A runner of mass 60.0kg runs around the edge of a horizontal turntable mounted o

Question

A runner of mass 60.0kg 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 2.50m/s . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.190rad/s relative to the earth. The radius of the turntable is 3.60m , and its moment of inertia about the axis of rotation is 81.0kg*m2 .

A)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.)
answer in rad/s please

I keep getting an answer of .07 but thats wrong pls help

Explanation / Answer

Angular speed of the turntable is 1 = -0.19 rad/s. Radius of the turntable is R = 3.6 m. Moment of inertia of the turntable is I = 81 kg m2. Mass of the runner is M = 60 kg. Magnitude of the runner's velocity relative to the earth is v1 = 2.5 m/s. ------------------------------------------------------------------------------------------------------------- From law of conservation of angular momentum     Mv1R + I 1 = (I + MR^2) 2 The final angular speed of the system is 2             2 = (Mv1R + I1 ) / (I+MR^2)                = (60 kg)(2.5 m/s)(3.6 m)  - (81 kgm2)(0.19 rad/s) / [ 81 kg m2 + (60 kg) (3.6 m)^2]                             =  (540 - 15.4) / 858.6                = 0.611 rad/s Angular speed of the turntable is 1 = -0.19 rad/s. Radius of the turntable is R = 3.6 m. Angular speed of the turntable is 1 = -0.19 rad/s. Radius of the turntable is R = 3.6 m. Moment of inertia of the turntable is I = 81 kg m2. Mass of the runner is M = 60 kg. Mass of the runner is M = 60 kg. Magnitude of the runner's velocity relative to the earth is v1 = 2.5 m/s. ------------------------------------------------------------------------------------------------------------- From law of conservation of angular momentum     Mv1R + I 1 = (I + MR^2) 2 The final angular speed of the system is 2             2 = (Mv1R + I1 ) / (I+MR^2)                = (60 kg)(2.5 m/s)(3.6 m)  - (81 kgm2)(0.19 rad/s) / [ 81 kg m2 + (60 kg) (3.6 m)^2]                             =  (540 - 15.4) / 858.6                = 0.611 rad/s

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