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 18kg?m2. She then tucks into a small ball, decreasing this moment of inertia to 3.6kg?m2. While tucked, she makes two complete revolutions in 1.0s .

If she hadn't tucked at all, how many revolutions would she have made in the 1.8s from board to water?

Answer 1

Concepts and reason

The concept to solve this problem is conservation of angular momentum.

Firstly, use conservation of angular momentum and find the expression of the initial angular velocity of the diver.

Then, substitute the values in the expression of angular velocity to find its value.

Finally, find the number of revolutions made by the diver by multiplying the angular velocity with total time.

Fundamentals

According to the conservation of angular momentum, if no external force acts on a system, then the angular momentum of the system remains same at all the times. The expression of conservation of angular momentum is as follows:

${I_1}{\omega _1} = {I_2}{\omega _2}$

Here, ${I_{\rm{1}}}$ is the initial moment of inertia of the system, ${I_2}$ is the final moment of inertia of the system, ${\omega _1}$ is the initial angular velocity of the system, and ${\omega _2}$ is the final angular velocity of the system.

(A)

The expression of the angular momentum of a system is as follows:

$L = I\omega$

Here, I is the moment of inertia of the system and $\omega$ is the angular velocity of the system.

The expression of conservation of angular momentum is as follows:

${I_1}{\omega _1} = {I_2}{\omega _2}$

Ange the above expression for ${\omega _1}$ .

${\omega _1} = \frac{{{I_2}{\omega _2}}}{{{I_1}}}$

The diver makes two revolutions per second. The angular velocity is measured in revolutions per second. The final angular velocity is 2 rev/s.

Substitute 2 rev/s for ${\omega _2}$ , $3.6{\rm{ kg}} \cdot {{\rm{m}}^2}$ for ${I_2}$ , and $18{\rm{ kg}} \cdot {{\rm{m}}^2}$ for ${I_1}$ in the above expression.

$\begin{array}{c}\\{\omega _1} = \frac{{\left( {3.6{\rm{ kg}} \cdot {{\rm{m}}^2}} \right)\left( {2{\rm{ rev/s}}} \right)}}{{\left( {18{\rm{ kg}} \cdot {{\rm{m}}^2}} \right)}}\\\\ = 0.4{\rm{ rev/s}}\\\end{array}$

The angular velocity when the diver has not tucked is the initial angular velocity. The angular velocity of the diver when she has not tucked is as follows:

$\omega = 0.4{\rm{ rev/s}}$

The number of revolutions that could have been made by the diver if her angular velocity was $\omega$ is as follows:

$N = \omega \left( t \right)$

Here, t is the time duration for which the diver had revolved.

Substitute 0.4 rev/s for $\omega$ , and 1.8 s for t in the above expression.

$\begin{array}{c}\\N = \left( {0.4{\rm{ rev/s}}} \right)\left( {1.8{\rm{ s}}} \right)\\\\ = 0.72{\rm{ revolutions}}\\\end{array}$

Ans:

The number of revolutions are 0.72 revolutions.