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[centripetal acceleration] a disk-shaped space station 200 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center. what should the period (the time to complete one revolution) of this disk be so that the centripetal acceleration of all points on its rim is g (



Answer :

The time period of the disk-shaped space station which spins uniformly about an axis perpendicular to the plane of the disk is 20.07 s.

Given that,

Diameter d = 200 m

Radius r = d/2 = 200/2 = 100 m

As the body is in circular motion, centripetal force can be equated to the weight.

m* g = m* r* ω²

r* ω² = g

ω² = g/r

Frequency ω = √g/r = √(9.8/100) = 0.313 rad/s

We know the relation between ω and T as

T = 2π/ω = (2*π)/0.313 = 20.07 s

Thus, the time period of the disk is 20.07 s.

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