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Angular speed of rotation problem

  1. Apr 2, 2006 #1
    Could anyone please help me figure out how to approach this problem?

    Space stations have been proposed to accommodate the surplus population of the Earth. The initial design is for a hollow, uniform, cylindrical space station of diameter 3.15 km, length 10.35 km, and total mass of 1.19 x 10^10 metric tons. The space station is to be spun about the symmetry axis coincident with the axis of the cylindrical shape.

    (a) What angular speed of rotation is needed to simulate the magnitude of the local acceleration due to gravity (9.81 m/s2) for objects on the perimeter of the space station?

    (b) What is the rotational kinetic energy of the space station?


    I beleive that for part (b) i can use the equation KErot = (1/2)*I*(omega)^2 while finding I using I=(1/2)*m*R^2 so with the numbers given I calculate I to be (1/2)*(1.19*10^13 kg)*(1575m)^2 = 1.476*10^19

    I still am unsure as to how to calculate part (a), which I need for part (b) as well. The only equation I could find that relates acceleration to angular speed is a=r*(omega)^2 which would give me 9.81 = (1575)*(omega)^2 which gives omega=0.0789 but i don't think this is correct. Could anyone give me a hint as to what equation(s) i need for finding omega (angular acceleration).

    Thanks!
     
  2. jcsd
  3. Apr 2, 2006 #2

    nrqed

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    Your omega looks right to me!

    Patrick
     
  4. Apr 2, 2006 #3
    oh actually it was right i was just typing it in wrong, thanks for the help!

    the last part of the same problem is also giving me some trouble

    (c) Small, thrusting rockets mounted tangential to the circular cross section are to set the space station in rotational motion. Starting from rest, the spacecraft reaches the angular speed calculated in part (a) after 1.10 y. Each rocket is capable of exerting a force of magnitude 1000 N continuously over the year. How many thrusting rockets are needed?

    I tried finding this answer by first using the equation:
    angular acceleration = (omega final - omega initial)/t or
    angular acceleration = (.0795 - 0)/(1.1years*365days*24hrs*60min*60sec = 34689600sec) = 2.27*10^-9 m/s^2
    then i used the equation:
    torque = I*angular accel. or
    torque = (2.9519*10^19)*(2.27*10^-9) = 6.715*10^10
    and then divided this by 1000N and got 6.715*10^7 rockets
    but this is not the correct answer so i must be attempting this problem the wrong way

    i have already confirmed that omega = 0.0789 and that I = 2.9159*10^19 so this means that i am probably just using the wrong equations as opposed the wrong numbers.

    if anyone could help me out, i'd really appreciate it.

    thanks again!!
     
  5. Apr 2, 2006 #4

    BobG

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    I thought you'd already found the moment of inertia was 1.475 x 10^19 kg-m^2.

     
  6. Apr 2, 2006 #5
    yes but i realized this was incorrect because that formula is assuming the cylinder is solid, but in this problem we are to assume it is a cylindrical shell or hoop so you do not divide by 2 so I = 2.95*10^19.

    sorry, i should have mentioned that in my post!
     
  7. Apr 2, 2006 #6

    nrqed

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    just a comment: the angular acceleration is in radian per second^2, not m/s^2
    your mistake is to take the torque and divide it by the tangential force. A torque is not a force. You have (for a force applied tangentially) torque = force times the radius. So you must first divide the torque by the radius to find the total tangential force and *then* divide by the 1000 N.

    Patrick
     
  8. Apr 2, 2006 #7
    aha!

    you're completely right, so it needs 42629 rockets (42628.38 rounded up to a whole number). i figured that was the place where i was making the incorrect assumptions. that makes more sense now :smile:

    thank you!
     
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