The Spinning Figure Skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center . When his hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass 8.0 kg. When outstretched, they span 1.8 m; when wrapped, they form a cylinder of radius 25 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to .40 kg*m^2. If his original angular speed is 0.40 rev/s, what is his final angular speed?

Read about and apply the law of consewvation of angular momentum. After all of the help you have been getting here, I was hoping you would start showing some of your own work or thought process.

I or others will be happy to critique your work, once shown.

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What's the point of this website if all the "help" is pure crap?

To solve this problem, we can start by considering the law of conservation of angular momentum, which states that the initial angular momentum of a system is equal to the final angular momentum of that system, assuming no external torques act on it.

The angular momentum of an object is given by the product of its moment of inertia (I) and its angular velocity (ω), which can be expressed as L = Iω.

Let's start by finding the initial angular momentum (L_initial). The initial angular speed is given as 0.40 rev/s, which we can convert to radians per second by multiplying by 2π, since one revolution is equal to 2π radians.

ω_initial = 0.40 rev/s * 2π = 0.40 * 2π rad/s

Now, to find the moment of inertia (I) of the hands and arms, we can use the formula for the moment of inertia of a hollow cylinder:

I = (1/2) * M * R^2,

where M is the mass, and R is the radius of the cylinder. In this case, M is the combined mass of the hands and arms, given as 8.0 kg, and R is the radius of the cylinder, given as 25 cm or 0.25 m.

I = (1/2) * 8.0 kg * (0.25 m)^2 = 0.25 kg*m^2

Since the moment of inertia of the rest of the body is constant and equal to 0.40 kg*m^2, the total initial moment of inertia (I_initial) is the sum of the moment of inertia of the hands and arms plus the moment of inertia of the rest of the body:

I_initial = I + 0.40 kg*m^2 = 0.25 kg*m^2 + 0.40 kg*m^2 = 0.65 kg*m^2

Now, using the law of conservation of angular momentum, we can set the initial angular momentum equal to the final angular momentum and solve for the final angular velocity (ω_final):

L_initial = L_final

I_initial * ω_initial = I_final * ω_final

0.65 kg*m^2 * (0.40 * 2π rad/s) = I_final * ω_final

Simplifying this equation, we can solve for ω_final:

ω_final = (0.65 kg*m^2 * (0.40 * 2π rad/s)) / I_final

Now, we have all the information we need to substitute the values and calculate the final angular speed. Note that the mass and radius of the hands and arms are not needed for computing the final angular speed.

Please plug in the values into the equation to get the final angular speed (ω_final).