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Rotation of a turntable

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2.05 kg, 33.64 cm diameter turntable rotates at 365 rpm on frictionless bearings. Two 280 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick. What is the turntable's angular velocity, in rpm, just after this event?

    2. Relevant equations

    I= m(r^2)/2

    3. The attempt at a solution

    I changed the 365 rpm to 38.22 rad/s
    Then i found that the Ki=mt(r^2)(wi^2)/2
    Then i found that the Kf=(mt+mb1+mb2)(r^2)(wi^2)/2)

    Then i solved for wf and got the equation mt*wi^2/(the mass of the turntable and blocks). Then the square root of that and got 33.87 rad/s

    Then turned 33.87 rad/s to rpm and got 323.46.....Did i use the wrong I maybe? I used the I from above. and combined it with the Kinetic energy.
  2. jcsd
  3. Jun 1, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You cannot assume that kinetic energy is conserved. But something else is.
  4. Jun 5, 2009 #3
    momentuem.....ahhh i see. stupid mistake on my part.
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