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Ans\begin{aligned}\omega & =6 \mathrm{rad} / \mathrm{s} \\2 & =15 \mathrm{rad} / \mathrm{s}^{2} \\m & =4 \mathrm{~kg} . \\\left(a_{B}\right)_{n} & =\left(\mathrm{rBC}_{B C}\right) \cdot \omega^{2} \\& =0.1 \times 6^{2}=3.6 \mathrm{~m} / \mathrm{s}^{2} \rightarrow \\\left(A_{B}\right) t & =\left(r_{B}\right) \cdot \alpha=0.1 \times 15=1.5 \mathrm{~m} / \mathrm{s}^{2}+\alpha\end{aligned}a_{A}=a_{G}+a_{A G}\begin{array}{l}\alpha=\sin ^{-}\left(\frac{B C}{A B}\right) \\\alpha=30^{\circ}\end{array}\begin{aligned}\Rightarrow G A & =(1.5+3.6)+(0.2 \alpha<\theta) \\+\rightarrow & \Rightarrow 0=3.6-(0.2 \alpha) \cos \theta \\\alpha & =\frac{3.6}{0.2 \cos 30}=20.78746 \mathrm{rod} / \mathrm{s}^{2}\end{aligned}\text { Sma }=\Sigma(m a) e r r\overline{R_{A}}(0.2 \cos 30)=\bar{I} \cdot \alpha \text {. }(\alpha \bar{I})=\frac{1}{12} m l^{2} \alpha=\underbrace{\frac{1}{I}(4) \times 0.4^{2}}_{\bar{I}} \cdot(20.78746)R_{A}=\frac{1.108664}{0.2 \cos 30^{\circ}}R_{A}=6.400877 \mathrm{~N} \text { Ays } ...