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PHYSICS ---HELP!

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A 3.6 m diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s. Its total moment of inertia is 1700 kg\m^2. Four people standing on the ground, each of mass 62 kg, suddenly step onto the edge of the merry-go-round.

1. What is the angular velocity of the merry-go-round now?

2. What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

  • PHYSICS ---HELP! - ,

    What do you think the answer is?

  • PHYSICS ---HELP! - ,

    I keep getting 0.68 ?????

  • PHYSICS ---HELP! - ,

    I get something close to that.

  • PHYSICS ---HELP! - ,

    Total kinetic energy
    = (1/2)Iω²

    ω1=0.78 rad/s
    I1=1700 kg-m²
    I2=1700+4*(62*1.8²)=2503.5
    (1/2)I1ω1²=(1/2)I2ω2²
    Solve for
    ω2
    =√(I1/I2)ω1
    =0.64 rad/s.
    Check my calculations.

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