Answer
Ans:
$$
\begin{aligned}
\omega_B r_B=21.9 \mathrm{rad} / \mathrm{s} \curvearrowleft
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
\int_{15 \mathrm{rad} / \mathrm{s}}^{\omega_A} \omega d \omega & =\int_0^{\theta_A}\left(2+0.006 \theta^2\right) d \theta \\
\left.\frac{\omega^2}{2}\right|_{15 \mathrm{rad} / \mathrm{s}} ^{\omega_A} & =\left.\left(2 \theta+0.002 \theta^3\right)\right|_0 ^{\theta_A} \\
\frac{\omega_A^2}{2}-\frac{15^2}{2} & =2 \theta_A+0.002 \theta_A^3 \\
\omega_A & =\sqrt{0.004 \theta_A^3+4 \theta+225} \mathrm{rad} / \mathrm{s} \\
\theta_A=10(2 \pi)= & 20 \pi \mathrm{rad}, \\
\omega_A & =\sqrt{0.004(20 \pi)^3+4(20 \pi)+225} \\
& =38.3214 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
gear $B$ is meshed with gear $A$
$$
\begin{aligned}
\omega_B r_B=\omega_A r_A ; \quad \omega_B(175) & =38.3214(100) \\
\omega_B & =21.8979 \mathrm{rad} / \mathrm{s} \\
\omega_B & =21.9 \mathrm{rad} / \mathrm{s} \curvearrowleft
\end{aligned}
$$
