R.Cos α = m.g-----(1)

towards the center
F = ma
R.Sinα = m(v^2)/r-----(2)

(2)/(1) Tanα = (v^2)/r.g
α = tan^(-1)⁡ { (v^2)/r.g }
α = tan^(-1)⁡ 1/4
α = 14 degrees

Answer 2

if you consider the inclined coin similar to an inclined curve, consider these forces

Ncos teta -mg=0 ....eq 1

Nsin teta= mv^2/R....eq....2

if you divide eq 2 by eq 1

you find

tan teta= v^2/Rg

an then teta= tan^-1 (v^2/Rg)



This the only way I know to solve this problem...hope it's good

P.A.

Answer 3

You seem to have all the answers, so why do you ask us lot - are you testing us all or something?

Get a life!