Question

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?

Answer 1

Answer:

tucked angular velocity is 3.5times greater than straight angular veocity.

EXPLANATION:

Assumptions:

✓Any external force acting on the driver is neglected.(both air. Resistance and gravity)

✓based on this the angular momentum of the driver when she's straight and when she's tucks is constant.

I(straight) ×ω(straight)= I(tucked)× ω(tucked)

Is×ωs=It× ωt

Is×ωs/It=ωt

ωt=ls×ωs/lt=ls×ωs /lt

ωt=14/4ωs

ωt=3.5ωs

From above calculation, tucked angular velocity is 3.5times greater than straight angular velocity.

Answer 2

Answer:

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)