We have mentioned that the inner surface of a rotatin space station can be used

Question

We have mentioned that the inner surface of a rotatin space station can be used for artifical gravity. Suppose a nearly hollw 150,000kg cylinder with a diameter of 25m is used for thr artificial gravity this way.
a.) Find the angular velocity necessary to match Eart's apparent gravity at the surface.
b.) Find the kinetic energy of the space station about its center line.
c.) Suppose the station is brought from rest to this state after rockets fire for two hours. Find the angular acceleration and the total torque necessary to do this.

Explanation / Answer

Okay, this means that the cylinder , while rotating , should pull a body inside with an acceleration equal to 10 m/s^2 ( Assuming g= 10 units)
The centripetal acceleration of the cylinder = r * 2

Putting this equal to g.

10 = 25/2 * 2

Hence 2 = 20/25

Hence = 0.8944 rad/sec

Now Moment of inertia = m*r2 = 150000 * 25*25/4 = 23437500 kg-m^2

Hence Kinetic enrgy of the space station = 1/2 * I *  2 = 1/2 * 23437500 * 0.8944 * 0.8944 = 9.28 * 10^6 Joule

Time to make final angular velocity zero = 2 hours = 2* 60 * 60 = 7200 seconds

Hence * t = initial

= 0.8944/7200 = 1.24 * 10^-4 rad/sec^2

= I * =  23437500 * 1.24 * 10^-4 = 2911.45 N-m

Hope you get the solution.

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