A rigid body of mass 1.8 kg oscillates about a horizontal pivot 0.28 m from its centre of mass. Twenty small-amplitude oscillations take 32.4 s. Use g = 9.81 m s⁻². Starting with τ = I_p α and restoring torque τ = −mgd sin θ, derive the small-angle period and determine the moments of inertia about the pivot and a parallel centre-of-mass axis. State a procedural check on the small-angle assumption.

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T = 2π√(I_p/mgd); I_p = 0.3287 kg m²; I_cm = 0.1876 kg m². Check that measured period is unchanged for smaller amplitudes. For small angles measured in radians, sin θ ≈ θ. Thus I_p d²θ/dt² = −mgdθ, the differential equation of simple harmonic motion about the stable hanging equilibrium. Comparing with θ″ = −ω²θ gives ω² = mgd/I_p and T = 2π√(I_p/mgd). The observed period is the total time divided by 20, T = 1.62 s. Rearrange to I_p = mgdT²/(4π²) = 0.3287 kg m². Use the distance from pivot to centre of mass, not the body's overall length. By the parallel-axis theorem I_p = I_cm + md². Therefore I_cm = 0.1876 kg m², positive and smaller than I_p as expected. Repeat the timing with a smaller release angle and compare periods within uncertainty. Consistent periods support the small-angle model; timing many cycles reduces the relative influence of starting and stopping reaction times.

Infer rotational inertia from a physical pendulum · set 8

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