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 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.

Answer 1

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

Answer 2

The factor by which her angular velocity when tucked greater than when straight is; 3.5

What is the Angular Velocity?

We are given;

Straight moment of inertia; I₁ = 14 kgm²

Tucked moment of inertia; I₂ = 4 kgm²

Straight angular velocity; ω₁

Tucked angular velocity; ω₂

In the given system of divers change in position we see that the angular momentum is conserved. Thus;

I₁*ω₁ = I₂*ω₂

where;

ω₁ is straight angular velocity

ω₂ is tucked angular velocity

Thus;

14 * ω₁ = 4 * ω₂

ω₂ = (14/4)ω₁

ω₂ = 3.5ω₁

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