Physics: The rotational inertia of a uniform thin rod about its end is ML2/3, where M is the mass and L is the length. Such a rod is hung vertically from one end and set into small amplitude oscillation. If L = 1.0 m this rod will have the same period as a simple pendulum of length:

The rotational inertia of a uniform thin rod about

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23.01.2022

0.67m

Explanation:

∑τ= Iα = - (L/2) mgsin(Θ)

(The net torque equals the negative horizontal component of gravity at the center of mass)(it's negative because the force is oppose of displacement)

sin(Θ) ≈ Θ , at small angles

α (1/3ML^2) = - (L/2) mg θ

2nd order differential equation

Θ = Asin(ωt)

^take the derivative twice

α = -Aω^2sin(ωt) = -θω^2

Go back to α (1/3ML^2) = - (L/2)mg θ

(-θω^2)(1/3ML^2) = - (L/2)mgθ

Cancel stuff out and solve for ω

$Frequency=\sqrt{\frac{3g}{2l} }\\Period = 2(pi)\sqrt{\frac{3g}{2l} }=2(pi)\sqrt{\frac{3}{2} }\sqrt{\frac{g}{l} }$

ω $=\sqrt{\frac{3g}{2l} }$

T = $2\pi \sqrt{\frac{2l}{3g} }$

Plug in L = 1.0m

T = 1.639sec.

Simple Pendulum:

$Period=2(pi)\sqrt{\frac{l}{g} }$

1.639 = $2\pi\sqrt{\frac{l}{g} }$

L = 0.67m

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