Skip to main content

Moment of inertia by torsional pendulum

Short answer

Suspend the body from a wire, measure period T₁. Add a body of known moment of inertia I₀, measure new period T₂. From T₁² = 4π²I/C and T₂² = 4π²(I+I₀)/C, eliminate C to get I = I₀ T₁²/(T₂² − T₁²).

Derive a relation to find the moment of inertia of a rigid body about an axis passing through its center of gravity using the torsional pendulum.

A rigid body is hung by a wire so that the axis (through its centre of gravity) coincides with the wire. Twisting it makes it oscillate as a torsional pendulum.

   /////////
      |  wire, torsional constant C
   ===O===  body, I (unknown)

Period of the torsional pendulum

Restoring torque = −Cθ, so I d²θ/dt² = −Cθ, which is angular SHM:

T = 2π √(I/C)   →   I = C T²/4π²

C is not known directly, so use a body of known moment of inertia.

Method

  1. Suspend the unknown body (moment of inertia I) and measure its period T₁:
T₁² = 4π² I / C                       ... (1)
  1. Place on it (or attach to the wire) a regular body, e.g. a ring or disc, with known moment of inertia I₀ about the same axis. Measure the new period T₂:
T₂² = 4π² (I + I₀) / C                ... (2)
  1. Divide (1) by (2):
T₁²/T₂² = I/(I + I₀)
I T₂² = (I + I₀) T₁²
I (T₂² − T₁²) = I₀ T₁²

I = I₀ T₁² / (T₂² − T₁²)

Measure T₁, T₂, and calculate I₀ from the mass and dimensions of the standard body (for a disc I₀ = ½MR²; for a ring I₀ = MR²) to obtain I of the body about the axis through its centre of gravity.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Engineering Physics chapter 1 questions →