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Mass of cylinder using ang.velocity

  1. Nov 10, 2007 #1
    A light string is wrapped around the outer rim of a solid uniform cylinder of radius .375m 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 When the system is released from rest, you determine that the stone reaches a speed of 3.50 m/s after having fallen 2.50m .
    What is the mass of the cylinder?

    Ive solved for:
    F=M*a
    3kg*9.8=29N
    29N*2.5m=73.5J

    i know its a solid cylinder so I=1/2 Mr^2 so i need to solve for the inertia. from here i got stuck i know i have to use

    Kf=1/2mv^2+1/2Iw^2 i dont know where to go from here. do i solve for the angular velocity(w)? or Inertia?

    i know v=rw so would w=v/r i could use 3.5m/s/.375m to get 9.3rad/s?
     
  2. jcsd
  3. Nov 11, 2007 #2
    anyone know if im doin this one rite?
     
  4. Nov 11, 2007 #3
    i used the Ki + Ui=Kf + Uf

    Ki=0
    Ui=mgy=73.5J
    Kf=1/2mv^2+1/2Iw^2
    Uf=0

    i got
    73.5=1/2Iw^2+18.375 so

    1/2(I)w^2=55.125J

    im stuck on Kf where i have to solve for the two unknowns I* w* i used v=rw to solve for w is that correct?
     
    Last edited: Nov 11, 2007
  5. Nov 11, 2007 #4
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