High-speed radial cam — motion program and pressure angle: 22-Mec-A2 Kinematics and Dynamics of Machines, Undated paper, Q3

Question 3: High-speed radial cam — motion program and pressure angle (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. Radial (in-line, translating) roller-follower cam. Follower displacement $s(\theta)=h\sin\theta$ with $h=10$ mm over the rise–return interval $\theta\in[0^\circ,180^\circ]$; cam speed $N=3600$ rpm (constant), i.e. $\omega=N\cdot 2\pi/60$.

Find. (I) $s,v,a$ program with peak values; (II) a base-circle radius $R_b$ that limits the maximum pressure angle to $20^\circ$.

Approach. Differentiate $s(\theta)$ once with respect to time (using $\dot\theta=\omega$) for velocity and twice for acceleration; then apply the in-line roller-follower pressure-angle relation $\tan\phi=(ds/d\theta)/(R_b+s)$ and size $R_b$ for $\phi_{\mathrm{max}}\le20^\circ$.

Cam angular speed. $\displaystyle \omega=\frac{2\pi N}{60}=\frac{2\pi(3600)}{60}=\boxed{376.99\ \text{rad/s}}.$
Displacement. $s(\theta)=10\sin\theta$ mm; it rises to $s_{\mathrm{max}}=10$ mm at $\theta=90^\circ$ and returns to $0$ at $\theta=180^\circ$.
Velocity. $\displaystyle v=\frac{ds}{dt}=\frac{ds}{d\theta}\,\omega=10\,\omega\cos\theta$. Peak at $\theta=0$ (and $180^\circ$): $$v_{\mathrm{max}}=10\,\omega=10(376.99)=3769.9\ \text{mm/s}=\boxed{3.77\ \text{m/s}}.$$
Acceleration. $\displaystyle a=\frac{d^2s}{dt^2}=-10\,\omega^2\sin\theta$. Peak magnitude at $\theta=90^\circ$: $$a_{\mathrm{max}}=10\,\omega^2=10(376.99)^2=1.421\times10^{6}\ \text{mm/s}^2=\boxed{1421\ \text{m/s}^2\ (\approx145\,g)}.$$ The very large acceleration is the reason the problem stresses “high-speed” — inertia loading dominates the cam design.
Pressure angle. For an in-line roller follower, $\tan\phi=\dfrac{ds/d\theta}{R_b+s}=\dfrac{10\cos\theta}{R_b+10\sin\theta}$ (with $ds/d\theta$ in mm/rad). Since the numerator is largest and the denominator smallest at $\theta=0^\circ$ (and $180^\circ$), the maximum pressure angle is $$\phi_{\mathrm{max}}=\arctan\!\frac{10}{R_b}.$$
Size the base circle for $\phi_{\mathrm{max}}\le20^\circ$. Set $\arctan(10/R_b)=20^\circ$: $$R_b=\frac{10}{\tan 20^\circ}=\boxed{27.5\ \text{mm}}.$$ For comparison a $30^\circ$ limit would allow $R_b=10/\tan30^\circ=17.3$ mm. So if the chosen base circle gives $\phi_{\mathrm{max}}>20^\circ$, the corrective action is to increase the base-circle (prime-circle) radius to $\ge27.5$ mm (equivalently add follower offset, or spread the same lift over a larger cam angle); enlarging $R_b$ reduces $\phi$ everywhere without changing the lift program.

Fig. 3. Follower $s$–$v$–$a$ diagrams over the $0^\circ$–$180^\circ$ program: displacement a half-sine hump, velocity a cosine (zero at mid-lift, peak at the ends), acceleration a negative half-sine (peak magnitude at mid-lift).

Quantity	Value
Cam speed $\omega$	376.99 rad/s
Max lift $s_{\mathrm{max}}$	10 mm (at 90°)
Max velocity $v_{\mathrm{max}}$	3.77 m/s (at 0°, 180°)
Max acceleration $a_{\mathrm{max}}$	1421 m/s² ≈ 145 g (at 90°)
Base circle for $\phi_{\mathrm{max}}=20^\circ$	$R_b=27.5$ mm

Topic. Cam displacement diagrams (SVAJ) and the base-circle–pressure-angle trade-off for a translating roller follower.

Key relations. $v=\omega\,ds/d\theta$, $a=\omega^2\,d^2s/d\theta^2$ (constant $\omega$); in-line roller pressure angle $\tan\phi=(ds/d\theta)/(R_b+s)$.

Why this works. Because the cam turns at constant speed, time derivatives are just $\omega$-scaled angle derivatives, so the $v$ and $a$ shapes follow directly from differentiating $s(\theta)$. The pressure angle measures how far the contact-force line tilts from the follower’s motion; a larger base circle “flattens” the cam locally, shrinking $ds/d\theta$ relative to the radius and hence $\phi$.

Common pitfall. Looking for $\phi_{\mathrm{max}}$ at mid-lift (where $a$ peaks) — it is actually at the ends, where $ds/d\theta$ is largest and $s$ is smallest. A flat-faced follower would have $\phi\equiv0$, so the pressure-angle question presumes a roller follower. Keep $ds/d\theta$ in mm per radian, not per degree.

Source. Norton, Design of Machinery, §8.3–8.6 (SVAJ diagrams, pressure angle, base-circle sizing).

Question 3 of 7: High-speed radial cam — motion program and pressure angle

Question 3: High-speed radial cam — motion program and pressure angle (20 marks)