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VladimirAG [237]
4 years ago
13

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?
Physics
1 answer:
pav-90 [236]4 years ago
5 0

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

T = \dfrac{58.1}{28}

T =2.075\ s

we know,

T = 2\pi\sqrt{\dfrac{I}{k}}

I = k (\dfrac{T}{2\pi})^2

I =0.618\times \dfrac{T^2}{4\pi^2}

I =0.618\times \dfrac{2.075^2}{4\pi^2}

      I = 0.0674 kg.m²

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

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