Oscillations L10 Physical Compound Pendulum, Torsional Pendulum & Intro to Rolling SHM скачать в хорошем качестве

Oscillations L10 Physical Compound Pendulum, Torsional Pendulum & Intro to Rolling SHM

In Oscillations Lecture 10, we give a compact introduction to three important angular SHM systems: Physical (compound) pendulum – any rigid body oscillating about a horizontal axis. Torsional pendulum – body suspended by a twisting wire. Rolling SHM – a solid cylinder attached to a spring and rolling without slipping. The aim is to show how the same SHM pattern appears again and again: Restoring torque proportional to angle, Angular acceleration alpha = minus omega squared theta, Time period T = 2 pi by omega. 1. Physical (Compound) Pendulum Definition: Any rigid body suspended from a fixed horizontal axis and allowed to oscillate in a vertical plane. Distance from pivot to centre of mass = L (small L) Mass = m Moment of inertia about pivot = I (capital I) For small angular displacement theta: Gravitational torque: tau = minus m g L sin theta approximately equal to minus m g L times theta. Rotational equation: I d2theta by dt2 = tau. So: d2theta by dt2 = minus (m g L divided by I) theta Hence omega = square root of (m g L divided by I) T = 2 pi square root of (I divided by m g L) Using the parallel axis theorem: I = I cm plus m L squared, with I cm = m k squared (k = radius of gyration about CM) We get: T = 2 pi square root of (k squared plus L squared divided by L g) Define equivalent length of a simple pendulum: L eq = k squared divided by L plus L So T = 2 pi square root of (L eq divided by g) Minimum period occurs when L equals k: T min = 2 pi square root of (2 k divided by g) Example (metre stick): Uniform rod length 1 m, pivot at 75 cm mark. Distance CM to pivot d = 0.25 m I cm = m L rod squared by 12 I about pivot = I cm + m d squared Use T = 2 pi square root of (I divided by m g d). 2. Torsional Pendulum Rigid body suspended by a wire. When twisted by angle theta: Restoring torque: tau = minus k theta (k = torsion constant of wire) Rotational equation: I d2theta by dt2 = minus k theta So d2theta by dt2 = minus (k divided by I) theta Hence: omega = square root of (k divided by I) T = 2 pi square root of (I divided by k) No small angle restriction is needed as long as the wire remains within its elastic limit. Example (disc on wire): Uniform disc: m = 0.200 kg, r = 0.050 m I = m r squared by 2 = 2.5 times 10 power minus 4 kg m squared Given T = 0.20 s From T = 2 pi square root of (I divided by k): k = 4 pi squared I by T squared approximately 0.25 kg m squared per s squared. 3. Rolling SHM: Solid Cylinder Attached to a Spring Solid cylinder: mass m, radius R Spring constant k, horizontal surface, rolling without slipping Small rotation theta: Centre displacement x = R theta Spring extension x = R theta Spring force F s = k x = k R theta (towards mean) Equations used (not fully derived in detail): Translation: F s minus friction f = m a Rotation about CM: f R = I cm alpha Rolling without slipping: a = R alpha For solid cylinder: I cm = (1 by 2) m R squared From these: alpha = minus (2 k divided by 3 m) theta So omega = square root of (2 k divided by 3 m) Time period: T = 2 pi square root of (3 m divided by 2 k) This shows how combined translation plus rotation can still give simple harmonic motion. Key Takeaways Physical pendulum: T = 2 pi square root of (I divided by m g L), with I = I cm + m L squared and equivalent length L eq = k squared divided by L plus L. Torsional pendulum: tau = minus k theta, T = 2 pi square root of (I divided by k). Rolling SHM (solid cylinder plus spring, pure rolling): T = 2 pi square root of (3 m divided by 2 k). In all these cases, the SHM form comes from: Restoring torque proportional to angular displacement, Angular SHM equation d2theta by dt2 plus omega squared theta equals zero.