Answer
Ans:
$$
\begin{aligned}
a_B=29.0 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
& \int_{5 \mathrm{rad} / \mathrm{s}}^{\omega_A} \omega d \omega=\int_0^{\theta_A} 0.8 \theta d \theta \\
& \left.\frac{\omega^2}{2}\right|_{5 \mathrm{rad} / \mathrm{s}} ^{\omega_A}=\left.\left(0.4 \theta^2\right)\right|_0 ^{\theta_{\Lambda}} \\
& \frac{\omega_A^2}{2}-\frac{5^2}{2}=0.4 \theta_A^2 \\
& \omega_A=\left\{\sqrt{0.8 \theta_A^2+25}\right\} \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
At $\theta_A=3(2 \pi)=6 \pi \mathrm{rad}$,
$$
\begin{aligned}
& \omega_A=\sqrt{0.8(6 \pi)^2+25}=17.585 \mathrm{rad} / \mathrm{s} \\
& \alpha_A=0.8(6 \pi)=4.8 \pi \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$
Since pulleys $A$ and $C$ are connected
$$
\begin{aligned}
& \omega_C r_C=\omega_A r_A ; \quad \omega_C(40)=17.585(50) \\
& \omega_C=21.982 \mathrm{rad} / \mathrm{s} \\
& \alpha_C r_C=\alpha_A r_A ; \quad \alpha_C(40)=(4.8 \pi)(50) \\
& \alpha_C=6 \pi \mathrm{rad} / \mathrm{s}^2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \left(a_B\right)_t=\alpha_C r_B=6 \pi(0.06)=1.1310 \mathrm{~m} / \mathrm{s}^2 \\
& \left(a_B\right)_n=\omega_C^2 r_B=\left(21.982^2\right)(0.06)=28.9917 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
$$
\begin{aligned}
a_B=\sqrt{\left(a_B\right)_t^2+\left(a_B\right)_n^2} & =\sqrt{1.1310^2+28.9917^2} \\
a_B & =29.01 \mathrm{~m} / \mathrm{s}^2=29.0 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
