Question 5 of 8: Radial cam‑follower motion design (rise–dwell–fall)

Question 5 — Radial cam‑follower motion design (rise–dwell–fall) (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.

(a) Displacement programs, cam speed and SVAJ maxima.

Find. Cam speed; the motion programs that minimise (rise) peak acceleration and (fall) peak velocity while obeying the fundamental law; the $s$‑$v$‑$a$‑$j$ maxima; base‑circle / pressure‑angle discussion.

1. Cam speed and segment angles. One motion cycle = one revolution in $T=0.030$ s, so $$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.030}=\boxed{209.4\ \text{rad/s}}=2000\ \text{rpm}.$$ The segment angles are $\beta=\omega\,\Delta t$: rise $\beta_r=209.4(0.015)=\pi=\boxed{180^\circ}$, dwell $60^\circ$, fall $\beta_f=209.4(0.010)=\boxed{120^\circ}$ (sum $360^\circ$ ✓).
2. Fundamental law ⇒ full‑rise/full‑fall smooth programs. The fundamental law of cam design requires $s$, $v$ and $a$ continuous (finite $j$) across the whole cycle, so at each rise–dwell and dwell–fall boundary $v=0$ and $a=0$. Double‑dwell programs (3‑4‑5 polynomial, cycloidal, modified trapezoid, modified sine) all satisfy this. The objective then selects which:
   • Rise — minimise peak acceleration ⇒ MODIFIED TRAPEZOIDAL. Of the smooth double‑dwell curves it has the lowest acceleration factor $C_a=4.888$ (vs. 5.77 for 3‑4‑5, 6.28 for cycloidal). (Parabolic is lower at 4.0 but has infinite jerk — it violates the fundamental law — so it is disqualified.)
   • Fall — minimise peak velocity ⇒ MODIFIED SINE. It has the lowest velocity factor $C_v=1.760$ (vs. 1.875 for 3‑4‑5, 2.00 for cycloidal/modified‑trapezoid).
3. Maxima from the standard factors $v_{\mathrm{max}}=C_v h\omega/\beta$, $a_{\mathrm{max}}=C_a h\omega^2/\beta^2$, $j_{\mathrm{max}}=C_j h\omega^3/\beta^3$.
   Rise (modified trapezoid, $\beta_r=\pi$): $C_a=4.888,\ C_v=2.000,\ C_j=61.4$ $$v_{\mathrm{max}}=\boxed{5.33\ \text{m/s}},\quad a_{\mathrm{max}}=\boxed{869\ \text{m/s}^2},\quad j_{\mathrm{max}}\approx\boxed{7.28\times10^{5}\ \text{m/s}^3}.$$ Fall (modified sine, $\beta_f=2\pi/3$): $C_a=5.528,\ C_v=1.760,\ C_j=69.5$ $$v_{\mathrm{max}}=\boxed{7.04\ \text{m/s}},\quad a_{\mathrm{max}}=\boxed{2211\ \text{m/s}^2},\quad j_{\mathrm{max}}\approx\boxed{2.78\times10^{6}\ \text{m/s}^3}.$$ The fall values are larger because $\beta_f<\beta_r$ (the same lift is done in a smaller angle, and $a\propto1/\beta^2$, $j\propto1/\beta^3$).
4. Sketch. Displacement diagram over one revolution below; $v$, $a$, $j$ follow the standard modified‑trapezoid (rise) and modified‑sine (fall) shapes, each returning to zero at the segment ends so the curves connect smoothly through the dwell.

(b) Base circle, pressure angle, improvements. For a roller follower the pressure angle is $\phi=\arctan\!\big[(ds/d\psi)/(R_b+s)\big]$; it peaks where the follower velocity is greatest (mid‑rise / mid‑fall) and must be kept below $\approx30^\circ$ to avoid side thrust and jamming. Because the fall is fast ($\beta_f=120^\circ$) its peak $\phi$ is the binding one. If the computed $\phi_{\mathrm{max}}>30^\circ$, the only effective remedy that does not change the motion program is to enlarge the base circle $R_b$ (which reduces $\phi$ everywhere); a larger roller and adequate return‑spring preload (to keep the follower on the cam against the high $a_{\mathrm{max}}=2211$ m/s$^2$ inertia during the fall) complete the check. For a flat‑faced follower the pressure angle is identically $0^\circ$ and the base circle is instead limited by convexity, $R_b+s+s''>0$ everywhere.

Cam speed $\omega=209.4$ rad/s (2000 rpm). Programs chosen to satisfy the fundamental law and the stated objectives (min accel on rise, min velocity on fall).