Explanation:
A point on the rim of the wheel went from rest to 15 m/s so we can solve for the acceleration as
[tex]v = v_0 + at \Rightarrow a = \dfrac{v}{t} = \dfrac{15\:\text{m/s}}{8.0\:\text{s}}[/tex]
or
[tex]a = 1.9\:\text{m/s}^2[/tex]
We also know that the angular acceleration [tex]\alpha[/tex] is
[tex]a = r\alpha \Rightarrow \alpha = \dfrac{a}{r}[/tex]
Using r = 0.30 cm and a = 1.9 m/s^2, we get
[tex]\alpha = \dfrac{a}{r} = \dfrac{1.9\:\text{m/s}^2}{0.30\:\text{m}}[/tex]
[tex]\;\;\;\;=6.3\:\text{rad/s}^2[/tex]
