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

Answer:

1.3  rev/s

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