Question 15.SP.5: Collars A and B are pin-connected to bar ABD and can slide a...

MODELING and ANALYSIS: Model bar ABD as a rigid body. From kinematics you know

\mathbf{v}_B=\mathbf{v}_A  +  \mathbf{v}_{B / A}=\mathbf{v}_A  +  \boldsymbol{\omega} \times \mathbf{r}_{B / A}

Substituting in known values (Fig. 1) and assuming  \boldsymbol{\omega}=\omega \mathbf{k}  gives you

\begin{aligned}v_B \cos 60^{\circ} \mathbf{i}  +  v_B \sin 60^{\circ} \mathbf{j}=(0.9)  \mathbf{i}+\\\omega \mathbf{k} \times\left[\left(0.3 \cos 30^{\circ}\right) \mathbf{i}  +  \left(0.3 \sin 30^{\circ}\right) \mathbf{j}\right] \\0.500 v_B \mathbf{i}  +  0.866 v_B \mathbf{j}=(0.9  –  0.15 \omega) \mathbf{i}  +  0.260 \omega \mathbf{j}\end{aligned}

Equating components,

\begin{aligned}\mathrm{i}: &=0.500 v_B=0.9  –  0.15 \omega \\\mathrm{j}: &=0.866 v_B=0.260 \omega\end{aligned}

Solving these equations gives you  v_B = 0.900 m/s and ω = 3.00 rad/s

\omega=3.00  \mathrm{rad} / \mathrm{s}\Lsh

Velocity of D. The relationship between the velocities of A and D is

\mathbf{v}_D=\mathbf{v}_A  +  \mathbf{v}_{D / A}=\mathbf{v}_D  +  \omega \times \mathbf{r}_{D / A}

Substituting in values from above gives

\begin{aligned}&\mathbf{v}_D=(0.9)  \mathbf{i}  +  3.00  \mathbf{k} \times\left[\left(0.6 \cos 30^{\circ}\right) \mathbf{i}  +  \left(0.6 \sin 30^{\circ}\right) \mathbf{j}\right] \\&\mathbf{v}_D=(0.9  –  0.9) \mathbf{i}  +  1.559  \mathbf{j}\end{aligned}

\mathbf{v}_D=1.559  \mathrm{~m} / \mathrm{s} \uparrow

REFLECT and THINK: The velocity of point D is straight up at this instant in time, but as the bar continues to rotate counterclockwise, the direction of the velocity of D will continuously change.