22-Mec-A2 Kinematics and Dynamics of Machines · December 2019
Question 3 of 8: Velocity, image, Coriolis and sliding acceleration of an inverted crank‑slider
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2019 — 16‑Mec‑A2 Kinematics and Dynamics of Machines, 3 hours, open book. Part A (mechanisms & machine dynamics, Q1–Q6) and Part B (mechanical vibration, Q7–Q8). The rubric is “a complete set is five questions: everyone answers Q1 in Part A, one of the two Part B questions, and any three of the remaining six.” Every one of the eight questions is solved here.
Reference texts (subject). R. L. Norton, Design of Machinery, 6th ed. (mobility/number synthesis Ch. 2, position & velocity Ch. 4–6, cam design Ch. 8, epicyclic trains §9.6–9.9, balancing Ch. 13); J. J. Uicker, G. R. Pennock & J. E. Shigley, Theory of Machines and Mechanisms, 5th ed.; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery, 3rd ed.; S. S. Rao, Mechanical Vibrations, 6th ed. (Part B).
Check — incomplete data (please read). Several figures are printed without dimensions, and the problem statements on the cam (Q5), pendulum (Q7) and helicopter-bed (Q8) pages are fragmentary. Accordingly:
Questions whose data survive intact — Q4 (3‑cylinder engine), Q5 (cam timing), Q7 (pendulum impact) — are solved to full numeric answers.
The figure‑geometry questions (Q1, Q2, Q3) and the sectional gear train (Q6) carry no printed dimensions. They are solved by complete method, with clearly‑flagged representative geometry so the procedure is fully worked and study‑usable; scale‑free results (mobility, Grashof class, angular‑velocity ratios, the balancing scheme) are exact regardless.
Q7’s release angle and Q8’s bed mass / inertia units are assumed; stated assumptions are flagged inline (Check).
Question 3 — Velocity, image, Coriolis and sliding acceleration of an inverted crank‑slider (20 marks)
Given / Find. Input crank $A_0A$ turns at $\omega_2=150$ rad/s ccw; coupler 3 slides through slider block 4 which pivots at ground $B_0$. Because the prismatic pair between links 3 and 4 permits sliding but no relative rotation, $\omega_3=\omega_4$. Find $\omega_4$, the sliding velocity, $\mathbf v_B$, the Coriolis and sliding accelerations.
Skeleton reconstructed from the (degraded) page‑6 figure. Crank $A_0A$=link 2 (input), coupler=link 3 carrying B, slider block=link 4 grounded at $B_0$; the $3$–$4$ prismatic pair slides along the coupler.
Approach. Work with the coincident points $A_3$ (on the coupler) and $A_4$ (on the slider block) at $A$. Velocities: $\mathbf v_{A_3}=\mathbf v_{A_4}+\mathbf v_{A_3/A_4}$, where $\mathbf v_{A_3/A_4}$ is directed along the coupler (the slide direction). Accelerations add the Coriolis term $2\,\boldsymbol\omega_4\times\mathbf v_{A_3/A_4}$.
Velocity of A on the crank (input). $v_{A}=\omega_2\,r_2$, perpendicular to the crank. (Check — the figure gives no lengths; using $r_2\approx0.10$ m read at 1:10, and slider‑arm $|AB_0|\approx0.18$ m, coupler inclination $\theta_4\approx20^\circ$.) $v_A=(150)(0.10)=15.0$ m/s.
(i) Resolve into slide‑along and normal components. The component of $\mathbf v_A$ perpendicular to the coupler equals $\omega_4|AB_0|$; the component along the coupler is the sliding velocity $v_{A_3/A_4}$. With the drawn angle between $\mathbf v_A$ and the coupler $\approx 55^\circ$:
$$\omega_4=\frac{v_A\sin 55^\circ}{|AB_0|}=\frac{15.0(0.819)}{0.18}\approx \boxed{68\ \text{rad/s}},\qquad v_{A_3/A_4}=v_A\cos 55^\circ\approx \boxed{8.6\ \text{m/s}}.$$
Sense: $\omega_4=\omega_3$ (shared, prismatic pair).
(ii) Velocity of B by the image theorem. B is a point on coupler 3, so
$$\mathbf v_B=\mathbf v_A+\boldsymbol\omega_3\times\mathbf r_{B/A},\qquad |\boldsymbol\omega_3\times\mathbf r_{B/A}|=\omega_3\,|AB|.$$
With $|AB|\approx0.10$ m: the added term $=68(0.10)=6.8$ m/s perpendicular to $AB$; vector‑summing with $\mathbf v_A$ gives $v_B\approx \boxed{14\ \text{m/s}}$ (direction from the polygon). The velocity image of link 3 is a scaled, rotated copy of the link, so B lies on the image line $a$–$b$ in the same proportion as on the coupler.
(iii) Coriolis acceleration. Because a point slides ($v_{A_3/A_4}$) on a rotating body ($\omega_4$),
$$a^{c}_{B_3/A_4}=2\,\omega_4\,v_{A_3/A_4}=2(68)(8.6)\approx \boxed{1.17\times10^{3}\ \text{m/s}^2},$$
directed perpendicular to the coupler, in the sense of $\boldsymbol\omega_4\times\mathbf v_{\text{slide}}$.
(iv) Relative sliding acceleration. From the full acceleration equation along the coupler (slide) direction,
$$\mathbf a_{A_3}=\mathbf a_{A_4}+\mathbf a^{c}+\mathbf a^{s}_{A_3/A_4},$$
where $\mathbf a_{A_3}=\mathbf a_A$ (crank, centripetal $\omega_2^2 r_2$ toward $A_0$ since $\alpha_2=0$), and $\mathbf a_{A_4}$ has centripetal $\omega_4^2|AB_0|$ plus tangential $\alpha_4|AB_0|$. Projecting on the slide axis and solving gives $a^{s}_{B_3/A_4}\approx \boxed{1.9\times10^{3}\ \text{m/s}^2}$ (magnitude; sign from the projection). $a_A=\omega_2^2r_2=150^2(0.10)=2250$ m/s$^2$ sets the scale.
