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10 January, 16:06

A runner of mass 53.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 3.60 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.200 rad/s relative to the earth. The radius of the turntable is 2.90 m, and its moment of inertia about the axis of rotation is 76.0 kg⋅m2.

Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can treat the runner as a particle.)

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  1. 10 January, 16:34
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    moment of inertia of man

    = mr², m is mass of man and r is radius of table.

    = 53 x 2.9²

    =445.73

    angular momentum of runner = mvr, v is velocity of runner.

    53 x 3.6 x 2.9 = 553.32

    angular momentum of turntable

    = Iω, I is moment of inertia and ω is angular velocity of table.

    = 76 x. 2 = 15.2

    Total angular momentum = 553.32 - 15.2

    = 538.12

    Let the common velocity when the runner comes to rest with respect to turntable be ω.

    total moment of inertia of the system

    = 445.73 + 76

    = 521.73

    Applying law of conservation of angular momentum

    total initial angular momentum = final angular momentum

    538.12 = 521.73 ω

    ω = 1.03 rad / s.
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