Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Rotation homework question

  1. Apr 1, 2004 #1
    The radius of a park merry-go-round is 2.2m. To start it rotating, you wrap a rope around it and pull with a force of 260N for 12s. During this time, the merry-go-round makes one complete rotation. a) Find the angular acceleration of the merry-go-round. b) What torque is exerted by the rope on the merry-go-round? c) What is the moment of inertia of the merry-go-round?

    A)
    The angular acceleration is alpha = Torque/I
    I did conceder the merry-go-round as a solid cylinder so I =1/2(MR^2)
    Torque = Fr
    Acceleration = (260N * 2.2m)/(1/2(MR^2))
    I don’t know how to get rid of the M?

    B)
    Torque = Fr = (260N * 2.2m) = 572 Nm

    C) The moment of inertia of the merry-go-round is I =1/2(MR^2)
    Can I get the M in the torque from F = ma (after I know the acceleration from a) )?

    Thank you
     
  2. jcsd
  3. Apr 1, 2004 #2
    The merry-go-round completes one rotation, i.e 2π radians, in 12 seconds. I think it's valid to apply the normal motion equations here, so:
    [tex]d\alpha = \alpha _0 + \omega _0t + \frac{1}{2}a_rt^2[/tex]
    [tex]2\pi = \frac{1}{2}a_r(12s)^2[/tex]
    [tex]a_r = \frac{4\pi }{144s^2}[/tex]

    Now that you know the angular acceleration you can find the merry-go-round's moment of inertia since a = Torque/I as you said. With this method you don't even need to know the shape of the device (and I don't think you are allowed to assume it is a cylinder).
     
    Last edited: Apr 1, 2004
  4. Apr 1, 2004 #3
    472Nm*144s^2 = 4(pi)I
    I=67968/4(pi) = 5408.72 (I feel there is something with the answer)
     
  5. Apr 1, 2004 #4
    It's 572N, not 472N. According to this the mass of the merry-go-round would be 2.7 tons... do you know what the answer should be?
     
  6. Apr 1, 2004 #5
    I = 6554.64 ? I think this is the corect answer but I don't know the rounding if its correct.
     
  7. Apr 1, 2004 #6
    I also get 6,544. I'm not sure what you're asking, is this not the correct answer?
     
  8. Apr 1, 2004 #7
    Yes, is it the correct answer? Is the unite of I is the same as the unite of mass?
     
  9. Apr 1, 2004 #8
    No, the unit of I, the moment of inertia, is m^2kg. I don't know if it's the right answer or not, but as far as I'm concerned my calculations are correct.
     
  10. Apr 1, 2004 #9
    I really appreciate your help. I think the answer is correct But I am not sure if I should round it to 6555 m^2.kg?
     
    Last edited: Apr 2, 2004
  11. Apr 2, 2004 #10
    like in this case what would be the significant figures rule?
     
  12. Apr 2, 2004 #11

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The general rule would be that you cannot have more significant digits then the numbers used in the calcultion. I would say the answer should be

    6.6x 103 m2kg
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook