--- title: "Circular disk as physical pendulum" subject: "Engineering Physics" chapter: 1 key: "semester_1/engineering_physics:disk-physical-pendulum-rim" source: https://www.bctengineer.com/q/semester_1/engineering_physics/disk-physical-pendulum-rim site: BCT Engineering --- # Circular disk as physical pendulum Engineering Physics · Chapter 1: Oscillation - Asked in: 2067 Asar, 5 marks - Video: [Physical Pendulum Problems - Moment of Inertia - Physics](https://www.youtube.com/watch?v=scIVIhChL1I) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/disk-physical-pendulum-rim - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1 ## Question A uniform circular disk whose radius R is 12.6 cm is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance r < R is there a pivot point that gives the same period? ## Short answer Period when pivoted at rim: T = 2π√((3/2)MR²/(MgR)) = 2π√(3R/(2g)) ≈ 0.873 s for R=0.126 m. For a pivot at distance r from centre givng same period, solv r + R²/(2r) = 3R/2 → r = R/2 = 6.3 cm. ## Answer Given: R = 12.6 cm = 0.126 m, g = 9.8 m/s². The disc is pivoted at a point on its rim, with the axis perpendicular to the disc. #### (a) Period 1. Moment of inertia about the centre: I_G = ½MR², so k² = R²/2. 2. About the rim (parallel-axis theorem): I = ½MR² + MR² = (3/2)MR². 3. Distance from the pivot to the centre: *l* = R. For a physical pendulum T = 2π√(I/(Mgl)): ```text T = 2π √( (3/2) M R² / (M g R) ) = 2π √(3R/2g) T = 2π √(3 × 0.126 / (2 × 9.8)) = 2π √(0.01929) = 2π × 0.1389 ``` **Answer (a): T ≈ 0.873 s.** #### (b) Pivot at distance r from the centre giving the same period For a pivot at distance r from the centre, the equivalent simple-pendulum length is L = r + k²/r. It must equal that of the rim pivot, L = 3R/2: ```text r + R²/(2r) = 3R/2 2r² − 3R r + R² = 0 (2r − R)(r − R) = 0 → r = R/2 or r = R ``` The required point with r < R is r = R/2 = 0.063 m. **Answer (b): r = R/2 = 6.3 cm from the centre.** This agrees with interchangeability: the centre of oscillation lies at k²/R = R/2 from the centre, on the side opposite the pivot. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the moment of inertia of a uniform disc of radius R about a perpendicular axis through a point on its rim? A. ½MR² B. MR² C. 2MR² D. (3/2)MR² Answer: D. I = ½MR² + MR² (parallel-axis theorem with distance R) = (3/2)MR². (Key point: Parallel-axis theorem) 2. Which expression gives the period of the disc pivoted at its rim? A. 2π√(3R/2g) B. 2π√(R/g) C. 2π√(2R/3g) D. 2π√(3R/g) Answer: A. T = 2π√(I/(Mgl)) with I = (3/2)MR² and l = R gives 2π√(3R/2g) ≈ 0.873 s. (Key point: Period formula) 3. To find another pivot at distance r from the centre with the same period, which equation is set up? A. r = 3R/2 B. r + R²/(2r) = 3R/2 C. r² = R²/2 D. r + R/2 = R Answer: B. The equivalent simple-pendulum length r + k²/r with k² = R²/2 must equal 3R/2. (Key point: Equal-period condition) 4. Solving 2r² - 3Rr + R² = 0, the required point with r < R is A. r = R/2 B. r = R/3 C. r = 2R/3 D. r = R/4 Answer: A. The roots are r = R/2 and r = R; the smaller one is R/2 = 6.3 cm. 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