Question

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs account for 80%. We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 60.0 kg skater is 1.60 m tall, has arms that are each 74.0 cm long (including the hands), and a trunk that can be modeled as being 37.0 cm in diameter. If the skater is initially spinning at 72.0 rpm with her arms outstretched, what will her angular velocity ?2 be (in rpm ) after she pulls in her arms and they are at her sides parallel to her trunk? Assume that friction between the skater and the ice is negligble.

Answer 1

Mass of the trunk = 0.8 x 60 = 48 kg

Mass of each arm = (0.13/2) x 60 = 3.9 kg

So, mass of the head = 60 - 48 - 2(3.9) = 4.2 kg

Moment of Inertia of a solid cylinder about its cylindrical axis is: I = (1/2)mr²

Moment of inertia of a solid cylinder rotating about one of its ends is: I = (1/3)mL²

Moment of Inertia when the hands are stretched is:

(2/3)mL² + (1/2)(M+m')R² = 2.317 kgm²

Moment of Inertia when the hands are parallel to trunk is:

(1/2)(M+m+m')R² = 1.027 kgm²

Now now conservation of Angular Momentum

This gives the Angular velocity when the hands are parallel to the trunk to be:

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