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What is the rotational inertia of a solid iron disk of mass 43.00 kg, with a thickness of 7.310 cm and radius of 19.40 cm, about an axis through its center and perpendicular to it?

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Khoso123

Answer:

I=0.81 kg m²

Explanation:

Given Data

Mass=43.00 kg

thickness=7.310 cm

radius=19.40 cm

To

Rotational Inertia

Solution

I have attached a picture regarding the derivation of

Rotational Inertia. From that we know

I(Inertia)=1/2 MR²

First we need to convert radius cm to meter

So

r=19.40/100

r=0.1940 m

I=1/2×(43.00kg)×(0.1940m)²

I=0.81 kg m²

Ver imagen Khoso123

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