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 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.

Answer 1

Answer:

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:

1. The spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.
2. friction between the skater and the ice is negligible.

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

1. Each rod has moment of inertia = $\frac{1}{3} mL^{2}$

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 =  $\frac{1}{3}$ x 4.81 x $0.7^{2}$ = 0.79 kg-m   for each arm

2. Her body as a cylinder has moment of inertia =  $\frac{1}{2} mr^{2}$

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 = $\frac{1}{2}$ x 64.38 x $0.175^{2}$ = 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 = $\frac{2\pi }{60}$ 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 ÷ $\frac{2\pi }{60}$  = 176.38 rpm