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Rotational Inertia and Motion problems

  1. Apr 24, 2014 #1
    Hello everyone, I am trying to solve some homework but I am not entirely sure what formulas to use.

    Problem) There are 3 kids that weigh 25kg each sitting in the center of a merry go round (Disk). The merry-go-round weighs 400 kg and has a diameter of 3 m. It is initially spinning at 50 rpm.

    A) What is the initial rotation?

    I tried using the formula 1/2 mr^2. .5(400kg)(1.5)^2 = 450kg*m/s

    B) What is the initial kinetic energy of the merry go round with the kids on it?
    I tried 1/2 IW^2
    1/2 (450)(5.2359878)^2 which equals 6168.5 N

    W would be the 50 rpm converted to radians/second which would be 5.23rad/s and I was = 450 from earlier.

    C) If all the kids move to the edge of the merry go round, what will be the new angular velocity in rad/s?

    I got a little lost on this, would the proper formula be L = iw. 50 = 450w and w be the answer?

    D) What is the magnitude of the linear velocity of each child as they sit on the edge and it spins at the rate round in C?

    V = rω
    V = 1.5(Answer from C in radians/sec)

    E) The kids now want to stop. they drag their feet so they each exert a force F that is tangent to the disk in a direction that causes the merry-go-round to stop. How much net work must be done by the 2 kids in order to bring the disk with them on the edge to a stop.?

    Not sure how to do this one.

    F) If the disk stops after 6 seconds, what is the deceleration of the disk?
    Wouldn't this ω = ωo + (alpha)t
    5.23(rad/s) = 0 + (alpha)(6)
    5.23/6 = alpha?

    G) What is the torque on the disk?
    Not sure how to solve this one

    H) Magnitude of the average Force F each kid needs to exert?
    Not sure again how to solve this

    I) How many revolutions will the disk need to turn before coming to a stop?
     
    Last edited: Apr 24, 2014
  2. jcsd
  3. Apr 24, 2014 #2

    BiGyElLoWhAt

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    Gold Member

    A) I don't know what you mean by initial rotation
    B) ##\frac{1}{2}I\omega ^2## is the correct formula, and the kids are at the center (i.e. the axis of rotation, so they don't contribute to the moment of inertia) so you're good there.
    C)Hint, something in this situation is conserved, and something changes when the children move outwards.
    D)Right equation, but wrong numbers. See C (HA!)
    E) Your looking for work, so energy, sooo...?
    F) No, you have to find a different omega. Again, See C.
    G) ##T= R\ (cross)\ F##
    H) Use what you get from F
    J) Use the rotational analogue for your displacement kinematic equation

    Hope this helps
     
  4. Apr 24, 2014 #3
    It did help some, Thank you.

    I got part A/B/C/D done I believe. I did A as finding I, (I = 1/2 mr^2).

    For A I got 450 kg* m^2 /s
    B: 6168.5 J
    C:4.40925
    D: 6.613875


    For G, I am not sure what you mean. The only three formulas I have that involve T are T = 1/f (frequency)
    T = la and T = Delta L / Delta time.
     
    Last edited: Apr 24, 2014
  5. Apr 24, 2014 #4

    BiGyElLoWhAt

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    Gold Member

    T as in torque, not as in Period.

    Torque is equal to radius cross the force.
     
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