A space station shaped like a giant wheel has a radius of r = 100m and a moment

Question

A space station shaped like a giant wheel has a radius of r = 100m and a moment of inertia of 5 x 108 kgxm2. A crew of 150 people of average mass 65 kg is living on the rim, and the station's rotation causes the crew to experience an apparent free-fall acceleration of g. A research technician is assigned to perform an experiment in which a ball is dropped at the rim of every station every 15 minutes and the time interval for the ball to drop a given distance is measured as a test to make sure the apparent value of g is correctly maintained. One evening, 100 average people move to the center of the station for a union meeting. The research technician, who has already been performing his experiment for an hour before the meeting, and his mood sours even further by his boring experiment in which every time interval for the dropped ball is identical for the entire evening.

Find the tangential speed. Find the new free-fall acceleration experienced by those remaining at the rim of the station.

Explanation / Answer

Radius of the wheel, R = 100 m

    Mass of each person, M = 65 kg

    Moment of inertia of the wheel, I = 5.0 X 108 kg . m2

Case I :

    No. of persons at the rim = 150

    Total mass of 150 persons = 150 * 65

                                            = 9750 kg

    Moment of inertia of the wheel + persons, I1 = I+ M R 2

                                                                       = 5.0 X 10 8 + ( 9750 * 100 2 ) kgm^2

    Centripetal acceleration = g

    2 R = g

    Angular velocity, 1 = ( g / R)

                                   = ( 9.8 / 100 )

                                   = 0.313 rad/s

Case II :

    No. of persons at the rim = 50

    No. of persons at the center = 100

    Total mass of 50 persons = 50 * 65

                                            = 3250 kg

    Total mass of 100 persons = 100 * 65

                                            = 6500 kg

    Moment of inertia of the wheel + persons, I2 = I+ M R 2

                                                                       = 5.0 X 10 8 + ( 3250 * 100 2 ) + 6500* 0

                                                                       = 5.0 X 10 8 + ( 3250 * 100 2 ) kg m^2

    Centripetal acceleration = a

    2 R = a

    Angular velocity, 2 = ( a / R)

                                   = ( a / 100 )


    From the law of conservation of angularmomentum,

    I1 1 = I2 2

    [ 5.0 X 10 8 + ( 9750 * 1002 ) ] * 0.313 = [ 5.0 X 10 8 + ( 3250 * 1002 ) ] * ( a / 100)

    187017500= *532500000 ( a / 100)

    0.351 = ( a / 100 )

    0.1232 = a / 100

    New Acceleration, a = 12.32 m/s^2 or 1.257g
tangential speed v = r                                =   31.3 m/s             

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