A person with mass mp = 78 kg stands on a spinning platform disk with a radius of R = 1.5 m and mass md = 198 kg. The disk is initially spinning at w = 1.4 rad/s. The person then walks 2/3 of the way

1) The moment of inertia of the spinning platform can be calculated using the formula I = 1/2 m R^2, where m is the mass of the disk and R is its radius. Using this formula, the moment of inertia of the disk is I_d = 1/2 (198 kg) (1.5 m)^2 = 222.75 kg m^2. The moment of inertia of the person can be \approx imated as I_p = m_p R^2, since the person is standing at the edge of the disk. Therefore, I_p = (78 kg) (1.5 m)^2 = 175.5 kg m^2. The total moment of inertia of the system is the sum of the moments of inertia of the disk and person, or I_total = I_d + I_p = 222.75 kg m^2 + 175.5 kg m^2 = 398.25 kg m^2. 2) When the person moves 2/3 of the way toward the center of the disk, their moment of inertia changes. The new moment of inertia of the person can be \approx imated using the parallel axis theorem: I_p' = I_p + m_p (R/3)^2, where R/3 is the new distance of the person from the center of the disk. Plugging in the values, we get I_p' = 175.5 kg m^2 + (78 kg) (0.5 m)^2 = 192.75 kg m^2. The new total moment of inertia of the system is the sum of the moments of inertia of the disk and person, or I_total' = I_d + I_p' = 222.75 kg m^2 + 192.75 kg m^2 = 415.5 kg m^2. 3) The law of conservation of angular momentum states that the initial angular momentum of the system is equal to the final angular momentum of the system, assuming no external torques act on the system. Using this principle, we can calculate the final angular velocity of the disk. Initially, the angular momentum of the system is L_initial = I_total ? = (398.25 kg m^2) (1.4 rad/s) = 557.55 kg m^2/s. When the person moves 2/3 of the way toward the center of the disk, the total moment of inertia of the system increases to 415.5 kg m^2. Therefore, the final angular velocity of the disk can be calculated using the equation L_final = I_total' ?_final, where ?_final is the final angular velocity of the disk. Solving for ?_final, we get ?_final = L_initial / I_total' = 557.55 kg m^2/s / 415.5 kg m^2 = 1.34 rad/s. 4) The initial total kinetic energy of the system is given by K_initial = 1/2 (I_total ?^2 + m_p R^2 ?^2), where the first term represents the kinetic energy of the spinning disk and the second term represents the kinetic energy of the person standing on the rim. Plugging in the values, we get K_initial = 1/2 [(398.25 kg m^2) (1.4 rad/s)^2 + (78 kg) (1.5 m)^2 (1.4 rad/s)^2] = 1197.01 J. When the person walks 2/3 of the way toward the center of the disk, their kinetic energy can be \approx imated as K_p' = 1/2 m_p v^2, where v is the velocity of the person at their final position. We can use conservation of energy to find v: initially, the total energy of the system is K_initial, and when the person reaches their final position, the total energy of the system is K_final = 1/2 (I_total' ?_final^2 + K_p'). Equating K_initial and K_final and solving for v, we get v = sqrt{[2 (K_initial - (1/2) I_total' ?_final^2)] / m_p} = 2.04 m/s. Therefore, the kinetic energy of the person at their final position is K_p' = 1/2 (78 kg) (2.04 m/s)^2 = 159.25 J. The final total kinetic energy of the system is K_final = 1/2 (I_total' ?_final^2 + m_p R^2 ?_final^2 + K_p') = 1/2 [(415.5 kg m^2) (1.34 rad/s)^2 + (78 kg) (1.5 m)^2 (1.34 rad/s)^2 + 159.25 J] = 904.27 J. Therefore, the change in the total kinetic energy of the system is ?K = K_final - K_initial = 904.27 J - 1197.01 J = -292.74 J, which means the energy decreased. 5) The centripetal acceleration of an object moving in a circular path can be calculated using the formula a = v^2 / r, where v is the velocity of the object and r is the radius of the circle. At R/3, the radius of the circle is 0.5 m / 3 = 0.1667 m. Using the velocity calculated in part 4, we get a = (2.04 m/s)^2 / 0.1667 m = 25.15 m/s^2. 6) When the person walks back to the rim of the disk, their moment of inertia returns to the original value of I_p = (78 kg) (1.5 m)^2 = 175.5 kg m^2. Therefore, the new total moment of inertia of the system is I_total'' = I_d + I_p = 222.75 kg m^2 + 175.5 kg m^2 = 398.25 kg m^2. We can again use conservation of angular momentum to calculate the final angular velocity of the disk. Initially, the angular momentum of the system is L_initial = I_total' ?_final = (415.5 kg m^2) (1.34 rad/s) = 556.29 kg m^2/s. When the person walks back to the rim of the disk, the total moment of inertia of the system decreases to 398.25 kg m^2. Therefore, the final angular velocity of the disk can be calculated using the equation L_final = I_total'' ?_final', where ?_final' is the final angular velocity of the disk. Solving for ?_final', we get ?_final' = L_initial / I_total'' = 556.29 kg m^2/s / 398.25 kg m^2 = 1.40 rad/s, which is the same as the initial angular velocity of the disk.