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

Answer 1

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