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A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.290 Hz. The pendulum has a mass of 2.40 kg, and the pivot is located 0.300 m from the center of mass. Determine the moment of inertia of the pendulum about the pivot point.

Respuesta :

Answer:

1.4584 kg[tex]m^2[/tex]

Explanation:

Time period of a physical pendulum is given by [tex]T=2Π\sqrt{\frac{I}{mgd}}[/tex]

Here f=0.290 so [tex]T=\frac{1}{F}=\frac{1}{0.29}=3.44827[/tex]

Mass =2.40 kg

d=0.300 m

g =9.8 m[tex]sec^2[/tex]

So [tex]3.448=2\times \pi \sqrt{\frac{I}{2.4\times 9.81\times 0.300}}=1.4584[/tex]  kg-[tex]m^2[/tex]

So the moment of inertia of the pendulum about the pivot point will be 1.4584 kg-[tex]m^2[/tex]

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