An ice skater can be modeled by a cylinder of radius R and l

An ice skater can be modeled by a cylinder of radius R and length L (trunk and legs together), 2 small thin cylinders of radius r and length l (arms), and a sphere also of radius R (head). The head has mass m_h, the trunk m_t, and the arms m_a. The ice skater rotates about her central axis. Find the moment of inertia of the ice skater with arms outstretched horizontally and then with arms vertical.

Solution

When ice skater strectched her arms horizontally outwrads then her arms spin arounds her central axis which can be taken as a thin rod about an axis through center perpendicular to the length. Length of this rod is equal to twice the length of the thin cylinder from which arms are modeled.

Thus, moment of inertia of ice skater with her arms outstretched horizontally can be given as

I_out = I_h +I_t +I_a

I_out = 2/5 (m_hR²) +1/2 (m_t R²) +1/12(m_a)(2l)²

I_out =R²(2/5 m_h+1/2 m_t)+1/3(m_al²)

Moment of inertia of ice skater is equal to sum of moment of inertia of sphere (head), moment of inertia of cylinder (trunk) and (arms) spinning around same central axis.

Thus, moment of inertia of ice skater with her arms vertical can be given as

I_in = I_h +I_t +I_a

I_out = 2/5 (m_hR²) +1/2 (m_t +m_a) R²

I_out =R²[2/5 m_h+1/2 (m_t +m_a)]