Respuesta :
Answer:
Explanation:
Given
mass of objects m is 2.8 kg
objects is [tex]r=1 m[/tex] away from axis of rotation
angular velocity [tex]\omega _1=0.75 rad/s[/tex]
moment of inertia of stool and student is[tex]=3 kg-m^2[/tex]
New distance of objects [tex]r'=0.34 m[/tex]
conserving Angular momentum
Initial Angular momentum [tex]L_1=I\omega _1[/tex]
[tex]I_1=3+2\times 2.8\times 1^2[/tex]
[tex]I_1=3+5.6=8.6 kg-m^2[/tex]
[tex]L_1=8.6\times 0.75=6.45[/tex]
For Second case
[tex]I_2=3+2\times 2.8\times 0.34^2=3.32 kg-m^2[/tex]
[tex]L_2=3.32\times \omega _2[/tex]
[tex]L_1=L_2[/tex]
[tex]6.45=3.32\times \omega _2[/tex]
[tex]\omega _2=1.94 rad/s[/tex]
