Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be modeled as a physical pendulum as a rod oscillating around one end. By what percentage will the period change if the temperature increases by 10°C? Assume the length of the rod changes linearly with temperature, where L=L0 (1+αΔT) and the rod is made of brass (α=18*10-6°C-1).

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum, the angular momentum remains constant. In other words:

ζ = Δ (Iω) / Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque (no rotational force applied)

ζ=0 → Δ (Iω) = 0 → I₂ω₂ - I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ = ω₁/ω₂, (2) represents final state and (1) initial state

we know also that ω=2π/T, where T is the period of the pendulum

I₂/I₁ = ω₁/ω₂ = (2π/T₁) / (2π/T₂) = T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁², I₂ = 1/3 ML₂² → I₂/I₁ = (1/3 ML₂²) / (1/3 ML₁²) = L₂²/L₁² = (L₂/L₁) ²

since

L₂ = L₁ (1+αΔT), L₂/L₁=1+αΔT, where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁ = (L₂/L₁) ² = (1+αΔT) ²

finally

%change in period = (T₂-T₁) / T₁ = T₂/T₁ - 1 = (1+αΔT) ² - 1

%change in period = (1+αΔT) ² - 1 = [ 1+18*10⁻⁶ °C⁻¹ * 10 °C]² - 1 = 3,6 * 10⁻⁴ = 3,6 * 10⁻² % = 0,036 %