Answer :
Answer: [tex]2.29\ rad/s[/tex]
Explanation:
Given
Mass of Platform M=150 kg
the radius of Platform R=2 m
Moment of inertia of Platform [tex]I=300\ kg.m^2[/tex]
Mass of student [tex]m=60\ kg[/tex]
The initial angular speed of the student is [tex]\omega _1=1.4\ rad/s[/tex]
Initially, the moment of inertia of platform and student is
[tex]I_1=\dfrac{1}{2}MR^2+mR^2\\\\I_1=300+60\times 4=540\ kg.m^2[/tex]
The final moment of inertia
[tex]I_2=\dfrac{1}{2}MR^2+mr^2\\\\I_2=300+60\times 0.7^2=329.4\ kg.m^2[/tex]
Conserving angular momentum
[tex]I_1\omega_1=I_2\omega _2\\\\540\times 1.4=329.4\times \omega_2\\\\\omega_2=\dfrac{756}{329.4}=2.29\ rad/s[/tex]