## Engineering Mechanics: Statics & Dynamics (14th Edition)

$\omega=-19.245rad/s$ $\alpha=-183.04 rad/s^2$
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: $\alpha=-183.04 rad/s^2$