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 1.50 m tall, has arms that are each 66.0 cm long (including the hands), and a trunk that can be modeled as being 33.0 cm in diameter. If the skater is initially spinning at 69.0 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.

Answer 1

Answer:

Her angular velocity is 439.41 rpm

Explanation:

The moment of inertia when her arms are at her side is

$I_{f} =\frac{mr^{2} }{2}$

Where m=70 kg, r=33/2=0.165 m

The moment of inertia when her arms are stretched out is

$I_{i} =\frac{m_{b}r^{2} }{2} +2(\frac{m_{h}L^{2} }{12}+m_{h} (\frac{L}{2}+r)^{2} )$

Where L=66 cm=0.66 m

mb= mass of body except hands=0.87*70=60.9 kg

mh= mass of hands+arms=0.13*70=9.1 kg

The conservation of angular momentum is:

Li=Lf

Ii*wi=If*wf

Clearing wf:

$w_{f} =\frac{I_{i}w_{i} }{I_{f} } =\frac{w_{i}(\frac{m_{b}r^{2} }{2}+2(\frac{m_{h}L^{2} }{12}+m_{h}(\frac{L}{2}+r)^{2})) }{\frac{mr^{2} }{2} }$

Replacing values:

$w_{f} =\frac{69*(\frac{60.9*0.165^{2} }{2}+2(\frac{9.1*0.66^{2} }{12}+9.1(\frac{0.66}{2}+0.165)^{2} }{\frac{70*0.165^{2} }{2} } =439.41rpm$