Answer:
Angular velocity, [tex]N_f = 242.36 rpm[/tex]
Explanation:
The mass of the skater, M = 74.0 kg
Mass of each arm, [tex]m_{a} = 0.13 * \frac{M}{2}[/tex] ( since it is 13% of the whole body and each arm is considered)
[tex]m_{a} = 0.13 * 37\\m_a = 4.81 kg[/tex]
Mass of the trunk, [tex]m_{t} = M - 2m_{a}[/tex]
[tex]m_t = 74 - 2(4.81)\\m_{t} = 64.38 kg[/tex]
Total moment of Inertia = (Moment of inertia of the arms) + (Moment of inertia of the trunks)
[tex](I_{T} )_i = 2(\frac{m_{a}L^2 }{12} + m_a(0.5L + R)^2) + 0.5 m_t R^2[/tex]
[tex](I_{T} )_i = 2(\frac{4.81 * 0.7^2 }{12} + 4.81(0.5*0.7 + 0.175)^2) + 0.5 *64.38* 0.175^2\\(I_{T} )_i = 3.052 + 0.986\\(I_{T} )_i = 4.038 kgm^2[/tex]
The final moment of inertia of the person:
[tex](I_{T} )_f = \frac{1}{2} MR^{2} \\(I_{T} )_f = \frac{1}{2} * 74*0.175^{2}\\(I_{T} )_f = 1.133 kg.m^2[/tex]
According to the principle of conservation of angular momentum:
[tex](I_{T} )_i w_{i} = (I_{T} )_f w_{f}\\w_{i} = 68 rpm = (2\pi * 68)/60 = 7.12 rad/s\\4.038 * 7.12 =1.133* w_{f}\\w_{f} = 25.38 rad/s\\w_{f} = \frac{2\pi N_f}{60} \\25.38 = \frac{2\pi N_f}{60}\\N_f = (25.38 * 60)/2\pi \\N_f = 242.36 rpm[/tex]
