A very thin symmetric ring

Question

A very thin symmetric ring with a radius of R was spun at an angular velocity around its central axis and set flat on a smooth horizontal surface. When the friction coefficient between the desk and the ring is, how long does it take for the ring to stop? How many times will the ring turn before it comes to a complete stop?

Answer 1

I = mR² for the ring at angular velocity, ω₀.

The equation of motion will be determined by the torque exerted by the friction force decelerating the spin tp 0:

-F_f R = Iα with F_f = Mgμ (force is negative because it acts against spin

MgμR = MR²α

Solve for α.

Then use rotational kinematics equations to solve for t and Δθ

ω = ω₀ + αt Δ

Δθ = ω₀t + 1/2 αt² or, easier once you have t, Δθ = 1/2 (-ω₀)/t