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Blocks on a Turntable

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Two identical blocks are tied together with a string and placed along the same radius of a turntable that is spinning about its center. The inner block is 4 cm from the center and the outer block is 5 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.71, and the string is taut.
    What is the maximum angular frequency such that neither block slides?

    2. Relevant equations
    a_c = (V^2)/R
    w = 2pi/T
    a_c = (w^2)R
    F = ma


    3. The attempt at a solution
    Sum of force on inner block (known to be zero0
    0 = µmg - T + m(w^2)R1
    Sum of force on outer block (also zero)
    0 = µmg + T + m(w^2)R2

    I then took the two tensions to be equal, solved for one, then substituted in the other an solved for w, the angular frequency.
    -T = µmg + m(w^2)R2
    inserted
    0 = µmg +(µmg + m(w^2)R2) + m(w^2)R1
    masses cancel
    0 = 2µg + (w^2)(R2+R1)
    w = 1.24

    Where did I go wrong?
     
  2. jcsd
  3. Feb 8, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Check your math again. Remember your units are cm = .01 m

    Btw, you missed a sign along the way.

    Another way to think about it is that the outward acceleration needs to overcome the frictional maximum resistance.

    That implies directly that 2m*μ *g = ω²(r1 + r2)
     
  4. Feb 8, 2009 #3
    Yup, it was my units. Not sure where I dropped the sign, but I got it to work out. Thank you.
     
  5. Feb 8, 2009 #4
    So, continuing this problem, they give the mass of the blocks to be 32g (.032kg) and ask for the tension of the string connecting the blocks. I thought that I could take either of my equations, insert everything that I now have (w=12.43 rad/s) and just solve for tension. But that didn't work. It seems it should, but I must be missing something.
     
  6. Feb 9, 2009 #5

    LowlyPion

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    Homework Helper

    Look at the inner block in isolation.

    You have the maximum frictional resistance μ*m*g and that is being offset by 2 forces
    1) centrifugal acceleration m*ω²*r1 and
    2) the Tension necessary to make up the difference.
     
  7. Feb 9, 2009 #6
    That did it. I had assumed that the wrong direction for the centripital force to act. That was also what confused my signs in the first part of the problem. Thanks again.
     
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