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Solid pulley & mass on an inclined plane

  1. Jun 25, 2013 #1
    1. The problem statement, all variables and given/known data

    See the image.

    2. Relevant equations

    T=Iα, F=ma, I=0.5*m*r^2, Friction=u*N

    3. The attempt at a solution

    I used the torque equation to get Tension = mcylinder*a/2, and then plugged that into a force equation for the block.

    mblockgsin(30)-Tension-Friction=mblock*a

    so

    mblockgsin(30)-mcylinder*a/2-μmblockgcos(30)=mblock*a

    I ended up with μ=0.36. Just wanted to check my answer because I never ended up using the radius of the cylinder (it canceled out), so I was wondering why it was given.
     

    Attached Files:

    Last edited: Jun 25, 2013
  2. jcsd
  3. Jun 26, 2013 #2
    I agree with you.

    I agree with you. The torque on the pulley is supplied by the tension, so
    [itex]\tau = T \times R_{pulley}[/itex].​
    A point along the circumference must have a linear acceleration equal to the acceleration of the block, so
    [itex]\alpha_{pulley} R_{pulley} = a_{block}[/itex].​
    And the moment of inertia for a solid cylinder is
    [itex]I=\frac12 MR^2[/itex]​
    So
    [itex]\tau = I\alpha = \left(\frac12 M R^2\right)\left(\frac a R\right) = T \times R[/itex].​
    If we solve for the tension, all the R's indeed drop out:
    [itex]T=\frac12 M a[/itex],​
    exactly like you found.
     
    Last edited: Jun 26, 2013
  4. Jun 26, 2013 #3
    Yup. I just did this problem again using energy conservation - if the acceleration is 1.6 m/s^2, then in 1 second the block gains 1.6m/s of speed and travels 0.8m while the pulley attains an angular speed of 32 rad/s. Energy conservation gives me the same answer for u so maybe that's why the radius was given, in case you wanted to solve it that way.

    Thanks.
     
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