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Rotational / Linear kinematic problem

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img691.imageshack.us/img691/2702/drawing1.png [Broken]

    A 1kg block is suspended from a rope that wraps around a pulley where it is pulled by a 10N force. The pulley is a disk with mass 50g and has a diameter of 4cm. Assume enough friction that the rope does not slip but neglect friction in the bearing.

    2. Relevant equations
    This is actually one of the questions. I am figuring the three equations would be:

    [tex]\tau=I\alpha[/tex]

    [tex]F_{net}=10N - m_{b}g[/tex]

    [tex]a=\alphar[/tex]

    3. The attempt at a solution
    Calculate the acceleration of the block using your equations above.

    Well, I tried to start with the torque equation, which became:

    [tex]\frac{m_{p}r^{2}a}{2r}[/tex]

    [tex]\frac{m_{p}Fr}{2m}[/tex]

    Although, I have no idea what the "m" is, which came about through a=f/m . I am completely stumped at this point and making no progress. Any suggestions?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 8, 2009 #2

    Pythagorean

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    Gold Member

    1) Draw a free body diagram for the pulley that shows the torques on it, and the block with the forces on it. The torques will be dependent on the force and the radius at which the force is applied. There will be two torques on the pulley.

    2) Chose a direction to be positive and make equations out of your free-body diagram (including the some of the torques for the pulley and the sum of the forces for the block). You basically need to figure out the angular acceleration of the pulley (via sum of the torques) to see how fast it pulls the block up. Because it has mass, it will resist accelerations through it's moment of inertia, I.
     
  4. Nov 8, 2009 #3
    I had already drawn the FBD's. They really didn't help, as my issue was a result of something else.

    In my algebraic substitutions, I tried to get rid of angular acceleration as quickly as possible, when in reality that's what I should have solved for. Solving for [tex]\alpha[/tex] first is much much easier.

    Here is what I have

    [tex]\alpha = \frac{\tau}{I} = \frac{10r - m_{b}gr}{.5m_{p}r^{2}} = \frac{10-m_{b}g}{.5m_{p}r} = 400 \frac{rad}{s^{2}}[/tex]

    [tex]a = alpha * r = 8 \frac{m}{s^{2}}[/tex]

    This seems a little too quick of acceleration though... is this correct?
     
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