When does it stop?

Due to proportionality, $-a = W * k ... (1)$ where $a$ is the deceleration due to friction, $W$ is the angular velocity, $k$ is the constant of proportionality.

since $a = \frac{dW}{dt} ... (2)$

& $W = \frac{dθ}{dt} ... (3)$

we get from (1) & (2)

$dW = - dθ * k$ on integrating,

$W = k * θ + C$

since initially $θ = 0$ ,

$W^0 = C ... (4)$ ( $W^0$ is initial angular velocity. )

Now, after n revolutions

$\frac{W^0}{2} = - k *(2n*π) + W^0$ ( since after 1 revolution is equal to 2π radian )

$\frac{W^0}{2} = k * (2n*π )$

we get, $k = \frac{W^0}{4n*π} ... (5)$

Finally when the flywheel comes to rest, $W = 0$ $W^0 = -k*θ + W^0$

$0 = -\frac{W^0}{4n*π} * θ' + W^0$ ( $θ'$ = Total angular displacement )

$θ' = 4n*π$

$\frac{θ'}{2π} =$ Total number of revolutions $= 2n$

Since $n$ have already occurred, after n more revolutions the flywheel will come to stop.