Question

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?

Answer 1

Her angular speed (in rev/s) when her arms and one leg open outward is
`1.56(rev)/(s)`

Step-by-step 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(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=(I_iw_i)/(I_f)=(0.9*5.2 )/(3.0)(rev)/(s)=1.56(rev)/(s)`

Thus her angular speed (in rev/s) when her arms and one leg open outward is
`1.56(rev)/(s)`