a) See explanation
b) [tex]K_2>K_1[/tex]
Explanation:
a)
The angular speed of an object in rotation is the rate of change of its angular displacement:
[tex]\omega=\frac{\Delta \theta}{\Delta t}[/tex]
where
[tex]\Delta \theta[/tex] is the angular displacement
[tex]\Delta t[/tex] is the time elapsed
The angular momentum of an object in rotation is given by
[tex]L=I\omega[/tex]
where I is the moment of inertia of the body.
The moment of inertia of the athlete decreases as we move from figure 1 to figure 2: this is because the athlete pulls his arms and legs towards the body. Since the athlete is an isolated system, the angular momentum [tex]L[/tex] must remain constant; and therefore, since [tex]I[/tex] decreases, [tex]\omega[/tex] (angular speed) must increase.
On the other hand, when we move from figure 2 to figure 3 the moment of inertia of the athlete increases again, and therefore, since [tex]L[/tex] must remain constant, the angular speed will decrease.
b)
The rotational kinetic energy of an object in rotational motion is given by
[tex]K=\frac{1}{2}I\omega^2[/tex]
where
I is the moment of inertia
[tex]\omega[/tex] is the angular speed
Using
[tex]L=I\omega[/tex]
we can rewrite the rotational kinetic energy as:
[tex]K=\frac{1}{2}L\omega[/tex]
In part a), we said that the angular momentum L remains constant, however the angular speed [tex]\omega[/tex] increases as we move from figure 1 to figure 2. Since the rotational kinetic energy is proportional to both the angular momentum and the angular speed, but the angular momentum remains constant, this means that the rotational kinetic energy also increases as we move from figure 1 to figure 2.
So the answer is
[tex]K_2>K_1[/tex]
