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 accou
nt 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.
1 answer:
Answer:
176.38 rpm
Explanation:
The proportion of mass for arms and legs is 13%.
For legs and trunk, it's 80% of the total mass.
Additionally, the head accounts for 7% of the total mass.
Overall, the mass of the skater is 74.0 kg.
Each arm length is 70 cm, or 0.7 m.
The skater's height is 1.8 m, and the trunk diameter measures 35 cm, or 0.35 m.
Starting angular momentum is 68 rpm.
We make the following assumptions:
- The skater is modeled as a vertical cylinder (head, trunk, and legs), with arms extending horizontally as two uniform rods.
- Friction between the skater and the ice is considered negligible.
To analyze her body, we divide it into two parts, treating the arms as spinning rods.
1. Each arm (rod) has a moment of inertia of .
The arms comprise 13% of 74 kg, calculated as 0.13 x 74 = 9.62 kg.
Each arm is then evaluated as 9.62/2 = 4.81 kg.
Let L represent the length of each arm.
Thus,
I = x 4.81 x
= 0.79 kg-m for each arm.
2. The body, treated as a cylinder, has a moment of inertia of .
For the body, the radius r is half of the trunk diameter: r = 0.35/2 = 0.175 m.
The mass of the trunk amounts to (80% + 7%) of 74 kg, which calculates to 0.87 x 74 = 64.38 kg.
Therefore, I = x 64.38 x
= 0.99 kg-m.
Two cases are considered:
case 1: Body spinning with arms extended.
Total moment of inertia equals the combined moments of inertia of both arms and the trunk.
I = (0.79 x 2) + 0.99 = 2.57 kg-m.
The angular momentum is given by Iω.
Here, ω = angular speed = 68.0 rpm = x 68 = 7.12 rad/s.
The angular momentum then becomes 2.57 x 7.12 = 18.29 kg-rad/m-s.
case 2: Arms drawn in alongside the trunk.
The moment of inertia is attributed solely to the trunk. This is 0.91 kg-m.
The angular momentum equals Iω.
= 0.99 x ω = 0.91ω.
By the principle of conservation of angular momentum, the two angular momentum quantities are equal. Therefore,
18.29 = 0.99ω.
Solving gives ω = 18.29/0.99 = 18.47 rad/s.
This leads to 18.47 ÷ = 176.38 rpm.
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