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jeka94
3 years ago
13

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.0 kg⋅m2 and for arms and legs in is 0.90

kg⋅m2 . If she starts out spinning at 5.2 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?
Physics
1 answer:
babymother [125]3 years ago
7 0

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is 1.56\frac{rev}{s}

Explanation:

Initial moment of inertia when arms and legs in is I_i=0.90 kg.m^{2}

Final moment of inertia when her arms and on leg open outward, I_f=3.0 kg.m^{2}

Initial angular speed w_i=5.2\frac{rev}{s}

Let the final angular speed be w_f

Since external torque on her is zero so we can apply conservation of angular momentum

\therefore L_f=L_i

=>I_fw_f=I_iw_i

=>w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}

Thus her angular speed (in rev/s) when her arms and one leg open outward is 1.56\frac{rev}{s}

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