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4 years ago

11

Divers change their body position in midair while rotating about their center of mass. In one dive, the diver leaves the board w

ith 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?

2 answers:

shepuryov [24]4 years ago

7 0

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.

OlgaM077 [116]4 years ago

5 0

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)

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