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Disc cam with flat-faced follower numerical

Short answer

For SHM rise: v_max = ω·(πh/(2β)) = 1.257 m/s, a_max = ω^2·(π^2 h/(2β^2)) = 157.9 m/s^2. Profile radius of curvature ρ = r_b + s + d^2s/dθ^2, minimum 10 mm at end of rise. Face width: contact offset x = ds/dθ max 20 mm; add 5 mm each side → 30 mm (50 mm if symmetrical about axis).

A disc cam with a radial flat-faced follower has a base circle of radius 30 mm and rotates at 600 rpm. The follower rises 20 mm with simple harmonic motion during 90 degrees of cam rotation, then dwells. Find (a) the maximum velocity and acceleration of the follower during the rise, (b) the minimum radius of curvature of the cam profile during the rise, and (c) the minimum width of the follower face if the face must extend 5 mm beyond the extreme contact point on each side.

Theory

For a radial flat-faced follower, the contact point lies at a distance x=dsdθx = \dfrac{ds}{d\theta} from the follower axis, and the radius of curvature of the cam profile is

ρ=rb+s+d2sdθ2\rho = r_b + s + \frac{d^2 s}{d\theta^2}

where rbr_b is the base circle radius. If ρ\rho becomes zero or negative, the profile has a cusp (undercutting).

Data

h=20h = 20 mm, β=90∘=π/2\beta = 90^\circ = \pi/2 rad, rb=30r_b = 30 mm, ω=2π×60060=62.83\omega = \dfrac{2\pi \times 600}{60} = 62.83 rad/s.

SHM during rise:

s=h2(1−cos⁡πθβ),dsdθ=πh2βsin⁡πθβ,d2sdθ2=π2h2β2cos⁡πθβs = \frac{h}{2}\left(1 - \cos\frac{\pi\theta}{\beta}\right),\quad \frac{ds}{d\theta} = \frac{\pi h}{2\beta}\sin\frac{\pi\theta}{\beta},\quad \frac{d^2s}{d\theta^2} = \frac{\pi^2 h}{2\beta^2}\cos\frac{\pi\theta}{\beta}

(a) Maximum velocity and acceleration

(dsdθ)max=πh2β=π(20)2(π/2)=20.0 mm/rad\left(\frac{ds}{d\theta}\right)_{max} = \frac{\pi h}{2\beta} = \frac{\pi(20)}{2(\pi/2)} = 20.0\ \text{mm/rad} vmax=ω(dsdθ)max=62.83×0.0200=1.257 m/sv_{max} = \omega\left(\frac{ds}{d\theta}\right)_{max} = 62.83 \times 0.0200 = 1.257\ \text{m/s} amax=ω2(d2sdθ2)max=ω2 π2h2β2=(3948)(0.040)=157.9 m/s2a_{max} = \omega^2\left(\frac{d^2s}{d\theta^2}\right)_{max} = \omega^2\,\frac{\pi^2 h}{2\beta^2} = (3948)(0.040) = 157.9\ \text{m/s}^2

where π2h2β2=π2(20)2(π/2)2=40\dfrac{\pi^2 h}{2\beta^2} = \dfrac{\pi^2 (20)}{2(\pi/2)^2} = 40 mm.

(b) Minimum radius of curvature

Put x=πθ/βx = \pi\theta/\beta (from 0 to π\pi):

ρ=30+10(1−cos⁡x)+40cos⁡x=40+30cos⁡x mm\rho = 30 + 10(1 - \cos x) + 40\cos x = 40 + 30\cos x\ \text{mm}
xxθ\thetaρ\rho (mm)
0070
π/2\pi/245 deg40
π\pi90 deg10

The minimum is at the end of the rise:

ρmin=40−30=10.0 mm\rho_{min} = 40 - 30 = 10.0\ \text{mm}

It is positive, so the profile is not undercut.

(c) Face width

The contact point is at a distance x=ds/dθx = ds/d\theta from the follower axis. During the rise it moves from 0 out to a maximum of 20.020.0 mm (at mid-rise) and back, always on the same side of the axis. So the contact travel is 20.020.0 mm. Adding 5 mm at each end:

face width=5+20.0+5=30 mm\text{face width} = 5 + 20.0 + 5 = 30\ \text{mm}

(If the face is made symmetrical about the follower axis, as in a usual mushroom follower, or if a similar return puts contact on the other side, the width needed is 2(20+5)=502(20 + 5) = 50 mm.)

Answer: (a) vmax=1.257v_{max} = 1.257 m/s, amax=157.9a_{max} = 157.9 m/s2^2; (b) ρmin=10.0\rho_{min} = 10.0 mm; (c) face width = 30 mm (50 mm if the face is symmetrical about the axis).

✓ Answer checked by Claude Opus 5.5 (high effort)

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