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Find kinetic energy from potential energy and moment of inertia?

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A solid cylinder of mass 0.600kg is released from rest and allowed to roll down a board that has an incline of 30degrees to the horizontal floor. If the solid cylinder is released 0.60m above the floor, what is the cylinder's linear speed when it reaches the lowest end of the board? Take into account translational and rotational kinetic energy. Assume no sliding of the cylinder/no friction)

    2. Relevant equations

    Rotational KE= (1/2)*I*ω^2
    Translational KE= (1/2)*m*v^2
    PE1 + KE1 = PE2 + KE2

    3. The attempt at a solution

    PE1 + KE1 = PE2 + KE2
    mgh + 0 = 0 +1/2 mv^2
    mgh=1/2mv^2
    √2gh=v
    √2(9.8m/s2)(0.60m)
    =3.43 m/s

    I don't know if this is right though because the question throws me off when it says to take translational and rotational KE into account.
     
  2. jcsd
  3. Nov 28, 2011 #2
    You have forgotten the rotational KE.
     
  4. Nov 28, 2011 #3
    Because the object is rolling, that means it has translation as well as rotation. Therefore the potential energy is shared between two types of kinetic energy. What you have done only applies to a block sliding down a slope with no friction. You must account for the rotational KE. You have the formula for it. Omega can be related to linear velocity.....
     
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