SHM follower motion with a disc cam

Derivation

In SHM the follower moves like the projection of a point moving on a circle of diameter equal to the lift $h$. Let the cam turn through $\theta$ during the rise angle $\beta$. The point on the auxiliary circle turns through $\pi\theta/\beta$ while the cam turns $\theta$.

$$s = \frac{h}{2}\left(1 - \cos\frac{\pi\theta}{\beta}\right)$$

With $\theta = \omega t$, differentiating with respect to time:

$$v = \frac{ds}{dt} = \frac{\pi h\omega}{2\beta}\sin\frac{\pi\theta}{\beta}$$ $$a = \frac{dv}{dt} = \frac{\pi^2 h\omega^2}{2\beta^2}\cos\frac{\pi\theta}{\beta}$$

Maximum velocity is at mid-rise ($\theta = \beta/2$) and maximum acceleration is at the start ($\theta = 0$).

Numerical

Data: $h = 0.040$ m, $\beta = 120^\circ = 2.0944$ rad, $\omega = \dfrac{2\pi \times 300}{60} = 31.416$ rad/s, $\theta = 45^\circ$.

$$\frac{\pi\theta}{\beta} = \frac{\pi \times 45}{120} = 1.1781\ \text{rad} = 67.5^\circ$$

At $\theta = 45^\circ$:

$$\begin{aligned} s &= 0.020(1 - \cos 67.5^\circ) = 0.020(1 - 0.3827) = 12.35\ \text{mm} \\ v &= \frac{\pi(0.040)(31.416)}{2(2.0944)}\sin 67.5^\circ = 0.871\ \text{m/s} \\ a &= \frac{\pi^2(0.040)(987.0)}{2(2.0944)^2}\cos 67.5^\circ = 17.00\ \text{m/s}^2 \end{aligned}$$

Maximum values:

$$v_{max} = \frac{\pi h\omega}{2\beta} = 0.942\ \text{m/s}, \qquad a_{max} = \frac{\pi^2 h\omega^2}{2\beta^2} = 44.41\ \text{m/s}^2$$

Answer: at $45^\circ$: $s = 12.35$ mm, $v = 0.871$ m/s, $a = 17.00$ m/s$^2$; $v_{max} = 0.942$ m/s, $a_{max} = 44.41$ m/s$^2$.