Answer
a) $a=2.5m/s^2$
b) $\phi=17.7^o$
Work Step by Step
We have $\alpha=+2rad/s^2$, initial angular velocity $\omega_0=+1.5rad/s$ and $r=0.38m$
a) At $t=0.5s$, the blade acquires a velocity of $$\omega=\omega_0+\alpha t=+2.5rad/s$$
At the same time,
- The blade's centripetal acceleration $a_c=r\omega^2=2.38m/s^2$
- The blade's tangential acceleration $a_T=\alpha r=0.76m/s^2$
Because these two accelerations are perpendicular with each other, the magnitude of the total acceleration is $$a=\sqrt{a_c^2+a_T^2}=2.5m/s^2$$
b) The angle $\phi$ between $\vec{a}$ and $\vec{a_c}$ is the angle the total acceleration makes with the radial axis. So we have $$\tan\phi=\frac{a_T}{a_c}=0.319$$ $$\phi=17.7^o$$
