[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..

Answer 1

Model the center of mass of the coin in a FBD. There are two forces acting on the coin:

gravity=m*g

and centripetal
=m*v^2/r

if the coin is inclined towards the center of the radius of rotation, the components of the forces are

m*g*cos(th)

and m*v^2/r*sin(th)

since these must balance,

tan(th)=g*r/(v^2)
th=atan(9.81*.4)

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Answer 2

in case you think approximately the vulnerable coin equivalent 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 eq a million you detect tan teta= v^2/Rg an then teta= tan^-a million (v^2/Rg) This the only way i understand to remedy this concern...desire it particularly is sturdy P.A.

Answer 3

Use the formula:
tan (angle) = (v^2)/ rg

The answer will differ depending on the value of g that you are using, either g = 10ms^(-2) or g = 9.8ms^(-2).

tan (angle) = 1^(2)/ (0.4 x 10)
tan (angle) = 0.25
angle = 14.04 degrees

OR

tan (angle) = 1^(2)/ (0.4 x 9.8)
tan (angle) = 0.26
angle = 14.57 degrees