Spring force and contact force on cam follower

Why a spring is needed

A disc cam can push the follower, but it cannot pull it back. During the second half of the rise the follower decelerates, and its inertia tends to carry it forward, away from the cam. A spring (or a positive-drive groove) must keep it in contact; otherwise the follower jumps and the intended motion is lost.

Follower acceleration (SHM rise)

$$a = \frac{\pi^2 h\omega^2}{2\beta^2}\cos\frac{\pi\theta}{\beta}$$

Data: $m = 0.8$ kg, $h = 0.025$ m, $\beta = 90^\circ = \pi/2$, $\omega = \dfrac{2\pi \times 400}{60} = 41.888$ rad/s, $\omega^2 = 1754.6$.

$$a_{max} = \frac{\pi^2 h\omega^2}{2\beta^2} = \frac{\pi^2 (0.025)(1754.6)}{2(\pi/2)^2} = 87.73\ \text{m/s}^2$$

(Here $\pi^2/(2\beta^2) = 2$, so $a_{max} = 2h\omega^2 = 2 \times 0.025 \times 1754.6$. The same value can be written as $\dfrac{h}{2}\left(\dfrac{\pi}{\beta}\right)^2\omega^2$.)

(a) Maximum inertia force

$$F_{I,max} = m\,a_{max} = 0.8 \times 87.73 = 70.18\ \text{N}$$

It occurs at the beginning of the rise (positive acceleration, so the cam must push) and at the end of the rise (negative acceleration, so the follower tends to leave the cam).

(b) Minimum spring stiffness

At the end of the rise the spring is compressed by the lift $h$, and its force there must be at least equal to the inertia force:

$$F_0 + k\,h \ge m\,a_{max}$$ $$k \ge \frac{70.18 - 15}{0.025} = 2207\ \text{N/m} \approx 2.21\ \text{N/mm}$$

In practice, a margin of 20 to 50 percent is added.

(c) Maximum force from the cam

At the start of the rise, the spring force is just the preload $F_0$, and the cam must supply the spring force plus the inertia force:

$$F_{cam} = m a_{max} + F_0 = 70.18 + 15 = 85.2\ \text{N}$$

At any angle, the cam force is $N = m a + F_0 + k s$. With the stiffness found above, this reduces to $N = 42.6(1 + \cos x)$ N, where $x = \pi\theta/\beta$. So the force falls from its maximum at the start to zero (just touching) at the end of the rise. The spring force at the end of the rise is 70.2 N, equal to the inertia force.

Answer: (a) $F_{I,max} = 70.18$ N; (b) $k_{min} = 2207$ N/m; (c) $F_{cam,max} = 85.2$ N (at the start of the rise).