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Newton's Second Law and Rotation

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A 0.70-kg disk with a rotational inertia given by MR^2/2 is free to rotate on a fixed horizontal
    axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass hangs
    from the free end. If the string does not slip, then as the mass falls and the cylinder rotates,
    the suspension holding the cylinder pulls up on the cylinder with a force of:
    A. 6.9N
    B. 9.8N
    C. 16N
    D. 26N
    E. 29N

    2. Relevant equations

    Newton's second law for rotating bodies

    3. The attempt at a solution

    I was able to get the acceleration of the disk + the mass by myself following the solution below. It's probably pretty obvious but I just can't seem to figure out how he got T = mg + M(g-at), specifically the M(g-at) part assuming that the suspension holding the cylinder pulls up on the cylinder with a force equal to the sum of the weights of the pulley and the block.

    This is the solution I am trying to make sense of:

    The suspension holding the cylinder pulls up on the cylinder with a force of

    T = mg + M(g-at), where m - disk, M - 2kg mass, at - the acceleration of a 2.0-kg mass. At the same time that is the tangential acceleration of the disk. Let's find it.

    J dω/dt = M(g-at) * R,

    mR²/2 * 1/R dV/dt = M(g-at) * R, where at dV/dt = at.

    m at/2 = M(g-at)

    at = 2g/(m/M+2) = 2*9.8/(0.7/2 + 2) = 8.34 m/s²

    T = mg + M(g-at) = 0.7*9.8 + 2*(9.8-8.34) = 6.86 + 2*1.46 = 9.8 N
     
  2. jcsd
  3. Nov 30, 2013 #2

    Andrew Mason

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    Draw free body diagrams. The forces acting on the axle must sum to 0 since the axle is not accelerating. The forces acting on the falling mass M must add to Mat.

    The forces acting on the axle are the tension in the axle support, Tx (upward), mg (downward) and the tension in the string (downward). The forces acting on the falling mass are: Mg (downward) and the tension in the string, Ts(upward).

    So:

    Tx - mg - Ts = 0

    and

    Mg - Ts = Mat

    These reduce to:

    Tx = mg + Mg - Mat

    AM
     
  4. Nov 30, 2013 #3
    By axle, do you mean the disk, or the thing that's supporting the disk+mass to the ceiling? Also (assuming by axle you mean the disk) how come it isn't accelerating? Isn't there a net torque caused by the tension in the string causing angular acceleration in the disk?
     
  5. Nov 30, 2013 #4

    Andrew Mason

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    You state that the disk is free to rotate about a fixed axis suspended from the ceiling. I appeared to me that this axis passed through the centre of the disk (radius R). So it appears to me that the disk rotates on a fixed horizontal axle through its centre of mass. How else would it rotate?

    The centre of mass of the disk does not accelerate because the axle is fixed. What you want to determine is the upward force on that axle.

    AM
     
  6. Nov 30, 2013 #5

    CWatters

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    Andrew means it doesn't have linear acceleration (it does have angular acceleration)

    Your assumption is incorrect or perhaps you need to think about the "weight of the block M". The mass M is accelerating downwards so the suspension isn't carrying it's full weight (eg it's not Mg).

    Consider..

    If the disc was locked so that mass M was just hanging there then the force on mass M would be Mg and the force on the suspension would be mg + Mg.

    If mass M was somehow in total free fall (eg the string had been cut) then then force due to mass M would be zero. In that case the force on the suspension would be just mg + 0

    So in your problem the force on the suspension is going to be somewhere between these two.

    What is the tension in the string?
     
  7. Nov 30, 2013 #6
    OK, so to summarize:
    - the disk has angular acceleration and the block has linear acceleration
    - the disk has no linear acceleration, so by Newton's Second Law (not for rotation), it has no (linear) acceleration so the sum of the forces = 0
    - the tension in the string causes a torque on the disk so it has the same tangential acceleration as the hanging mass

    Hmm, I think those are the key ideas to take away. Am I missing something. I think in the end, once I realized that why the disk is not accelerating, the whole solution became much more clear.
     
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