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
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
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
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
%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 %
i believe the answer is a
it has to be b, about 97 percent.