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KE of Rotation & Moment of Inertia

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    A light string is wrapped around the outer rim of a solid uniform cylinder of diameter 75.0 cm that can rotate without friction about an axle through its center. A 3.00 kg stone is tied to the free end of the string, as shown in the figure . When the system is released from rest, you determine that the stone reaches a speed of 3.50 m/s after having fallen 2.50 m.

    What is the mass of the cylinder?


    2. Relevant equations

    KE=1/2(1/2MR^2)Ī‰^2

    v=rĪ‰

    3. The attempt at a solution

    v=rw
    3.50=.375w
    w=9.3

    ke=1/2(1/2M(.375)^2)(9.3)^2

    I am doing this right and
    is 30406kg correct?
     

    Attached Files:

  2. jcsd
  3. Nov 4, 2011 #2

    I like Serena

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    Hi again valeriex0x! :smile:

    What did you do with the kinetic energy of the stone?

    Btw, the mass you found looks a bit large doesn't it?
     
    Last edited: Nov 4, 2011
  4. Nov 4, 2011 #3
    i got it
     
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