Answer
$\omega_C=1.68rad/s$
$\theta_C=1.68rad$
Work Step by Step
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 $\theta_C=1.68rad$
