Question

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most of her body as a uniform cylinder. Suppose the skater has a mass of 64 kg. One-eight of that mass is in her arms, which are 60 cm long and 20 cm from the vertical axis about which she rotates. The rest of her mass is approximately in the form of a 20-cm-radius cylinder. a. Estimate the skater's moment of inertia to two significant figures. b. If she were to hold her arms outward, rather than at her sides, would her moment of inertia increase, decrease, or remain unchanged? Explain.

Answer 1

Answer:

a) I = 1.44 Kg.m²

b) I = 3.12 Kg.m²

As rarm increases, her moment of inertia will increase.

Explanation:

Given

The skater's mass: M = 64 Kg

Mass of both arms: m = (M/8) = 64 Kg / 2 = 8 Kg

Mass of one arm: = marm = m/2 = 8 Kg / 2 = 4 Kg

Mass of her rest body: Mcyl = M - 2m = 64 Kg - 8 Kg = 56 Kg

rarm = 0.20 m

Rcyl = 0.20 m

Larm = 0.60 m

a) We apply the equation (Using the Steiner Theorem)

I = Icyl + Iarms

I = (Mcyl*Rcyl²/2) + 2*marm*rarm²

I = (56 Kg*(0.20 m)²/2) +2(4 Kg)(0.20 m)²

I = 1.44 Kg.m²

b) Suppose that the center of the mass of her outstretched arm is in the middle, so that the mass of the arm will be at a distance of 50 cm, then

rarm = 0.50 m

We apply the same equation, then

I = Icyl + Iarms

I = (Mcyl*Rcyl²/2) + 2*marm*rarm²

I = (56 Kg*(0.20 m)²/2) +2(4 Kg)(0.50 m)²

I = 3.12 Kg.m²

As rarm increases, her moment of inertia will increase.