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Roman55 [17]
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 14 kgâ‹…m2 and in a tucked position is 4.0 kgâ‹…m2 , by what factor is her angular velocity when tucked greater than when straight?Express your answer using two significant figures.
Physics
1 answer:
Anettt [7]4 years ago
8 0

Answer:

3.5

Explanation:

I_1 = Straight moment of inertia = 14 kgm²

I_2 = Tucked moment of inertia = 4 kgm²

\omega_1 = Straight angular velocity

\omega_2 = Tucked angular velocity

In this system the angular momentum is conserved

I_1\omega_1=I_2\omega_2\\\Rightarrow \omega_2=\dfrac{I_1\omega_1}{I_2}\\\Rightarrow \omega_2=\dfrac{14\omega_1}{4}\\\Rightarrow \omega_2=3.5\omega_1

The factor is 3.5

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