Well, it seems like the skater is quite the physics expert, doing calculations while spinning! Let's see if I can help with the final angular speed.
First, let's find the moment of inertia of the skater's hands and arms when they are outstretched. We can consider them as a slender rod rotating about an axis through its center. The moment of inertia of a slender rod is given by I = (1/3) * M * L^2, where M is the mass and L is the length. Plugging in the values, we get:
I_outstretched = (1/3) * 8.50 kg * (1.80 m)^2
Now, when the skater wraps his arms, they form a thin-walled hollow cylinder. The moment of inertia of a thin-walled hollow cylinder is given by I = (1/2) * M * R^2, where M is the mass and R is the radius. Plugging in the values, we get:
I_wrapped = (1/2) * 8.50 kg * (0.25 m)^2
Now, let's calculate the total moment of inertia of the skater by adding the moment of inertia of the remainder of his body (0.450 kg路m^2) to the moment of inertia of his wrapped arms:
I_total = I_outstretched + I_wrapped + 0.450 kg路m^2
Now, we can use the conservation of angular momentum to find the final angular speed. The angular momentum is given by L = I * 蠅, where 蠅 is the angular speed. Since the moment of inertia is constant, the initial angular momentum is equal to the final angular momentum:
I_outstretched * 蠅_outstretched = I_total * 蠅_final
Now we can rearrange the equation to solve for 蠅_final:
蠅_final = (I_outstretched * 蠅_outstretched) / I_total
Plugging in the values, we can calculate the final angular speed. I'll leave the actual calculation to you, but remember to convert the angular speed from rev/s to rad/s. Good luck!