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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?

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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