Answer
$\omega=8.7027rad/s$
$\alpha=-50.4968rad/s^2$
Work Step by Step
We can determine the required angular velocity and angular acceleration as follows:
As $cos\theta=\frac{-y}{0.3}$
$\implies y=-0.3cos\theta$
We know that
$v_y=\frac{dy}{dt}$
$\implies v_y=\frac{dy}{d\theta}\times \frac{d\theta}{dt}$
$\implies v_y=0.3sin\theta(\frac{d\theta}{dt})~~~$[eq(1)]
We plug in the known values to obtain:
$2=0.3sin50(\frac{d\theta}{dt})$
This can be rearranged as:
$\frac{d\theta}{dt}=\frac{2}{0.3sin50}$
$\implies \omega=8.7027rad/s$
After differentiating eq(1), we obtain:
$a_y=0.3cos\theta(\frac{d\theta}{dt})^2+0.3sin\theta\frac{d^2\theta}{dt^2}$
We plug in the known values to obtain:
$3=(0.3cos 50)\omega^2+(0.3sin50)\alpha$
As $\omega=8.7027rad/s$
$\implies \alpha=\frac{3-(0.3cos50\times (8.7027)^2)}{0.3sin50}$
This simplifies to:
$\alpha=-50.4968rad/s^2$
