A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most of her body as a uniform cylinder. Suppose the skater has a mass of 64 kg . One eighth of that mass is in her arms, which are 60 cm long and 20 cm from the vertical axis about which she rotates. The rest of her mass is approximately in the form of a 20-cm-radius cylinder.A) Estimate the skater's moment of inertia to two significant figures.B)If she were to hold her arms outward, rather than at her sides, would her moment of inertia increase, decrease, or remain unchanged?

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

I = [tex]I_{cm}[/tex] + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

[tex]I_{total}[/tex]=[tex]I_{body}[/tex] + 2 [tex]I_{arm}[/tex]

[tex]I_{body}[/tex] = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

M = 7/8 m total

M = 7/8 64

M = 56 kg

The mass of the arms is

m’= 1/8 m total

m’= 1/8 64

m’= 8 kg

As it has two arms the mass of each arm is half

m = ½ m ’

m = 4 kg

The arms are very thin, we will approximate them as a particle

[tex]I_{arm}[/tex] = M D²

Let's write the equation

[tex]I_{total}[/tex] = ½ M R² + 2 (m D²)

Let's calculate

[tex]I_{total}[/tex] = ½ 56 0.20² + 2 4 0.20²

[tex]I_{total}[/tex] = 1.12 + 0.32

[tex]I_{total}[/tex] = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase