Answer:
Her angular speed (in rev/s) when her arms and one leg open outward is [tex]1.56\frac{rev}{s}[/tex]
Explanation:
Initial moment of inertia when arms and legs in is [tex]I_i=0.90 kg.m^{2}[/tex]
Final moment of inertia when her arms and on leg open outward, [tex]I_f=3.0 kg.m^{2}[/tex]
Initial angular speed [tex]w_i=5.2\frac{rev}{s}[/tex]
Let the final angular speed be [tex]w_f[/tex]
Since external torque on her is zero so we can apply conservation of angular momentum
[tex]\therefore L_f=L_i[/tex]
=>[tex]I_fw_f=I_iw_i[/tex]
=>[tex]w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}[/tex]
Thus her angular speed (in rev/s) when her arms and one leg open outward is [tex]1.56\frac{rev}{s}[/tex]
