Answer :
To solve this exercise it is necessary to apply the kinematic equations of angular motion.
By definition we know that the displacement when there is constant angular velocity is
[tex]\theta= \theta_0 +\omega t[/tex]
From our given data we know that,
[tex]\omega = 76\frac{rev}{min}[/tex]
[tex]\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})[/tex]
[tex]\omega = 7.958rad/s[/tex]
Moreover we know that
[tex]\theta_0 = 0.47 rad[/tex]
Therefore for time t=8.1s we have,
[tex]\theta= \theta_0+ \omega t[/tex]
[tex]\theta= 0.47+(7.958)(8.1)[/tex]
[tex]\theta = 64.9298rad[/tex]
That number in revolution is:
[tex]\theta = 64.9298rad(\frac{1rev}{2\pi})[/tex]
[tex]\theta = 15.108 Revolutions[/tex]
Here, we see that there are 15 complete revolutions
And 0.108 revolutions i not complete, so the tunable rotation is
[tex]\theta_{net} = 0.108*2\pi=0.216\pi[/tex]
Therefore the angle of the speck at a time 8.1s is [tex]0.216\pi[/tex]