Solving a Figure Skater's Moment of Inertia

[Tina20]

Homework Statement

A 45.8 kg figure skater is spinning on the toes of her skates at 1.18 rpm. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (41.3 kg, 18.6 cm average diameter, 167.0 cm tall) plus two rod-like arms (2.24 kg each, 70.1 cm long) that each rotate about an axis through the inner end of the rod. The skater then raises her arms straight above her head, where she appears to be a 45.8 kg, 18.6 cm diameter, 199.0 cm tall cylinder. What is her new rotation frequency, in radians per second?

Homework Equations

I have no idea how to go about this question.

Some potentially useful equations:

Iiwi = Ifwf
where i = initial, f = final
I = moment of interia
w = angular velocity

I for a rod about end = 1/3ML^2
I for cylinder or disk about center = 1/2MR^2

The Attempt at a Solution

Iiwi = Ifwf
(Iarm + Iarm + Ibody)wi = (Ibody)wfinal
(2Iarm + Ibody)wi = (Ibody)wfinal
(2(1/3ML^2) + 1/2MR^2)wi = (1/2MR^2)wf

So I tried the above equation, but my answer is coming out wrong :( Any help please!

 

[anathema]

Thank you for your question. To solve this problem, we need to use the conservation of angular momentum equation, which states that the initial angular momentum of a system is equal to the final angular momentum of the system. In this case, the system consists of the figure skater and her rotating arms.

Let's start by defining the variables we will need:

m1 = mass of the figure skater (45.8 kg)
m2 = mass of each arm (2.24 kg)
L1 = length of each arm (70.1 cm = 0.701 m)
L2 = height of the figure skater (167.0 cm = 1.67 m)
wi = initial angular velocity (1.18 rpm = 0.123 rad/s)
wf = final angular velocity (what we are trying to find)

We can then write the conservation of angular momentum equation as:

Iiwi = Ifwf

where Ii is the initial moment of inertia and If is the final moment of inertia. We need to calculate these values for both orientations of the figure skater.

For the initial orientation, the figure skater can be modeled as a cylindrical torso (41.3 kg, 18.6 cm diameter, 167.0 cm height) plus two rod-like arms. The moment of inertia for a cylinder about its center is given by:

Icylinder = 1/2MR^2

where M is the mass of the cylinder and R is the radius. Plugging in the values, we get:

Icylinder,initial = 1/2(41.3 kg)(0.093 m)^2 = 0.181 kgm^2

The moment of inertia for a rod about its end is given by:

Irod = 1/3ML^2

where M is the mass of the rod and L is the length. Plugging in the values, we get:

Irod,initial = 1/3(2.24 kg)(0.701 m)^2 = 0.329 kgm^2

Now we can calculate the total initial moment of inertia using the parallel axis theorem, which states that the moment of inertia of a body about a point is equal to the moment of inertia of the body about a parallel axis through the center of mass plus the product of the mass and the square of the distance between the two axes.

Iinitial = Icylinder