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

Step-by-step 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) = ((Mv_(1)R + I\omega_(1)))/((I + MR^(2)))`

=
`(54 kg * 3.10 m/s * 2.8 m - 85 kg m^(2) * 0.15 rad/s)/(85 kg m^(2) + (54 kg) * (2.80 m)^2)`

=
`(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