Question 2 of 7: Velocity of link 6 by the graphical method

Question 2: Velocity of link 6 by the graphical method (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Crank 2 ($O_2B$, CCW at $\omega_2=50$ rad/s) drives rod 3, which passes through an oscillating trunnion block 4 pinned to ground at fixed pivot $C$; the rod end pins to slider 5 at $D$, which runs on the horizontal arm of the L-shaped follower 6 (guided to translate vertically at $E$). Joint coordinates measured on the figure (scale 1:5, $0.635$ mm per unit): $O_2(178,96),\,B(622,92),\,C(462,338),\,D(258,643)$, so $r_2=|O_2B|\approx0.282$ m, $|CB|\approx0.186$ m, $|BD|\approx0.419$ m.

Find. the linear velocity $v_6$ of follower 6, and the relocation of $D$ that reverses it.

Approach. Get $\mathbf v_B$ from the crank; the rod turns about the fixed trunnion $C$, so its perpendicular velocity component at $B$ fixes $\omega_3$; carry $\omega_3$ to the rod end $D$; the follower moves vertically, so $v_6$ is the vertical component of $\mathbf v_D$.

1. Velocity of the crank pin $B$. With $\omega_2=50$ rad/s and $r_2=0.282$ m, $$v_B=\omega_2\,r_2=50(0.282)=14.1\ \text{m/s},\quad\perp O_2B.$$
2. Angular velocity of rod 3. The rod always passes through the fixed pivot $C$; the component of $\mathbf v_B$ perpendicular to the rod equals $\omega_3\,|CB|$. Resolving, $$\omega_3=\frac{(\mathbf v_B)_{\perp\text{rod}}}{|CB|}=\frac{7.79}{0.186}=\boxed{41.8\ \text{rad/s (CCW)}}.$$
3. Velocity of the rod end $D$. Treating $B$ and $D$ as points on the rigid rod, $\mathbf v_D=\mathbf v_B+\boldsymbol\omega_3\times\mathbf r_{D/B}$, which evaluates to $\mathbf v_D=(14.5,\,4.43)$ m/s.
4. Linear velocity of follower 6. Slider 5 slides horizontally on the follower arm, so the follower’s (vertical) velocity equals the vertical component of $\mathbf v_D$: $$\boxed{v_6=(\mathbf v_D)_y\approx 4.4\ \text{m/s (upward)}.}$$
5. (b) Reversing the follower velocity. On the velocity image, every point of rod 3 has a velocity proportional to its distance from the pivot $C$ and perpendicular to the rod, and points on opposite sides of $C$ move in opposite senses. Since $v_6$ is the vertical component of the velocity of the slider point $D$, relocate the slider joint $D$ to the opposite side of the trunnion $C$ along the rod, at the same distance $|CD|$. This flips the sign of the rotational contribution while preserving its magnitude, so the follower velocity is reversed with the same magnitude.