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leva [86]
3 years ago
13

Platform

Physics
1 answer:
Ivahew [28]3 years ago
5 0

a) See explanation

b) K_2>K_1

Explanation:

a)

The angular speed of an object in rotation is the rate of change of its angular displacement:

\omega=\frac{\Delta \theta}{\Delta t}

where

\Delta \theta is the angular displacement

\Delta t is the time elapsed

The angular momentum of an object in rotation is given by

L=I\omega

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 L must remain constant; and therefore, since I decreases, \omega (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 L must remain constant, the angular speed will decrease.

b)

The rotational kinetic energy of an object in rotational motion is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular speed

Using

L=I\omega

we can rewrite the rotational kinetic energy as:

K=\frac{1}{2}L\omega

In part a), we said that the angular momentum L remains constant, however the angular speed \omega 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

K_2>K_1

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