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What is the speed when a disk has reached maximum extension?

  1. Jun 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Two identical disks with mass m and radius r are connected via a massless wire of length L which is winded up around both disks. Disk B is connected to the ceiling and is free to rotate around its axis. Disk A is besides disk B and will fall due to the gravitational force with acceleration of g.

    Visualization

    http://i.imgur.com/LZAk170.png

    While disk A falls it will unwind. Question 1 is to find the acceleration of disk A - assuming it moves straight down. Question 2 is to find the speed of disk A when the massless wire has reached its maximum extension. Question 3 is to explain what happens when the wire has reached its maximum extension.

    2. Relevant equations
    • Fg=mg, ΣF=ma
    • Torque=Fr, Torque=Iα (I is moment of inertia and α is angular acceleration)
    • I=½mr2
    • α=atan/r (rolling without slipping)
    • Conservation of energy: K1+U1=K2+U2
    • Kinetic energy(translational + rotational): K=½mvcm2 + ½Iω2
    • vcm=rω (rolling withouth slipping)
    • Potential energy: U=mgh

    3. The attempt at a solution
    I'm stuck at question 2 but to know if i'm correct I have also put my answer for question 1 here:

    For question one I divided the situation in two parts. First part is to determine the tangular acceleration of disk B. The force applied on disk B is Fg from disk A which is mg. By equalising both torque terms you get mgr = ½mr2*atan/r. Which gives atan=2g.
    The second part is to find the downwards acceleration of disk A. There are two forces applying on disk A and that is the tension of the rope T and the Fg. Force T is the only force applying torque on disk A. So we can write ΣF=mg-T=macm. To solve this first we equalise the torque equations again: ½mr2*acm/r = Tr. This gives T=½macm. Subbing this in the force equation you get: g-½acm=acm --> acm=2/3g.
    The total downwards acceleration of disk A is atan+acm=8/3g.

    For question 2 I started with the conservation of energy. The starting position is when both disks are at rest at the same height. There is no kinetic energy and only potential energy with height L. When disk A falls the potential energy will turn into kinetic energy of both translation and rotation. Disk B stays at the same height but will also have kinetic energy of only rotation. Because disk B stays at the same height the potential energy of disk B will cancel out. So we get: UA1 = KA2 + KB2.
    mgL = ½mvcm2 + ½IωA2 + ½IωB2. The speed at which disk A moves down due to the rotation of disk B is vtan=rωB. The speed at which disk A moves down due to its own rotation is vcm=rωA. After subbing everything in you get: gL=¾vcm2+¼vtan2.
    I know the total speed of disk A going down is vtan+vcm but because of those fractions it's not that easy to sub it in. I have a feeling the solution is really easy but I cannot see it. Any help would be much appreciated.

    For question 3 i have a feeling disk B will start the oscillate from right to left and disk A will move along with it as a pendulum.
     
  2. jcsd
  3. Jun 7, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Why? Check the force balance at disk A. Hint: disk A is accelerating.

    For question 2, you can use the result from question 1 once that is correct, but conservation of energy is possible as well.

    There is a toy that looks very similar to this system, with a hand instead of disk B. How does that behave?
    The rotations won't stop immediately.
     
  4. Jun 7, 2015 #3
    Ok, so the force balance at disk A is the gravitational force downwards and an equal tension force upwards. There will be torque clockwise because the tension force is excerted at distance r from the rotation axis. Just as I solved it earlier i get for acceleration downwards acm=2/3g. Then will the force applied on disk B be F=macm? If so then after subbing and using the same equations i get for atan=3/2g. Total downwards acceleration is now 2g. I don't know if this is correct but if so can I use the equation v2=v02+2aL to say that the final speed is v=√(4gL)?

    I believe you're referring to a yoyo. So disk A will roll back up and then down again? I thought this was too easy because the original questions says yoyo instead of disk so saying it will behave like a yoyo seemed too simple for me.
     
  5. Jun 7, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If the forces on A are balanced it does not move down at all.
    Yes.
     
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