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Compound pendulum with added mass

Short answer

Compound pendulum period T = 2π√((k² + l²)/gl). A small mass fixed at the suspension point adds zero moment of inertia about the axis and its weight produces zero torque because the lever arm is zero, so the period remains unchanged.

Obtain an expression for the time period of a compound pendulum and show that its time period is unaffected by the fixing of a small additional mass to it at its center of suspension.

Time period of a compound pendulum

Let M be the mass, O the point of suspension, G the centre of gravity at distance l from O, and k the radius of gyration about G.

  1. Small angular displacement θ: restoring torque = −Mgl sinθ ≈ −Mglθ.
  2. Moment of inertia about O: I = Mk² + Ml².
  3. I d²θ/dt² = −Mgl θ gives d²θ/dt² + [Mgl/I] θ = 0, SHM with ω² = Mgl/I.
T = 2π √(I/(Mgl)) = 2π √((k² + l²)/(g l))

Effect of a small mass fixed at the point of suspension

Fix a small mass m at O itself.

  1. Moment of inertia of m about the axis through O: m × (distance from O)² = m × 0² = 0. So the total moment of inertia is unchanged: I′ = I.
  2. Torque of its weight about O: mg × 0 = 0, because the lever arm is zero. So the total restoring torque is still Mglθ.
  3. Both the moment of inertia about the axis and the restoring torque are unchanged, hence
T′ = 2π √(I/(Mgl)) = T

The time period is unaffected by the small mass fixed at the centre of suspension.

✓ Answer checked by Claude Opus 5.5 (high effort)

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