A diver leaves the board with a moment of inertia of 12 kg m² and an angular velocity of 1.8 rad/s…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A diver leaves the board with a moment of inertia of $12\ \text{kg m}^2$ and an angular velocity of $1.8\ \text{rad/s}$ . In the air she tucks, and her moment of inertia drops to $3.0\ \text{kg m}^2$ .

What is her angular velocity in the tuck?
By what factor does her kinetic energy of rotation change?

Need a reminder? The key ideaShow

Angular momentum: $L = I\omega$ for a body on a fixed axis, and $\vec{L} = \vec{r} \times \vec{p}$ (size $mvr_\perp$ ) for a particle. With no external torque, $L$ is conserved: $I_1\omega_1 = I_2\omega_2$ . The kinetic energy $\dfrac{L^2}{2I}$ need not be.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

← All Problems

← A 0.25 kg ball on a string moves in a horizontal circle of radius 0.60 m at 1.5 revolutions per…A merry-go-round has a moment of inertia of 400 kg m² and turns freely at 1.0 rad/s… →

More onMechanics › Rotational Motion

Discussion (0)

Loading…

A diver leaves the board with a moment of inertia of 12 kg m² and an angular velocity of 1.8 rad/s…

The worked solution

Discussion (0)

Want to go further in Mechanics › Rotational Motion?