---
title: "Spring force and contact force on cam follower"
subject: "Theory of Mechanism and Machine I"
chapter: 8
key: "semester_6/theory_of_mechanism_and_machine_i:cam-follower-spring-force-jump"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/cam-follower-spring-force-jump
site: BCT Engineering
---

# Spring force and contact force on cam follower

Theory of Mechanism and Machine I · Chapter 8: Force Analysis of Mechanisms

- Asked in: year not given, 8 marks
- Video: [Mechanics of Machinery - Cam Dynamics Tutorial 22](https://www.youtube.com/watch?v=XTw6tHDQUPY) — sSs19
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/cam-follower-spring-force-jump
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-8

## Question

A disc cam rotating at 400 rpm moves a horizontal translating roller follower of mass 0.8 kg through a lift of 25 mm with simple harmonic motion in 90 degrees of cam rotation. A return spring is preloaded with a force of 15 N when the follower is at its lowest position. Neglecting friction and the mass of the roller, find (a) the maximum inertia force during the rise, (b) the minimum spring stiffness required to prevent the follower from losing contact with the cam (jump) at the end of the rise, and (c) the maximum force exerted by the cam on the follower during the rise, taking this stiffness. Explain why the spring is needed.

## Short answer

During SHM rise, maximum inertia force is m a_max = 70.18 N. Minimum spring stiffness to prevent jump: (70.18 - 15)/0.025 = 2207 N/m. Maximum cam force at start = 85.2 N. A spring is needed because a disc cam only pushes; deceleration inertia would otherwise cause follower to lose contact.

## Answer

#### 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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is a return spring needed for a disc cam with a translating follower?
   A. The cam can push the follower but cannot pull it, and during deceleration inertia tends to carry it away from the cam
   B. To increase the lift of the follower
   C. To reduce the cam speed during the rise
   D. Because the follower mass is neglected
   Answer: A. A cam only pushes. In the second half of the rise the follower decelerates and could leave the cam (jump), so a spring must hold it in contact. (Key point: Why a spring is needed)

2. For SHM rise with h = 25 mm, beta = 90 degrees and omega = 41.89 rad/s, the maximum follower acceleration is:
   A. 43.9 m/s^2
   B. 70.2 m/s^2
   C. 175.5 m/s^2
   D. 87.7 m/s^2
   Answer: D. a_max = (pi^2 h omega^2)/(2 beta^2) = 2 h omega^2 = 2 x 0.025 x 1754.6 = 87.7 m/s^2 when beta = pi/2. (Key point: Maximum SHM acceleration)

3. What condition gives the minimum spring stiffness to prevent jump at the end of the rise?
   A. k h = m a_max, ignoring preload
   B. F0 + k h >= m a_max, giving k >= (70.18 - 15)/0.025 = 2207 N/m
   C. F0 >= m a_max at the start of the rise
   D. F0 + k h >= m a_max, giving k >= (70.18 + 15)/0.025
   Answer: B. At the end of the rise the spring is compressed by h and its force F0 + k h must at least equal the inertia force, so k = (70.18 - 15)/0.025 = 2207 N/m. (Key point: Minimum spring stiffness)

4. Where does the maximum force exerted by the cam on the follower occur, and what is it?
   A. End of the rise, 70.2 N
   B. End of the rise, 85.2 N
   C. Start of the rise, 85.2 N (inertia force 70.18 N plus preload 15 N)
   D. Start of the rise, 70.2 N
   Answer: C. At the start the cam must supply the inertia force and the preload: 70.18 + 15 = 85.2 N. At the end the cam force falls to zero. (Key point: Maximum cam force)

## Related in Theory of Mechanism and Machine I

- [SHM follower motion with a disc cam](https://www.bctengineer.com/question.md?key=semester_6%2Ftheory_of_mechanism_and_machine_i%3Ashm-cam-follower-displacement-velocity) — Theory of Mechanism and Machine I, chapter 3
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