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 67.0 kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. If the skater is initially spinning at 76.0 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 is in rpm!

Answer 1

Mass of the skater, M = 67 kg

Mass of the hands and arms, m = 13% of total mass = 0.13 x M = 0.13 x 67 = 8.71 kg

Mass of the trunk, M₁ = M-m = 67 – 8.71 = 58.29 kg

Height of the skater, h = 1.80 m

Length of hands and arms, l = 70 cm = 0.7 m

Diameter of the trunk of the skater, d = 35 cm

Radius of the trunk, r = 35/2 = 17.5 cm = 0.175 m

Moment of inertia of the trunk when arms are stretched out

, I_t = M₁ r²/2 = 58.29 x 0.175 x 0.175/2 = 0.892 kgm²

Moment of inertia of the arms and hand, I_a = ml²/3 = 8.71 x 0.7 x 0.7/3 = 1.423 kgm²

Total moment of inertia of the skater with arms outstretched,

I₁ = I_t + I_a = 0.892 + 1.423 = 2.315 kgm²

Initial speed of the skater, ω_i = 76.0 rpm

Total moment of inertia of the skater with arms pulled and parallel to the trunk,

I₂ = Mr²/2 = 67 x 0.175 x 0.175/2 = 1.026 kgm²

Let ω_f be the final angular velocity. Using law of conservation of angular momentum,

I₁ω_i = I_fω_f

2.315 x 76 = 1.026 x ω_f

ω_f = 171.5 rpm

the final angular velocity of the skater, ω_f = 171.5 rpm