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 74.0 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 68.0 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.

176.38 rpm

Explanation:

mass percentage of arms and legs = 13%

mass percentage of legs and trunk = 80%

mass percentage of head = 7%

Total mass of the skater = 74.0 kg

length of arms = 70 cm = 0.7 m

height of skater = 1.8 m

diameter of trunk = 35 cm = 0.35 m

Initial angular momentum = 68 rpm

We assume:

We split her body into two systems, the spinning hands as spinning rods

1. Each rod has moment of inertia =

mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg

mass of each side will be assumed to be 9.62/2 = 4.81 kg

L = length of each arm

therefore,

I =   x 4.81 x = 0.79 kg-m   for each arm

2. Her body as a cylinder has moment of inertia =  

r = radius of her body = diameter/2 = 0.35/2 = 0.175 m

mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg

I = x 64.38 x = 0.99 kg-m

We consider each case

case 1: Body spinning with arm outstretched

Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk

I = (0.79 x 2) +  0.99 = 2.57 kg-m

angular momentum = Iω

where ω = angular speed = 68.0 rpm = x 68 = 7.12 rad/s

angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s

case 2: Arms pulled down parallel to trunk

The momentum of inertia will be due to her body trunk alone which is 0.91 kg-m

angular momentum =  Iω

=  0.99 x ω = 0.91ω

according to conservation of angular momentum, both angular momentum must be equal, therefore,

18.29 = 0.99ω

ω = 18.29/0.99 = 18.47 rad/s

18.47 ÷  = 176.38 rpm

be subject to a force andchange its motion