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Vertical Acceleration

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A solid uniform disk unrolls from a string which is wrapped around it. If tension and weight are the only important forces, find the vertical acceleration of the disk in terms of gravity.

    2. Relevant equations

    3. The attempt at a solution

    mg - T = ma

    a = mg - T /m

    I know it's not this simple, but I don't know what else to do.
  2. jcsd
  3. Nov 11, 2008 #2


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    HINT: The tension creates a torque on the disc.
  4. Nov 11, 2008 #3
    So far I have α = 2 * Tension force / mass*radius. I don't understand if the answer is supposed to be in translational or rotational, and I don't know how to get the tension force if the disk is accelerating at an unknown rate.
  5. Nov 11, 2008 #4
    the answer is supposed to be translational, since it does not say angular acc.
    also, if you consider the axis of rotation through the centre of mass, and equate torque to
    I*[angular acc], using a kinematical constraint, you can gat another equation containing tension. Hope i helped....
  6. Nov 11, 2008 #5
    idid equate torque to I (angular)

    torque tension = I / alpha

    rearranged to find alpha

    alpha = tension * 2 / m*r

    I don't know what you mean by a kinematical constraint. Could you please tell me how I can find the tension to solve this equation?
  7. Nov 11, 2008 #6
    well, a kinematical constraint is simply a relation between two kinematical variable, which remains true irrespective of their value. in this particular case,
    [linear acc] = [radius of disk]* [angular acceleration]
  8. Nov 11, 2008 #7
    Oh! I've never seen that. But after plugging 2f/mr into that, I end up with acc = 2*f / m. How can I use that to answer the question " vertical acceleration in terms of g?"
  9. Nov 11, 2008 #8
    since you have solved the problem, i guess i will give the solution:

    mg - T = ma.......( i )

    T*r = [0.5*mr^2][alpha].......( ii )

    using the kinematical constraint,

    a = r*[alpha]......( iii )

    so you have three equations three unknowns, just don't substitute the value of g here, to get the answer in terms of g :)
  10. Nov 11, 2008 #9
    I still dont understand...

    I just keep manipulating this in circles. I have a = 2 (g -a). Why did I just find angular velocity?
  11. Nov 11, 2008 #10
    angular velocity, energy conservation need not be applied here. you are asked to find out the acceleration, so i say stick to force dynamics.....:wink:
  12. Nov 11, 2008 #11
    OK. I guess I haven't understood what you've been saying. Starting over, with just force dynamics, I found
    mg - T = ma
    a = mg - T /m
    This was my first post.
  13. Nov 11, 2008 #12


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    Just because we're using force and dynamics doesn't mean that we can ignore rotational motion. Let's return to these three equations:
    Since we don't want the solution in terms of the angular acceleration, the first thing to do is eliminate [alpha]. Can you do this?

    Use the second and third equations
  14. Nov 11, 2008 #13
    T*r = 1/2 m*r^2 (a/r)

    T = 1/2 m a
    a = 2T/m
  15. Nov 11, 2008 #14


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    Good :approve:.

    So now can you use this to eliminate T from equation (i) and then solve for a?
  16. Nov 11, 2008 #15
    m*g - 1/2 ma = ma
    g - a/2 = 2a
    2g = a
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