Question

A space station shaped like a giant wheel has a radius of 1.00 x 102 m and a moment of inertia of 5.00 x 108 kgm2. A crew of 150 lives on the rim, and the station is rotating so that the crew experiences an apparent acceleration of 1.0g. When 120 people move to the center of the station for a union meeting, the angular speed changes. What apparent acceleration is experienced by the mangers remaining at the rim? Assume the average mass of a crew member is 65.0 kg.

Please show ALL work

Answer 1

Given radius   r = 100 m

   Moment of inertia of space station (empty)   I_S   =   5.27 * 10⁸   kg-m²

   Moment of inertia of one person                    I₁   =   m * r²

                                                                                          =   65 * 100²   

                                                                                 =   65 * 10⁴ kg-m²

   Hence moment of inertia of 150 persons      I₁₅₀ =   150 * 65 * 10⁴

                                                                                  =   9.75 * 10⁷ kg-m²

   Moment of inertia of (150 - 120 = 20) persons   I₃₉   =   20 * 65 * 10⁴

                                                                                  =   1.3 * 10⁷   kg-m²

Calculate the total initial moment of inertia   

I₀ =   I_S + I₁₅₀

     =    5.00 x 10⁸   +   9.75 * 10⁷

     =   5.975 * 10⁸   kg-m²

   Calculate the final moment of inertia

I   =   I_S + I₃₉

      =   5 * 10⁸   +   1.3 * 10⁷

    =   5.13 * 10⁸   kg-m²

   Calculat the Initial acceleration   

a₀ = r * ω₀²

1.00 * 9.81 =   100 * ω₀²

ω₀   =   √(9.81 / 100)

      =   0.313 rad/s

   According to the law of conservation of angular momentum

   I₀ * ω₀   =     I * ω

   Final angular speed   ω   =   5.975 * 10⁸ * 0.313 / 5.13 * 10⁸

                                          =   0.365 rad/s

Calculate the acceleration at the rim   

a   =   r * ω²

     =   100 * 0.365²

    =   13.3 m/s²

    =   13.3 / 9.8

    =   1.357 g

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