a. The disk starts at rest, so its angular displacement at time [tex]t[/tex] is
[tex]\theta=\dfrac\alpha2t^2[/tex]
It rotates 44.5 rad in this time, so we have
[tex]44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}[/tex]
b. Since acceleration is constant, the average angular velocity is
[tex]\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2[/tex]
where [tex]\omega_f[/tex] is the angular velocity achieved after 6.00 s. The velocity of the disk at time [tex]t[/tex] is
[tex]\omega=\alpha t[/tex]
so we have
[tex]\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}[/tex]
making the average velocity
[tex]\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}[/tex]
Another way to find the average velocity is to compute it directly via
[tex]\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}[/tex]
c. We already found this using the first method in part (b),
[tex]\omega=14.8\dfrac{\rm rad}{\rm s}[/tex]
d. We already know
[tex]\theta=\dfrac\alpha2t^2[/tex]
so this is just a matter of plugging in [tex]t=12.0\,\mathrm s[/tex]. We get
[tex]\theta=179\,\mathrm{rad}[/tex]
Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that
[tex]\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2[/tex]
Then for [tex]t=6.00\,\rm s[/tex] we would get the same [tex]\theta=179\,\rm rad[/tex].
