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 63.0 kg skater is 1.60 m tall, has arms that are each 64.0 cm long (including the hands), and a trunk that can be modeled as being 32.0 cm in diameter. If the skater is initially spinning at 85.0 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

I = Itrunk + I hand

Trunk

m = 87% = 0.87x63 = 54.81 kg

r = 16cm = 0.16m

Itrunk = mr^2/ 2 = 0.7 kgm^2 .... cylinder

Hands

m = 13% = 0.13 x 63 = 8.2 kg

L = 64cm = 0.64 m

I hand = ML^2 /3 + Mr^2 ... MI of rod about parallel axis

= 8.2 x 0.64^2 / 3 + 8.2 x 0.16^2= 1.33kgm^2

I = 2.03 kgm^2

L1 = I1 w1

= 2.03 x 85 .... i

When the arms are brought in, almost all mass moves along trunk as a cylinder

I = 63 x 0.16^2 / 2 = 0.806 kgm^2

L2 = 0.806 w2 ...ii

By law of conservation of angular momentum

I1w1 = I2w2

2.03 x85 = 0.806 x w2

w2 = 214 rpm

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