Question

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.80 kg⋅m2 . If she starts out spinning at 5.5 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

Answer 1

correct

Step-by-step explanation:

Answer 2

1.3 rev/s

Step-by-step explanation:

`I_(o)` = Moment of inertia when arms and one leg of the skater is out = 3.5 kgm²

`I_(i)` = Moment of inertia when arms and legs of the skater are in = 0.80 kgm²

`w_(i)` = Angular speed of skater when arms and legs of the skater are in = 5.5 rev/s

`w_(o)` = Angular speed of skater when arms and legs of the skater are out = ?

Using conservation of angular momentum

`I_(i)`
`w_(i)` =
`I_(o)`
`w_(o)`

(0.80) (5.5) = (3.5)
`w_(o)`

`w_(o)` = 1.3 rev/s