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

$148~rad/s$
The required angular velocity can be determined as follows: $\alpha_A r_A=\alpha_B r_ B$ $\implies 300\sqrt{t}\times 0.7=\alpha_B\times 1.4$ $\implies \alpha_B=150\sqrt{t}$ We know that $\alpha_B=\frac{d\omega_B}{dt}$ $\implies \alpha_B dt=d\omega_B$ $\implies 150\sqrt{t}dt=d\omega_B$ $\implies \int_0^{\omega_B}=\int_0^t 150\sqrt {t}dt$ This simplifies to: $\omega_B=(\frac{150}{1.5}t^{1.5})|_0^t$ We plug in the known values to obtain: $\omega_B=(100)(1.3)^{1.5}$ $\implies \omega_B=148.22rad/s$ As the angular velocity of the propeller is the same as that of the gear B, we have: $\omega_p=148~rad/s$