22-Mec-A2 Kinematics and Dynamics of Machines · May 2013
Question 2 of 6: Radial Cam — 3-4-5 Polynomial Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Examinations — 07-Mec-A2 Kinematics and Dynamics of Machines, May 2013. Open-book, 3-hour paper; answer FIVE of six questions, all of equal value (20 marks). Part A (mechanisms & machine dynamics, Q1–Q4) and Part B (mechanical vibration, Q5–Q6). All six questions are solved here.
Reference texts: R. L. Norton, Design of Machinery (6th ed., McGraw-Hill) — velocity analysis, cams, gear trains, engine dynamics; C. E. Wilson & J. P. Sadler, Kinematics and Dynamics of Machinery (3rd ed.); S. S. Rao, Mechanical Vibrations (6th ed., Pearson) — Part B; J. E. Shigley, Mechanical Engineering Design — gear geometry.
Note on the figure-based questions. Q1 (press) is a scaled graphical problem (“Scale 1:15”); its link geometry is read from the printed diagram, so the mechanical advantage carries a graphical tolerance of roughly ±5 %. Q3 (planetary box) is solved from the kinematic topology inferred from the sectioned schematic. Both interpretations are stated explicitly in a check callout so the reader can re-scale from the original drawing.
Given. Rise $h=25\ \text{mm}$ over $\beta=90^\circ=\tfrac{\pi}{2}\ \text{rad}$; cam speed $\omega=125\ \text{rad/s}$ (constant). Normalized cam angle $u=\theta/\beta$ (so $0\le u\le1$ during the rise).
Find. $s,v,a,j$ during the rise; $a_\text{max}$ and $j_\text{max}$; a suitable base circle and the flat-faced-follower pressure angles at three cam positions.
Approach. The 3-4-5 polynomial gives zero velocity and acceleration at both ends (no jerk spikes at the boundaries except finite jerk); differentiate with respect to time via $u$, then apply the standard maxima. The flat-faced follower makes the pressure-angle question a matter of definition, and the base circle follows from keeping the profile radius of curvature positive.
Fig. 2 — rise-period kinematics of the 3-4-5 cam: displacement s (0–25 mm), velocity v, acceleration a and jerk j vs cam angle. s, v, a all start and end at zero; jerk is finite and peaks at the ends.
Maximum acceleration. $a$ is extreme where $\mathrm dj=0$, i.e. $u=\tfrac12(1\pm1/\sqrt3)=0.211,\,0.789$, giving the standard coefficient $5.7735$:
$$a_\text{max}=5.7735\,\frac{h\omega^{2}}{\beta^{2}}=5.7735\cdot\frac{0.025\,(125)^{2}}{(1.5708)^{2}}=\boxed{914\ \text{m/s}^2}.$$
Maximum jerk. The jerk bracket is largest at the ends ($u=0,1$), value 60:
$$j_\text{max}=60\,\frac{h\omega^{3}}{\beta^{3}}=60\cdot\frac{0.025\,(125)^{3}}{(1.5708)^{3}}=\boxed{7.56\times10^{5}\ \text{m/s}^3}.$$
Flat-faced follower: base circle and pressure angle
Pressure angle. For a flat-faced follower the common normal at the contact point is always parallel to the follower’s axis of translation, so the pressure angle is identically zero at every cam angle. Hence at 45°, 135° and 270°:
$$\phi(45^\circ)=\phi(135^\circ)=\phi(270^\circ)=\boxed{0^\circ}.$$
This is exactly why flat-faced followers are chosen where side thrust must be avoided.
Base circle from radius of curvature. The governing constraint for a flat-faced follower is instead that the pitch-surface radius of curvature stay positive, $\rho=R_0+s+\dfrac{d^2s}{d\theta^2}>0$. The most negative value of $\big(s+s''\big)$ over the cycle is $-35.3\ \text{mm}$ (during the rise, at $u=0.789$), so
$$R_0>35.3\ \text{mm}\ \Rightarrow\ \text{choose }\boxed{R_0=40\ \text{mm}}.$$
Flat-face width. The contact point slides off the follower axis by $ds/d\theta$; the peak is $(ds/d\theta)_\text{max}=1.875\,h/\beta=29.8\ \text{mm}$ (rise). The face must overhang the axis by at least this amount on the appropriate side, so a face length of about $\pm32\ \text{mm}$ ($\approx64$ mm total) is required.
