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

$\omega_C=1.68rad/s$ $\theta_C=1.68rad$
We can determine the required angular velocity and the angular displacement as follows: As $\omega_A=(\omega_A)+\alpha_A t$ $\implies \omega_A=0+3t=3t$ At $t=2s$ $\omega_A=3(2)=6rad/s$ We know that $\omega=\frac{d\theta}{dt}$ $\implies 3t=\frac{d\theta}{dt}$ $\implies 3tdt=d\theta$ $\int_0^t 3tdt=\int_0^{\theta_A} d\theta$ This simplifies to: $\theta_A=\frac{3}{2}t^2$ At $t=2s$ $\theta_A=\frac{3}{2}(2)^2=6 rad$ Similarly, $\omega_B=\frac{\omega_A r_A}{r_B}$ We plug in the known values to obtain: $\omega_B=\frac{6(35)}{125}=1.68rad/s$ and $\theta_A r_A=\theta_B r_B$ $\implies \theta_B=\frac{\theta_Ar_A}{r_B}=\frac{6(35)}{125}=1.68rad$ The shaft C has the same angular displacement and the angular velocity as that of B. Therefore $\omega_C=1.68rad/s$ and $\theta_C=1.68rad$