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Angular Momentum of a playground ride

  1. Nov 12, 2006 #1
    A playground ride consists of a disk of mass M = 46 kg and radius R = 2.0 m mounted on a low-friction axle. A child of mass m = 19 kg runs at speed v = 2.3 m/s on a line tangential to the disk and jumps onto the outer edge of the disk.

    What is the angular speed?

    I found the total angular momentum and used L_rot = I*w. Didn't work =/ Since the disk does not move, does that mean that L_total = L_rot? Or am I not getting something about L_trans? Cuz I assumed L_trans is zero because the system isn't moving.
     
  2. jcsd
  3. Nov 13, 2006 #2

    Doc Al

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    Staff: Mentor

    Translational momentum is not conserved, since the axle of the disk is fixed. But angular momentum is conserved. What's the initial angular momentum of the child just as she jumps onto the disk? What's the total rotational inertia of the system (disk + child) after she jumps on? Use that to solve for the angular speed.
     
  4. Nov 13, 2006 #3
    Can you explain how I should be adding up the rotational inertias? Not sure what the end result should look like.
     
  5. Nov 13, 2006 #4

    Doc Al

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    Staff: Mentor

    The total rotational inertia of the system is just the sum of the individual rotational inertias of the parts:
    (1) the disk (use a standard formula to calculate its I)
    (2) the child (treat the child as a point mass a distance R from the center)
     
  6. Nov 13, 2006 #5
    Hint: Assuming the child is a point mass (unless you want to integrate over the child :biggrin: ) you can use the parrallel axis theorem which says that the I of the child is [tex]mr^2[/tex]. Add this to the I of the disk to find total I. You add I's just like you would add m's, nothing complicated there. Use that to find total [tex]\omega[/tex]
     
  7. Nov 13, 2006 #6
    So L_total would be 2*19*2.3 -> 87.4 kg m/s. And I would be 19*2^2 + .5*49*2^2. L_total / I should be my answer then? I get 0.502299. Thnx for the help btw =]
     
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