A space station is constructed in the shape of a hollow ring of mass 4.55 x 104

Question

A space station is constructed in the shape of a hollow ring of mass 4.55 x 104 kg and radius 115 m. The station has 200 crew members (each with average mass 70 kg), who can walk on a deck formed by the inner surface of the outer cylindrical wall of the ring. Artificial gravity is produced when the ring rotates (that is, the crew members experience a normal force from the floor pushing them inward). (a) Assume the station is initially not rotating, and assume that the entire mass of the space station is concentrated in the outer ring. Also assume that the crew is initially on the outer rim. Rockets on the outside of the outer rim are fired tangentially to get the station spinning up to a speed such that the crew experiences a centripetal acceleration equal to g. What is the total angular momentum of the space station (plus crew) when it is up to maximum speed? (b) How much work is done to get the space station spinning? 134500000 JX Your response differs from the correct answer by more than 100%. (c) How long must the rockets be fired if there are 8 rockets, and each exerts a thrust of 175 N? 1426 s (d) The rockets are no longer thrusting. The entire crew moves to the very center of the space station. What acceleration does an ant experience on the outer rim?

Explanation / Answer

(a) For a force of mg to be exerted, a centripetal force of mw^2r is required. So, g=r*w^2 (where w= angular speed)

Or, 115*w^2= 9,81

or, w= 0.292 rad/s

Now, total m0ment of inertia I = Mr^2+ 200*mr^2 {M being rims's mass and m each human's mass} = 7.86*10^8 kgm^2

so, total angular momentum= I*w = 7.86*10^8 * 0.292= 2.29*10^8 kgm^2/s

(b) Work done= kinetic energy provided to ring= 0.5*I*w^2= 0.5* 7.86*10^8 *0.292^2= 3.35*10^7 J

(c) Now, torque*duration for which torque is applied= I*change in angular speed (total toque= force * distance= 8*175*115)

Or, 8*175*115 * t = 7.86*10^8 * 0.292= 2.29*10^8

or, t= 2.29*10^8/  8*175*115 = 1423s

(d)Now, angular momentum of 2.29*10^8 is conserved when everyone reach the centr of ring. So, 2.29*10^8= new moment of inertia(which is Mr^2) * new agular momentum w'

or, w'= 2.29*10^8/ 4.55*10^4 *115^2 = 0.38 rad/s

So, aceleration an ant experience on the outer rim= r*w'^2= 115*0.38^2= 43.76m/s^2

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