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What is the acceleration of a point on the circumference of the disk?

  1. Apr 3, 2008 #1
    1. The problem statement, all variables and given/known data


    What is the accelaration of a point on the circumference of the disk?

    2. Relevant equations

    [tex]\tau[/tex] = F.R = I.[tex]\alpha[/tex]

    I = M.R²/2

    a = R.[tex]\alpha[/tex]

    F(z) = m.g

    3. The attempt at a solution

    The mass exerts a force F(z) = m.g on the disk. This force is what puts the disk into motion with [tex]\tau[/tex] = F(z).R = m.g.R (1)

    Also, as I = M.R²/2 and a = R.[tex]\alpha[/tex] and [tex]\tau[/tex] = I.[tex]\alpha[/tex]
    ==> [tex]\tau[/tex] = M.R.a/2 (2)

    Out of (1) & (2):

    m.g.R = M.R.a/2 <=> m.g = M.a/2 <=> a = 2m.g/M

    The acceleration asked for is a = 2m.g/M

    But the teacher's solution says: [tex]\frac{2m.g}{2m+M}[/tex]

    What have I done wrong?

    Attached Files:

    Last edited: Apr 3, 2008
  2. jcsd
  3. Apr 3, 2008 #2
    I can't view your problem or picture yet, but if this is a disk being accelerated by a hanging mass, the force exerted on the disk is not mg. The mass and disk are connected by a string, and it's the string tension which accelerates the disk. The string's tension and the weight cannot be the same, or else the hanging object would be at rest.
  4. Apr 3, 2008 #3
    You can't view the picture? What does it say when you click it. (EDIT: made visible)

    Also, it said that the rope was without mass. I'm not really familiar with string tension. There is no information given about the string tension coëfficient (if something like that exists, at least), its length, girth, etc... I don't see how it fits in.
    Last edited: Apr 3, 2008
  5. Apr 3, 2008 #4
    Draw a free-body diagram of the hanging mass. There are two forces acting on it. This will give you an expression for a in terms of tension T and other constants. Same goes for a torque expression for the disk. 2 equations with 2 unknowns.

    I can't see the picture yet because it must be approved by the forum moderators first.
  6. Apr 3, 2008 #5
    Oh I see... So the string tension is somewhat like the normal force of suspended objects. (probably a bad analogy, but yeah...)

    So the force acting on the disk is not the gravitational force nor the string tension, but the difference between the two. So:

    m.g - L = m.a <=> L = m.g - m.a = m(g-a) (1)

    And we also had the equation [tex]\tau[/tex] = L.R = I.[tex]\alpha[/tex]
    and we get (after substitution using the other equations):

    L = M.a/2 (2)

    Out of (1) & (2):

    m(g-a) = M.a/2 <=> m.g - m.a = M.a/2 <=> 2m.g = M.a + 2m.a <=> a = [tex]\frac{2m.g}{2m+M}[/tex]

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