91Ó°ÊÓ

Given rotation rates are:- Clevis rotating about the z-axis: \( \Omega = 4 \) rad/s- Shaft B rotating about axis OA: \( \omega_0 = 3 \) rad/s- Change in angle \( \gamma \): \( \frac{d\gamma}{dt} = -\frac{\pi}{4} \) rad/s.The angular velocity of shaft B, \( \omega \), will have components due to these rotations:

The angular velocity \( \omega \) can be expressed as:\[\omega = \Omega \hat{k} + \omega_0 \hat{u}\]Here, \( \hat{k} \) represents rotation about the z-axis, and \( \hat{u} \) is the unit vector along OA.Translate \( \hat{u} \) into the xyz system assuming OA is between the z-axis and the x-axis and identified with \( \gamma \):\(\hat{u} = \cos(\gamma) \hat{i} + \sin(\gamma) \hat{k}\)Substitute \( \gamma = 30^{\circ} = \frac{\pi}{6} \):\[\hat{u} = \cos\left(\frac{\pi}{6}\right) \hat{i} + \sin\left(\frac{\pi}{6}\right) \hat{k}\]\[ \hat{u} = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{k}\]Now substitute back:\(\omega = 4 \hat{k} + 3\left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{k} \right)\)Simplifying:\[\omega = \frac{3\sqrt{3}}{2} \hat{i} + \left(4 + \frac{3}{2} \right) \hat{k}\]\[\omega = \frac{3\sqrt{3}}{2} \hat{i} + \frac{11}{2} \hat{k}\]

The angular acceleration \( \alpha \) of shaft B will include both changes in \( \omega_0 \) and changes due to \( \frac{d\gamma}{dt} \). Since \( \omega_0 \) is constant, its rate of change does not contribute. Instead, consider:- Angular velocity about OA is changing because \( \gamma \) is decreasing.- The component\( \alpha = \dot{\omega}_0 \hat{u} + \omega_{0} \dot{\hat{u}} \)The change in \( \hat{u} \) is due to the rotation about its axis:\[\\dot{\hat{u}} = \frac{-\sin(\gamma)}{\sqrt{3}/2}\left(-\frac{\pi}{4}\right) \hat{i} + \frac{\cos(\gamma)}{\frac{1}{2}}\left(-\frac{\pi}{4}\right) \hat{k} \]Substitute \( \gamma = \frac{\pi}{6} \): \[ =-\frac{1}{2} \left(-\frac{\pi}{4} \right) \hat{i} + \frac{\sqrt{3}}{2} \left(-\frac{\pi}{4} \right) \hat{k}\]Evaluating derivatives:\( \dot{\hat{u}} = \frac{\pi}{8} \hat{i} - \frac{\pi \sqrt{3}}{8} \hat{k}\)Therefore the resultant \( \alpha = 3 \left ( \frac{\pi}{8} \right)\hat{i} + 3 \left( -\frac{\pi \sqrt{3}}{8} \right)\hat{k} \):\[\alpha = \frac{3\pi}{8} \hat{i} - \frac{3\pi \sqrt{3}}{8} \hat{k}\]

To find the magnitude of \( \alpha \), use the expression:\[|\alpha| = \sqrt{\left( \frac{3\pi}{8} \right)^2 + \left( -\frac{3\pi \sqrt{3}}{8} \right)^2 }\]This simplifies to:\[|\alpha| = \sqrt{\frac{9\pi^2}{64} + \frac{27\pi^2}{64}} \]\[|\alpha| = \sqrt{\frac{36\pi^2}{64}}\]\[|\alpha| = \frac{6\pi}{8} = \frac{3\pi}{4}\] rad/s².