Answer:
0.336 rad/s
Explanation:
[tex]\omega_1[/tex] = Angular speed of the turntable = -0.2 rad/s
R = Radius of turntable = 2.9 m
I = Moment of inertia of turntable = [tex]76\ kgm^2[/tex]
M = Mass of turn table = 53 kg
[tex]v_1[/tex] = Magnitude of the runner's velocity relative to the earth = 3.6 m/s
As the momentum in the system is conserved we have
[tex]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[/tex]
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
