Marble and cotton ball friction assignment
1 answer:
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Answer:
<em>176.38 rpm</em>
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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
<em>We assume:</em>
- <em>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.</em>
- <em>friction between the skater and the ice is negligible.</em>
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
<em>Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk</em>
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
<em>The momentum of inertia will be due to her body trunk alone</em> which is 0.91 kg-m
angular momentum = Iω
= 0.99 x ω = 0.91ω
<em>according to conservation of angular momentum, both angular momentum must be equal</em>, therefore,
18.29 = 0.99ω
ω = 18.29/0.99 = 18.47 rad/s
18.47 ÷ = <em>176.38 rpm</em>