Question

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

Answer 1

Answer:

0.336 rad/s

Explanation:

$\omega_1$ = Angular speed of the turntable = -0.2 rad/s

R = Radius of turntable = 2.9 m

I = Moment of inertia of turntable = $76\ kgm^2$

M = Mass of turn table = 53 kg

$v_1$ = Magnitude of the runner's velocity relative to the earth  = 3.6 m/s

As the momentum in the system is conserved we have

$Mv_1R+I\omega_1=(I + MR^2)\omega_2\\\Rightarrow \omega_2=\dfrac{Mv_1R+I\omega_1}{I + MR^2}\\\Rightarrow \omega_2=\dfrac{53\times 3.6-76\times 0.2}{76+53\times 2.9^2}\\\Rightarrow \omega_2=0.336\ rad/s$

The angular velocity of the system if the runner comes to rest relative to the turntable which is the required answer is 0.336 rad/s