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A 1.4-m-long massless rod is pivoted at one end and swings around in a circle on a frictionless table. A block with a hole through the center can slide in and out along the rod. Initially, a small piece of wax holds the block 20 from the pivot. The block is spun at 50 , then the temperature of the rod is slowly increased. When the wax melts, the block slides out to the end of the rod. What is the final angular velocity?

  • physics - ,

    The moment of inertia (I) for the block (the point mass, as we are not given its dimensions) is mr^2:
    I1 =m•r1^2 =m •0.2^2 =0.04•m,
    I2 = m•r1^2 =m •1.4^2=1.96•m,
    Acording to the law of conservation of angular momentum
    L1 = L2
    I1•ω1 = I2•ω2,
    ω2 = I1•ω1/I2 =0.04•m•50/1.96•m =
    =1.02 (units?)

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