A figure skater rotates about a vertical axis through her center with both arms and one leg extended. Her rotation rate is 0.65 rev/s. She then pulls her arms and leg closer to her body and begins to spin at a 1.85 rev/s. What is the ratio of her new moment of inertia to her old?

Step by step explanation please. Thank you

To find the ratio of the figure skater's new moment of inertia to her old moment of inertia, we need to understand the concept of moment of inertia and how it changes when the mass is redistributed.

Step 1: Understand moment of inertia
Moment of inertia is a measure of an object's resistance to rotational motion around a given axis. It depends on the mass distribution of the object and the axis of rotation. For a point mass rotating about an axis, the moment of inertia is given by the formula:

I = m * r^2

Where:
I is the moment of inertia,
m is the mass of the object, and
r is the distance of the mass from the axis of rotation.

Step 2: Determine the initial moment of inertia
In the initial state, the figure skater has both arms and one leg extended. Assuming that her body can be approximated as a uniform distribution of mass, we can calculate the initial moment of inertia. Since the mass is evenly distributed, we can assume that the moment of inertia can be calculated around her center of mass.

Step 3: Calculate the initial moment of inertia
For simplicity, let's assume the figure skater's mass is m and the distance from her center of mass to her arms and leg is r1. Since her arms and one leg are extended, we can consider them as three extended rods.

The moment of inertia of a rod rotating about its center perpendicular to its length is given by the formula:
I_rod = (1/12) * m_rod * L_rod^2

In this case, we have three rods (one for each arm and one for the leg), so the total moment of inertia contributed by the extended parts is:
I_extended = 3 * ((1/12) * m_rod * L_rod^2)

Now, let's calculate the moment of inertia contributed by the rest of the body (trunk) assuming it can be approximated as a cylinder.

The moment of inertia of a cylinder rotating about its central axis is given by the formula:
I_cylinder = (1/12) * m_cylinder * h_cylinder^2 + (1/4) * m_cylinder * r_cylinder^2

Since the figure skater's body is symmetrical, the height (h_cylinder) and radius (r_cylinder) of the cylinder are equal. Let's call this common value r2.

So, the moment of inertia contributed by the trunk is:
I_trunk = (1/12) * m_cylinder * r2^2 + (1/4) * m_cylinder * r2^2

Finally, the initial moment of inertia is the sum of the moment of inertia contributed by the extended parts and the trunk:
I_initial = I_extended + I_trunk

Step 4: Determine the final moment of inertia
When the figure skater pulls her arms and leg closer to her body, the mass distribution changes. Assume that when the figure skater pulls her arms and leg closer, the mass originally located at r1 is now at a distance r3.

Since her mass is still evenly distributed, we can calculate the final moment of inertia using the same formulas as before, but with the new distances.

The moment of inertia contributed by the new extended parts is:
I_extended_new = 3 * ((1/12) * m_rod * L_rod^2)

The moment of inertia contributed by the new trunk is:
I_trunk_new = (1/12) * m_cylinder * r3^2 + (1/4) * m_cylinder * r3^2

Finally, the final moment of inertia is the sum of the moment of inertia contributed by the new extended parts and the new trunk:
I_final = I_extended_new + I_trunk_new

Step 5: Calculate the ratio of the new moment of inertia to the old
Now that we have both the initial and final moment of inertia, we can find the ratio:

Ratio = I_final / I_initial

Substituting the expressions calculated for the initial and final moment of inertia, we can determine the value of the ratio.