A runner of mass 54.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The run

Question

A runner of mass 54.0 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 3.10 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.150 rad/s relative to the earth. The radius of the turntable is 2.80 m, and its moment of inertia about the axis of rotation is 85.0 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.)answer in rad/s please

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Ben Gia 4 years 2021-08-21T03:49:25+00:00 1 Answers 153 views 0

Answers ( )

  1. huyenthanh
    0
    2021-08-21T03:50:32+00:00
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    Explanation:

    Angular speed of the turntable is = -0.150 rad/s.

    Radius of the turntable is R = 2.80 m  

    Moment of inertia of the turntable is I = 85 kg m^{2}

    Mass of the runner is M = 54 kg

    Magnitude of the runner’s velocity relative to the earth is = 3.10 m/s.

    Now, according to the law of conservation of angular momentum

          Mv_{1}R + I \omega_{1} = (I + MR^{2}) \omega_{2}

    And, the final angular speed of the system is \omega_{2}

       \omega_{2} = \frac{(Mv_{1}R + I\omega_{1})}{(I + MR^{2})}

             = \frac{54 kg \times 3.10 m/s \times 2.8 m - 85 kg m^{2} \times 0.15 rad/s}{85 kg m^{2} + (54 kg) \times (2.80 m)^2}            

             = \frac{468.72 - 12.75}{508.36}

             = 0.896 rad/s

    Thus, we can conclude that  the final angular velocity of the system if the runner comes to rest relative to the turntable is 0.896 rad/s.

             = 0.611 rad/s

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