Question 15.SP.11: Two 20-in. rods AB and DE are connected as shown. Point D is...

MODELING and ANALYSIS: Locate the instantaneous center of rotation C of bar AB as the intersection of line AC perpendicular to  \mathrm{v}_A  and line BC perpendicular to  \mathrm{v}_B  (Fig. 1). Knowing the location of C, you can determine the direction of the velocity of D. From this direction, and the direction of E, you can find the instantaneous center, I, for bar DE (Fig. 1).

a. Angular velocity of DE. From geometry,  r_{A/C} = (20 \cos 30°) in.,  so

\omega_{A B}=\frac{v_A}{r_{A / C}}=\frac{12  \mathrm{in} . / \mathrm{s}}{20 \cos 30^{\circ} \mathrm{in} .}=0.6928  \mathrm{rad} / \mathrm{s} \circlearrowleft

Now you can find  v_D  since  r_{D/C}  = 10 in.

\begin{aligned}&v_D=\omega_{A B} r_{D / C}=(0.6928  \mathrm{rad} / \mathrm{s})(10  \mathrm{in} .)=6.928  \mathrm{in} . / \mathrm{s} \\&v_D=6.928 \text { in./s } \text{⦮} 30^{\circ}\end{aligned}

Now, since you know the directions of the velocities of D and E,  \mathrm{v}_E = v_E ⦨ 30°, you can find point I, which is the instantaneous center of bar DE. From geometry,  r_{D/I} = 20 \cos 30° in., and therefore

\omega_{D E}=\frac{v_D}{r_{D I I}}=\frac{6.928  \mathrm{n} / \mathrm{s}}{20 \cos 30^{\circ}  \mathrm{in} .}=0.400  \mathrm{rad} / \mathrm{s} \quad \omega_{D E}=0.400  \mathrm{rad} / \mathrm{s} \circlearrowleft

b. Velocity of E. Using this angular velocity, you can easily determine the velocity of E:

v_E=\omega_{D E} r_{E J}=(0.400  \mathrm{rad} / \mathrm{s})\left(20 \sin 30^{\circ} \text { in. }\right)=4.00  \mathrm{in} . / \mathrm{s}

\mathbf{v}_E=0.333  \mathrm{ft} / \mathrm{s} \text{⦨} 30^{\circ}

REFLECT and THINK: The direction of   \omega_{D E}  makes intuitive sense; you would expect it to be rotating counterclockwise at the instant shown. You could have also solved this problem using vector equations.