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An engineer has an odd-shaped 13.5 kg object and needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant κ = 0.618 N·m. If this torsion pendulum oscillates through 28 cycles in 58.1 s, what is the rotational inertia of the object?

Respuesta :

Answer:

I = 0.0674 kg.m²

Explanation:

given,

mass = 13.5 Kg

torsion constant = k = 0.618 N.m

number of cycle = 28

time = 58.1 s

Time of one cycle

[tex]T = \dfrac{58.1}{28}[/tex]

[tex]T =2.075\ s[/tex]

we know,

[tex]T = 2\pi\sqrt{\dfrac{I}{k}}[/tex]

[tex]I = k (\dfrac{T}{2\pi})^2[/tex]

[tex]I =0.618\times \dfrac{T^2}{4\pi^2}[/tex]

[tex]I =0.618\times \dfrac{2.075^2}{4\pi^2}[/tex]

      I = 0.0674 kg.m²

the rotational inertia of the object is equal to  I = 0.0674 kg.m²

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