Answer
$\omega=-19.245rad/s$
$\alpha=-183.04 rad/s^2$
Work Step by Step
The required angular velocity and angular acceleration can be determined as follows:
As $cos\theta=\frac{x/2}{0.3}$
$\implies x=2\times 0.3cos\theta=0.6cos\theta$
We differentiate the above equation to obtain:
$v_x=\frac{dx}{dt}$
$\implies v_x=\frac{dx}{d\theta}\times \frac{d\theta}{dt}$
$\implies v_x=-0.6sin\theta(\frac{d\theta}{dt})~~~$[eq(1)]
We plug in the known values to obtain:
$10=-0.6\omega sin60$
This can be rearranged as:
$\omega=\frac{-10}{0.6sin60}$
$\implies \omega=-19.245rad/s$
Differentiating eq(1), we obtain:
$a_x=\frac{dv_x}{dt}$
$\implies a_x=\frac{dv_x}{d\theta}\times \frac{d\theta}{dt}$
$\implies a_x=-0.6 cos(\frac{d\theta}{dt})^2-0.6 sin\theta \frac{d^2\theta}{dt^2}$
We plug in the known values to obtain:
$-16=-0.6cos60\omega^2-0.6sin60\alpha$
This can be rearranged as:
$\alpha=\frac{-16-0.6cos60(-19.245)^2}{-0.6sin60}$
This simplifies to:
$\alpha=-183.04 rad/s^2$
