Divers change their body position in midair while rotating about their center of mass. In one dive, the diver leaves the board with her body nearly straight, then tucks into a somersault position. If the moment of inertia of the diver in a straight position is 14kg⋅m2 and in a tucked position is 4.0kg⋅m2, by what factor is her angular velocity when tucked greater than when straight?

Her angular velocity when tucked is greater than when straight by a factor of 0.23

Explanation:

Moment of inertia (I) = mr^2 = mv^2/w^2

m is mass of the diver

v is diver's linear velocity

w is her angular velocity

When straight, I = 14 kg.m^2

mv^2/w^2 = 14

w^2 = mv^2/14

w = sqrt(mv^2/14) = 0.27sqrt(mv^2)

When tucked, I = 4 kg.m^2

w^2 = mv^2/4

w = sqrt(mv^2/4) = 0.5sqrt(mv^2)

Her angular velocity when tucked is greater than when straight by 0.23 (0.5 - 0.27 = 0.23)

prolly a beacuse the momentuem the car has increases while going down hill