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Rotational dynamics problem

  1. Mar 14, 2014 #1
    1. The problem statement, all variables and given/known data
    a) A 1.5 kg frog sits at rest on top of a solid disk. The disk's mass is 4 kg, its radius is 1.25 m, and it rotates on a frictionless axle. The frog jumps from the disk at 3.7 m/s at an angle of 50°. Find the angular momentum of the frog about the axle as it leaves the disk.
    b)How much time will it take the disk in this problem to rotate one complete revolution after the frog jumps?


    2. Relevant equations
    a) ∠=mvrsinσ

    b)v=d/t or 2∏r/v=t
    3. The attempt at a solution
    I was able to find the angular momentum of the frog just fine. ∠=-4.46kgm^2/s. My problem is finding the time it takes the disk to complete one revolution after the frog has jumped. I used t=d/v or 2∏r/v=2∏(1.25m)/3.7m/s. The answer is supposed to be 4.40 seconds and I am getting something different. Am I setting up this problem incorrectly?
     
  2. jcsd
  3. Mar 14, 2014 #2

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    Hello Chase11,

    Welcome to Physics Forums! :smile:

    But that method won't work. You don't know what the velocity of the edge of the disk is (you were assuming it's 3.7 m/s, but it's not. That's the frog's velocity, not the disk's) [Edit: by that I mean you don't know what the velocity is yet. In a sense, that is what you will end up solving for.]

    Besides, it doesn't even take the mass of the disk into account.

    Try to start part b) of the problem with conservation of angular momentum. (As an interim step, you'll need to find the disk's moment of inertia. :wink:)
     
    Last edited: Mar 14, 2014
  4. Mar 14, 2014 #3

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    By the way, that's one big frog! :tongue2:
     
  5. Mar 14, 2014 #4
    You've assumed that the disk is rotating the same speed as the frog. Imagine if the disk was the planet earth instead.
     
  6. Mar 14, 2014 #5
    Sorry I am still confused. I found the moment of inertia of the disk is I=1/2Mr^2 or (1/2)(4kg)(1.25m)^2=3.125kgm^2.

    So if I use the conservation of angular momentum I have Iiωi=Ifωf. I don't understand what is going to change here between the initial and final values for the moment of inertia. Wouldn't it be the same for both?
     
  7. Mar 14, 2014 #6
    Moment of inertia doesn't change and the angular momentum of the frog-disk system doesn't change either.
     
  8. Mar 14, 2014 #7
    So how do I use Conservation of angular momentum to solve for this?
     
  9. Mar 14, 2014 #8
    Before the frog jumped, what was the angular momentum of the system?
     
  10. Mar 14, 2014 #9
    Zero?
     
  11. Mar 14, 2014 #10
    Correct, so what then is the angular momentum of the system after the frog jumped?
     
  12. Mar 14, 2014 #11
    It must also be zero.
     
  13. Mar 14, 2014 #12

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    Correct! :approve:

    So you already know the angular momentum of the frog part of the frog-disk system (after the jump). What does that tell you about the disk part of the frog disk system (after the jump)? :wink:
     
  14. Mar 14, 2014 #13
    So is it 4.46 kgm^2/sec since the frog's was -4.46?
     
  15. Mar 14, 2014 #14

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    Right! Very nice. :approve:

    So now, how does the disk's rotational period fit in?
     
  16. Mar 14, 2014 #15
    Okay so if I have ∠of disc=4.46kgm^2/sec 4.46=mvr

    so 4.46=(4kg)ω(1.25m) I get ω=.892 m/s.

    Then T=2pi/ω I get T=7.04sec. This is still the wrong answer.
     
  17. Mar 14, 2014 #16
    What happened to your beautiful Iω that you spent all that time calculating?
     
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