Question

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of $\mathbf{18}{\mathbf{kgm}}^{\mathbf{2}}$. She then tucks into a small ball, decreasing this moment of inertia to $\mathbf{3}\mathbf{.}\mathbf{6}{\mathbf{kgm}}^{\mathbf{2}}$. While tucked, she makes two complete revolutions in 1.0 s. If she hadn’t tucked at all, how many revolutions would she have made in the 1.5 s from board to water?

Short answer

Hence, she made revolution in 1.5 s from board to water is $0.60\mathrm{rev}$.

Step-by-step solution

Conservation of angular momentum.

If a system experience no external torques from the environment, the total angular momentum of the system is conserved.

$\Delta \overrightarrow{L\to t}=0$

Apply the law of conservation of angular momentum to a system whose moment of inertia changes gives:

$I_i{\omega }_i=I_f{\omega }_f=\mathrm{constant}$

She made revolution from board to water.

Apply conservation of angular momentum to the diver.

The number of revolutions she makes in a certain time is proportional to her angular velocity.

The ratio of her untucked to tucked angular velocity is:

$\frac{3.6}{18}=0.2$

If she had tucked, then she made revolution in the last $1.0\mathrm{s}$as:

$2\frac{3.6}{18}=0.4\mathrm{rev}$

In the last $1.0\mathrm{s}$, so the revolution made in the total $1.5\mathrm{s}$is:

$0.4\frac{1.5}{1}=0.60\mathrm{rev}$

Hence, she made revolution in $1.5\mathrm{s}$is $0.60\mathrm{reV}$.