A 56.0 ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.00 turns each second. The distance from one hand to the other is 1.5 . Biometric measurements indicate that each hand typically makes up about 1.25 of body weight.
A.What horizontal force must her wrist exert on her hand? (N)
B.Express the force in part (a) as a multiple of the weight of her hand. (w)
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To solve parts A and B of the problem, we first need to find the mass of the skater's hand.
Step 1: Calculate the mass of the skater's hand.
The problem states that each hand typically makes up about 1.25% of body weight. Let's assume the skater's total body weight is W.
Mass of each hand = (1.25/100) * W = 0.0125 * W
Step 2: Calculate the horizontal force exerted by the wrist on the skater's hand.
The horizontal force required for circular motion can be calculated using the formula:
Force = (mass * angular velocity^2) / radius
Given:
Angular velocity = 2.00 turns/second = 2 * (2π) radians/second (since one turn is equal to 2π radians)
Radius = 1.5 m
Mass of each hand = 0.0125 * W
Using the formula, we can calculate the force:
Force = (0.0125 * W) * (2 * (2π))^2 / 1.5
Simplifying further, we get:
Force = (0.0125 * W) * 16 * π^2 / 1.5
So, the horizontal force exerted by the wrist on the skater's hand is (0.0125 * W) * 16 * π^2 / 1.5, and this is the answer to part A.
To solve part B, we need to express the force in part A as a multiple of the weight of her hand.
Force = (0.0125 * W) * 16 * π^2 / 1.5
Since the weight is given as a multiple of the hand's weight, we can express the force as:
Force = (0.0125 * W) * 16 * π^2 / 1.5 = (0.0125 * W) * (16π^2 / 1.5)
Therefore, the force in part A can be expressed as a multiple of the weight of her hand by the factor (16π^2 / 1.5), and this is the answer to part B.