Question

A diver comes off a board with arms straight up and legs straight down, giving her a moment of

Answer 1

Answer:

0.19 rev

Explanation:

We can solve the problem by using the law of conservation of angular momentum. In fact, if we assume there are no external torques acting on the diver, the angular momentum must be conserved:

$L=I\omega$

where

L is the angular momentum

I is the moment of inertia

$\omega$ is the angular velocity

- When the diver is tucked,

$I = 3.6 kg m^2$ is her moment of inertia

She makes 2 revolutions (so, $4 \pi rad$) in t = 1.0 s, so her angular velocity is

$\omega=\frac{4\pi}{1.0 s}=4.0 rad/s$

So her angular momentum is

$L=(3.6 kg m^2)(4.0 rad/s)=14.4 kg m^2 /s$

- When the diver is not tucked,

The angular momentum is conserved, so $L=14.4 kg m^2 /s$

the moment of inertia is $I=18 kg m^2$

So the angular velocity is

$\omega=\frac{L}{I}=\frac{14.4 kg m^2/s}{18 kg m^2}=0.8 rad/s$

So in a time of t = 1.5 s, the angular displacement is

$\theta=\omega t=(0.8 rad/s)(1.5 s)=1.2 rad$

Converting into revolutions,

$2 \pi rad : 1 rev = 1.2 rad : x$

$x=\frac{(1 rev)(1.2 rad)}{2\pi rad}=0.19 rev$