Answer:
[tex]\alpha=28.57\frac{rad}{s^2}[/tex]
Explanation:
We use the following rotational kinematic equation to calculate the angular acceleration of the rod:
[tex]\omega_f=\omega_0+\alpha t[/tex]
Here [tex]\omega_f[/tex] is the final angular speed, [tex]\omega_0\\[/tex] is the initial angular speed and [tex]\alpha[/tex] is the angular acceleration. The rod starts rotating from rest ([tex]\omega_0=0[/tex]):
[tex]\alpha=\frac{\omega_f}{t}(1)[/tex]
Recall that the angular speed is defined in function of the tangential speed (v) and the radius (r) of the circular motion:
[tex]w_f=\frac{v_f}{r}(2)[/tex]
In this case the radius is given by [tex]r=\frac{20*10^{-2}m}{2}=0.1m[/tex]. Replacing (2) in (1):
[tex]\alpha=\frac{v_f}{rt}\\\alpha=\frac{20\frac{m}{s}}{(0.1m)7s}\\\alpha=28.57\frac{rad}{s^2}[/tex]