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 70.0 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 77.0 rpm with her arms outstretched, what will her angular velocity ω 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.

please show step by step how to solve. I keep getting 249. Thank you.

Answer 1

let
M = 70 kg
mass of hands, m_hands = 0.13*M

= 0.13*70

= 9.1 kg

mass of the remaining paarts, m = 0.87*M

= 0.87*70

= 60.9 kg

diameter, d = 35 cm = 0.35 m
L = 70 cm = 0.7 m

w1 = 77 rpm
initial moment of inertia, I1 = I_hands + I_trunk

= m_hands*L^2/3 + 0.5*m*(d/2)^2

= 9.1*0.7^2/3 + 0.5*60.9*(0.35/2)^2

= 2.419 kg.m^2

final moment of inertia, I2 = I_hands + I_trunk

= m_hands*(d/2)^2 + 0.5*m*(d/2)^2

= 9.1*(0.35/2)^2 + 0.5*60.9*(0.35/2)^2

= 1.211 kg.m^2

w2 = ?

Apply conservation of angular momentum

I2*w2 =I1*w1

w2 = I1*w1/I2

= 2.419*77/1.211

= 154 rpm <<<<<<<<<-----------------Answer