Consider the 2-body mechanism where the complete FBD is drawn to show the rotation of each body. Find the mass center (point c as shown in Figure 2.20), velocity, and acceleration of body 2.

Question

Accepted Answer

The direction cosine matrix of body 2 with respect to R is

[latex]\begin{aligned}\bar{n}_m^2 &=\left[\begin{array}{ccc}\cos (\pi- heta) & 0 & -\sin (\pi- heta) \0 & 1 & 0 \\sin (\pi- heta) & 0 & \cos (\pi- heta)\end{array} ight]\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \-\sin \alpha & \cos \alpha & 0 \0 & 0 & 1\end{array} ight] \bar{n}_m \quad(m=1,3) \&=\left[\begin{array}{ccc}-\cos heta & 0 & -\sin heta \0 & 1 & 0 \\sin heta & 0 & -\cos heta\end{array} ight]\left[\begin{array}{ccc}\cos \alpha & \sin \alpha & 0 \-\sin \alpha & \cos \alpha & 0 \0 & 0 & 1\end{array} ight] \bar{n}_m (2.233)\&=\left[\begin{array}{ccc}-\cos heta \cos \alpha & -\cos heta \sin \alpha & -\sin heta \-\sin \alpha & \cos \alpha & 0 \\sin heta \cos \alpha & \sin heta \sin \alpha & -\cos heta\end{array} ight] \bar{n}_m\end{aligned}[/latex]

where [latex]\dot{\alpha}=\omega[/latex]. The position vector of point c is given by

[latex]\begin{aligned}\bar{p}_c=& b \bar{n}_1^1+\frac{1}{2} \ell \bar{n}_1^2=b\left(\cos \alpha \bar{n}_1+\sin \alpha \bar{n}_2 ight) \&+\frac{1}{2} \ell\left(-\cos heta \cos \alpha \bar{n}_1-\cos heta \sin \alpha \bar{n}_2-\sin heta \bar{n}_3 ight.\=&\left(b \cos \alpha-\frac{1}{2} \ell \cos heta \cos \alpha ight) \bar{n}_1 \&+\left(b \sin \alpha-\frac{1}{2} \ell \cos heta \sin \alpha ight) \bar{n}_2-\frac{1}{2} \ell \sin heta \bar{n}_3 (2.234)\end{aligned}[/latex]

Differentiation of [latex]\bar{p}_c[/latex] leads to the velocity [latex]\bar{v}_c[/latex]

[latex]\begin{aligned}\bar{v}_c=& \frac{d}{d t} \bar{p}_c=\left[-b \omega \sin \alpha+\frac{1}{2} \ell(\dot{ heta} \sin heta \cos \alpha+\omega \cos heta \sin \alpha) ight] \bar{n}_1 \&+\left[b \omega \cos \alpha+\frac{1}{2} \ell(\dot{ heta} \sin heta \sin \alpha-\omega \cos heta \cos \alpha) ight] \bar{n}_2-\frac{1}{2} \ell \dot{ heta} \cos heta \bar{n}_3\end{aligned}[/latex]

Differentiating the velocity further we get [latex]\bar{a}_c[/latex] :

[latex]\begin{aligned}\bar{a}_c=\frac{d^2}{d t^2} \bar{p}_c=& {[-b\left(\dot{\omega} \sin \alpha+\omega^2 \cos \alpha ight)+\frac{1}{2} \ell(\ddot{ heta} \sin heta \cos \alpha} \&+\dot{ heta}^2 \cos heta \cos \alpha-\dot{ heta} \omega \sin heta \sin \alpha+\dot{\omega} \cos heta \sin \alpha \&-\omega \dot{ heta} \sin heta \sin \alpha+\omega^2 \cos heta \cos \alpha] \bar{n}_1+[b\left(\dot{\omega} \cos \alpha-\omega^2 \sin \alpha ight)\&+\frac{1}{2} \ell (\ddot{ heta} \sin heta \sin \alpha+\dot{ heta}^2 \cos heta \sin \alpha+\dot{ heta} \omega \sin heta \cos \alpha-\dot{\omega} \cos heta \cos \alpha\&+\omega \dot{ heta} \sin heta \cos \alpha+\omega^2 \cos heta \sin \alpha)] \bar{n}_2-\frac{1}{2} \ell (\ddot{ heta} \cos heta-\dot{ heta}^2 \sin heta) \bar{n}_3\end{aligned}[/latex]

Question 2.13.2: Consider the 2-body mechanism where the complete FBD is draw...

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