Question

7. An ice skater can be modeled bv 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 mh, the trunk mt, and the arms ma. 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.

Answer 1

We know that the moment of inertia of cylinder is 1/2 MR² and moment of inertia of sphere is 2/5 MR² and the moment of inertia of spherical surface  is  2/3 MR² where M is the mass and R is the radius.

Using this formula, we can solve this question.

For the cylinder whose radius is R and length is L and mass is M, the moment of inertia will be = 1/2 MR². Similarly , head has mass m_h and trunk has mass m_t , the moment of inertia will be 1/2 m_hR² + 1/2 m_tR² .

If you add them , we will get the below results.

I_horizontal = 2/5m_hR² + 1/2 m_tR^{2 +} maR² + m_aR² = (2/5 m_h + 1/2 m_t + 2m_a) R² + 2/3 m_al² + 1/2 m_ar² + 2m_aRL

sililary you can get the value of I_vertical. . Here the area of arms will be 3m_aR² + 4m_aRR

(2/5 m_h + 1/2 m_t + 2m_a) R² + 3m_aR² + 4m_aRR.

Hence , the result.