Question Number 136593 by I want to learn more last updated on 23/Mar/21
Answered by mr W last updated on 24/Mar/21
Commented by otchereabdullai@gmail.com last updated on 25/Mar/21
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Commented by mr W last updated on 24/Mar/21
$$\omega_{{DB}} =\mathrm{10}\:{rad}/{s} \\ $$$${v}_{{B}} =\mathrm{150}\sqrt{\mathrm{2}}\omega_{{DB}} \:=\mathrm{1500}\sqrt{\mathrm{2}}\:{mm}/{s} \\ $$$${v}_{{Bx}} ={v}_{{B}} \mathrm{cos}\:\mathrm{45}°=\mathrm{1500}\:{mm}/{s}=\mathrm{1}.\mathrm{5}\:{m}/{s} \\ $$$${v}_{{By}} ={v}_{{B}} \mathrm{sin}\:\mathrm{45}°=\mathrm{1500}\:{mm}/{s}=\mathrm{1}.\mathrm{5}\:{m}/{s} \\ $$$${v}_{{Ax}} ={v}_{{Dx}} ={v}_{{Bx}} =\mathrm{1}.\mathrm{5}\:{m}/{s} \\ $$$${v}_{{Ay}} =\mathrm{0} \\ $$$${v}_{{Dy}} =\frac{\mathrm{150}}{\mathrm{150}+\mathrm{200}}×{v}_{{By}} =\frac{\mathrm{3}}{\mathrm{7}}×\mathrm{1}.\mathrm{5}=\frac{\mathrm{4}.\mathrm{5}}{\mathrm{7}}\:{m}/{s} \\ $$$$\Rightarrow{v}_{{A}} =\mathrm{1}.\mathrm{5}\:{m}/{s} \\ $$$$\Rightarrow{v}_{{D}} =\sqrt{\mathrm{1}.\mathrm{5}^{\mathrm{2}} +\left(\frac{\mathrm{4}.\mathrm{5}}{\mathrm{7}}\right)^{\mathrm{2}} }=\mathrm{1}.\mathrm{642}\:{m}/{s} \\ $$$$\Rightarrow\omega_{{AB}} =\frac{{v}_{{By}} }{\mathrm{150}+\mathrm{200}}=\frac{\mathrm{1500}}{\mathrm{350}}=\frac{\mathrm{30}}{\mathrm{7}}=\mathrm{4}.\mathrm{286}\:{rad}/{s} \\ $$$$\omega_{{CA}} =\frac{{v}_{{Ax}} }{\mathrm{150}}=\frac{\mathrm{1500}}{\mathrm{150}}=\mathrm{10}\:{rad}/{s} \\ $$
Commented by I want to learn more last updated on 24/Mar/21
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciste}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$