Answer
$\omega_B=31.714rad/s$ (counter-clockwise)
Work Step by Step
We can determine the required angular velocity as follows:
As $d\omega=\alpha dt$
$\int_{\omega_{\circ}^{\omega_A}}=\int_0^t (2t^3)dt$
This simplifies to:
$[\omega]^{\omega_A}_{15}=[\frac{1}{2}t^4]_0^t$
$\implies \omega_A=\frac{1}{2}t^4+15$
At $t=3s$
$\omega=\frac{1}{2}(3)^4+15=55.5rad/s$
We know that
$\omega_B r_B=\omega_A r_A$
This can be rearranged as:
$\omega_B=\frac{r_A}{r_B}\omega_A$
We plug in the known values to obtain:
$\omega_B=\frac{100}{175}(55.5)$
$\implies \omega_B=31.714rad/s$ (counter-clockwise)
