Answer:
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
[tex]Frequency=\sqrt{\frac{3g}{2l} }\\Period = 2(pi)\sqrt{\frac{3g}{2l} }=2(pi)\sqrt{\frac{3}{2} }\sqrt{\frac{g}{l} }[/tex]
ω [tex]=\sqrt{\frac{3g}{2l} }[/tex]
T = [tex]2\pi \sqrt{\frac{2l}{3g} }[/tex]
Plug in L = 1.0m
T = 1.639sec.
Simple Pendulum:
[tex]Period=2(pi)\sqrt{\frac{l}{g} }[/tex]
1.639 = [tex]2\pi\sqrt{\frac{l}{g} }[/tex]
L = 0.67m
The simple pendulum has a length of 0.67 m
For a physical pendulum, the period (T) is:
[tex]T=2\pi\sqrt{\frac{I}{mgh} }[/tex]
I = rotational inertia = ML²/3, L = 1 m, h = L/2 = 1/2
For a simple pendulum, the period (T) is:
[tex]T=2\pi\sqrt{\frac{L_o}{g} }[/tex]
I = rotational inertia = ML²/3, L = 1 m,
T = T, hence:
[tex]2\pi\sqrt{\frac{I}{mgh} }=2\pi\sqrt{\frac{L_o}{g} }\\\\{\frac{I}{mgh}} ={\frac{L_o}{g} }\\\\{\frac{I}{mh}} =L_o\\\\\\\frac{mL^2/3}{m(L/2)}=L_o\\\\ L_o=\frac{2L}{3}=\frac{2*1}{3} =0.67[/tex]
The simple pendulum has a length of 0.67 m
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