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Angular Momentum Question concerning a Merry Go Round

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data
    If the steel disk has mass of 200 kg and a radius of 2 meters you can make it spin by applying a force to the rim. This torque increases the angular momentum of the disk. Suppose the force is 20 newtons. How long would you have to apply it to get the wheel spinning 5 times a minute?

    What would happen to the rate of spin if you then jumped on the rim of the wheel with your mass of 60 kg?

    2. Relevant equations
    I=1/2mr2

    I1*w0/(i1+i2)

    3. The attempt at a solution



    I think I have all the variables figured out I am just having an issue solving the questions.

    I have the w=0.1rad/s , I =400 kg*m^2, t=40 N*m.



    For the second problem I got the w2=0.07rad/s which would mean the merry go rounds rate of spin would decrease by .03 rad/s when the mass of 60kg jumps on.

    (400)(0.1)/(320+240) = 40 / 560 =0.07rad/s



    Any help would be appreciated thanks!
     
  2. jcsd
  3. Feb 20, 2017 #2
    What does this represent? Hopefully not 5 revs/minute.
     
  4. Feb 20, 2017 #3
    No I believe that is my angular momentum before the 60kg person jumps on.
     
  5. Feb 20, 2017 #4
    It has units of angular velocity though.
     
  6. Feb 20, 2017 #5
    Maybe that is where I have messed up. Most of the problems my teacher provided were not setup in this manner.
     
  7. Feb 20, 2017 #6
    You probably want to convert the angular velocity of 5 rev/minute to rad/s.
     
  8. Feb 20, 2017 #7
    The first time I solved it I got these as my answers but wasn't for sure if they were correct.

    1. 300s

    2. The acceleration would then be reduced to 0.0625 rad/s^2.
     
  9. Feb 20, 2017 #8
    So 5 * 2pi? or 31.42 radians ?
     
  10. Feb 20, 2017 #9
    It's 31.42 radians/minute. Convert that to rad/s.
     
  11. Feb 20, 2017 #10
    Alright so that gives me .52 rad/s
     
  12. Feb 20, 2017 #11
    Yes, so then work with that to calculate the time it takes to go from ω=0 to ω=0.52 rad/s. That will be based on the angular acceleration you calculated from your torque, inertia, etc.
     
  13. Feb 20, 2017 #12
    I must be slow today lol
    How long would you have to apply it to get the wheel spinning 5 times a minute? I can't wrap my head around the equation I need to solve to get how long I would have to apply the force to get it spinning 5 times a minute.
     
  14. Feb 20, 2017 #13
    What equations do you have to work with? If you remember the linear equations, it is very easy to figure out the angular motion equations.
    Just substitute θ for x, ω for v, α for a, Γ for F, I for m. I think that covers it all.
     
  15. Feb 20, 2017 #14
    Yeah but what formula solves for how long to apply the force to get the wheel to turn 5 times a minute ?
     
  16. Feb 20, 2017 #15
    I didn't memorize any equation that does that. But I know how to get angular acceleration from torque and moment of inertia.
     
  17. Feb 20, 2017 #16
    So would that just be my torque which is 40 N * m and my I which is 400 kg*m^2

    40=400a so that gives me my a=0.1 rad/s ?
     
  18. Feb 20, 2017 #17
    Yes, but to be more accurate:
    α = 0.1 rad/s^2
     
  19. Feb 20, 2017 #18
    in which equation used uses α?
     
  20. Feb 20, 2017 #19
    5 times per minute would be 12 seconds per turn. so it should be .083/sec.
     
  21. Feb 20, 2017 #20
    Torque=I*a
     
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