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A copper block rests 30.cm from the center of a steel turntable. The coefficient of s

  1. Dec 13, 2006 #1
    A copper block rests 30.cm from the center of a steel turntable. The coefficient of static friction between the block and the surface is .53. The turntable starts from rest and rotates with a constant angular acceleration of .50 rad/s^2. After what time interval will the block start to slip on the turntable.
    Hint: The normal force in this case equals the weight of the block.



    Fc=u(mew)timesN



    Tangential Speed = .15m/s
    mew =.53
    N = mg
    N = m(9.81 m/s^2)

    How do I find the mass?


    Please and Thank you...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 14, 2006 #2
    Hi chenny,

    This is a rotational motion problem. By Newton's 2nd law, Net force= ma. F- Ff = ma. Since tangential acceleration = radius x angular acceleration, F- Ff = m (0.3m x 0.5 rad/s²). When block starts to slip, Ff = (meu)N = 0.53mg. Manipulate Newton's 2nd law, u get F = ma = m dv/dt and solve for t. But u cannot find the mass of the block, thus u cannot find the time when block starts to slip on turntable.

    How to find the mass of the block?? Can someone pls help us??
     
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