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.)

Question

Physics

Mayra Wiley

19 January, 05:22

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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Answers (1)

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Caitlyn Booker · Answer 1

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.