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Pulley with rotational consideration

  1. Feb 7, 2006 #1
    An object descends from a pulley.
    Use the conservation of mechanical energy to find the linear speed of the descending mass (m = 1.0kg) after it has descended a vertical distance of 2.0m from rest. (For the pulley, M = 0.30kg and R = 0.15m. Neglect friction and mass of the string)

    I solved this question not using the conservation of mechanical energy. At first I tried using it though and I wound up getting:
    vcm = square root of (4/3 mgh/M)

    However, I determined the acceleration using a pervious example:
    a = 2mg/(2m + M)
    and then used v2 = 2ax
    and got 5.8m/s which is the right answer...

    but I don't know how to solve this question using the conservation of mechanical energy.
  2. jcsd
  3. Feb 7, 2006 #2
    Let us take the body pulley Earth as our system. During the following event no external work is done on the system. So the mechanical energy of the system is conserved.
    As the block slides an height h, then there is a loss of mgh amount of potential energy. Since there has been no loss of mechanical energy, the following potential energy is converted into Kinetic enegy. So you equate the change in potential energy to the change in Kinetic energy.
  4. Feb 7, 2006 #3
    yes, i understand that part. is kinetic energy in this case equal to:
    1/2 Icmw2 + 1/2 Mvcm2
    I = 1/2MR2
    so substituting v/R for w
    mgh = 1/4 Mvcm2 + 1/2 Mvcm2

    but solving for vcm here does not get me 5.8m/s
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