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?

The moment of inertia of a body is a scalar, additive quantity, so we can add the moment of inertia of each part, if everything revolves around the same axis

The moment of inertia of a cylinder is

I = ½ m r²

The moment of total inertia is

I = I_body + 2 I_arm

I = ½ M r² + 2 m_arm r²

The most body without arms is

M = M - 1/8 M

M = 64 - 1/8 64 = 64 (1-1 / 8)

M = 56 kg

The mass of the arms is

m = 8 kg

Each arm has a mass of m_arm = 4 kg

Calculate

a) The moment of inertia with the arms attached to the body

I = ½ 56 0.2² + 8 0.2²

I = 1.44 kg m²

b) With the arms the moment of inertia changes

For the arms we use the parallel axes theorem

I_arm = [tex]I_{cm}[/tex] + m d²

Let's approach the arm with a thin stick

[tex]I_{cm}[/tex] = 1/12 m L²

The distance is

d = L / 2 + 0.20

d = 0.6 / 2 + 0.2

d = 0.50 m

I = ½ M r² + (1/12 m L² + 2 m_arm d²)

I = ½ 56 0.2² + (1/12 * 8 0.3² +2 4 0.5²)

I = 1.12 + (0.06 +2)

I = 3.18 kg m²

The moment of inertia increases