To solve this problem we will apply the kinematic equations of angular motion for which the square of the angular velocity change is described is equivalent to twice the angular acceleration by time. This is,
[tex]\omega_2^2 - \omega_1^2 = 2\alpha \theta[/tex]
Here,
[tex]\omega_{1,2}[/tex]= Final and initial Angular Velocity
[tex]\alpha[/tex]= Angular acceleration
[tex]\theta[/tex]= Angular displacement
Our values are given as,
[tex]D = 2.06cm=0.0206m \rightarrow r = \frac{d}{2} = 0.0103m[/tex]
[tex]r = 0.0103 m[/tex]
[tex]\omega_1 = 15rad/s[/tex]
[tex]\omega_2 = 0rad/s \rightarrow[/tex] When is coming to rest
[tex]\alpha = -1 rad/s^2[/tex]
Replacing we have that,
[tex]\omega_2^2 - \omega_1^2 = 2\alpha \theta[/tex]
[tex]\theta = \frac{(\omega_2^22 - \omega_1^2 )}{2\alpha}[/tex]
[tex]\theta= \frac{(0 - (15 rad/s)^2 )}{2(-1 rad/s^2)}[/tex]
[tex]\theta= 112.5 rad[/tex]
Through the arc length ratio the distance in meters is
[tex]x = r\theta[/tex]
[tex]x = (0.0103 m)(112.5 rad)[/tex]
[tex]x = 1.158 m[/tex]
Therefore the coin will roll 1.158m before coming to rest.
