--- title: "SHM follower motion with a disc cam" subject: "Theory of Mechanism and Machine I" chapter: 3 key: "semester_6/theory_of_mechanism_and_machine_i:shm-cam-follower-displacement-velocity" source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/shm-cam-follower-displacement-velocity site: BCT Engineering --- # SHM follower motion with a disc cam Theory of Mechanism and Machine I · Chapter 3: Cams and Followers - Asked in: year not given, 8 marks - Video: [07 Theory of Machines CAMS' Motion Design - Part 1- Dr Ahmed Sameh](https://www.youtube.com/watch?v=5vpnxu8jiCE) — Dr Ahmed Sameh - د. أحمد سامح - 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/shm-cam-follower-displacement-velocity - Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-3 ## Question A disc cam rotates at a uniform speed of 300 rpm and raises a knife-edge follower through 40 mm with simple harmonic motion during 120 degrees of cam rotation. Derive the expressions for displacement, velocity and acceleration of the follower during rise, and calculate their values when the cam has turned 45 degrees from the start of the rise. Also find the maximum velocity and acceleration of the follower. ## Short answer SHM follower: s = h/2(1-cos(πθ/β)), v = πhω/2β sin(πθ/β), a = π²hω²/2β² cos(πθ/β). Given h=40mm, β=120°, ω=31.416 rad/s, at θ=45°: s=12.35mm, v=0.871m/s, a=17.00m/s². Maximum v=0.942m/s at mid-rise, maximum a=44.41m/s² at start. ## Answer #### 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$^2$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For SHM rise, the follower displacement at cam angle theta (rise angle beta, lift h) is: A. s = h(1 - cos(pi*theta/beta)) B. s = (h/2)(1 - cos(pi*theta/beta)) C. s = (h/2) sin(pi*theta/beta) D. s = (h/2)(1 + cos(pi*theta/beta)) Answer: B. The follower moves like the projection of a point on a circle of diameter h, giving s = (h/2)(1 - cos(pi*theta/beta)). (Key point: SHM displacement expression) 2. In SHM rise, where do maximum velocity and maximum acceleration occur? A. Both at mid-rise B. Velocity at start, acceleration at mid-rise C. Velocity at mid-rise, acceleration at the start D. Both at the end of rise Answer: C. v is proportional to sin(pi*theta/beta), maximum at theta = beta/2. a is proportional to cos(pi*theta/beta), maximum at theta = 0. (Key point: Where maxima occur) 3. With h = 40 mm, beta = 120°, what is the displacement at 45°? A. 20.00 mm B. 24.65 mm C. 6.17 mm D. 12.35 mm Answer: D. pi*45/120 = 67.5°, so s = 20(1 - cos67.5°) = 20(1 - 0.3827) = 12.35 mm. (Key point: Displacement at 45°) 4. With h = 40 mm, beta = 120° and 300 rpm (omega = 31.416 rad/s), what is v_max? A. 0.942 m/s B. 0.871 m/s C. 44.41 m/s D. 1.884 m/s Answer: A. v_max = pi*h*omega/(2*beta) = pi(0.040)(31.416)/(2 x 2.0944) = 0.942 m/s. 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