Physical and Compound Pendulums
A rigid body pivoted a distance d from its centre of mass has small-angle equation I_O theta double-dot + Mgd theta = 0 and period 2 pi sqrt(I_O/(Mgd)).
Why this shows up in the exam
Pivoted rods and discs · Rod-disc compound pendulums · Hinged bodies partly immersed in fluids
Learn the idea
A rigid body's pendulum period depends on its pivot inertia and centre-of-mass distance. Gravity acts at the centre of mass, while the whole body's distribution resists angular acceleration through its moment of inertia about the pivot.
🧠 Memory hook: Torque uses centre-of-mass distance; inertia uses the pivot.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- omega = sqrt(Mgd/I_O) — small-angle angular frequency of a physical pendulum
- T = 2 pi sqrt(I_O/(Mgd)) — physical-pendulum period
- I_O = I_cm + Md² — parallel-axis relation for the pivot inertia
How to approach it
- 1Locate the combined centre of mass
- 2Compute inertia about the actual pivot
- 3Linearise the restoring torque and form the ratio
Common slip-ups that cost marks
- •Using I about the centre instead of the pivot
- •Using total length instead of centre-of-mass distance
- •Treating a freely spinning attached disc like a rigidly fixed disc
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A mass of 1 kg is attached to a spring of force constant 100 N/m. Find the angular frequency of small oscillations.
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