Modelling a Pirouette

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Unit: Energy and Momentum of Rotating Systems

Prerequisites

Conservation of Angular Momentum Rotational Work-Kinetic Energy Theorem

Later Topics

None

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Part 1

A figure skater is modeled as a central cylinder of mass $M$ and radius $R$ , with moment of inertia $I_{cyl} = \frac{1}{2}MR^2$ . Two arms are modeled as thin rods, each of mass $m$ and length $L$ , with moment of inertia about their center $I_{rod} = \frac{1}{12}mL^2$ .

Initially, the skater holds their arms extended horizontally outward from the edge of the cylinder.

Using the Parallel Axis Theorem, derive an expression for the initial total moment of inertia $I_i$ of the skater.

Correct!

Solution:

The total moment of inertia is the sum of the cylinder contribution and the contributions from both arms.

For the central cylinder about its symmetry axis:

I_{cyl} = \frac{1}{2}MR^2

For one arm (modeled as a thin rod), the moment of inertia about its own center of mass is:

I_{rod} = \frac{1}{12}mL^2

The arm’s center of mass is at distance $d = R + \frac{L}{2}$ from the rotation axis, so by the Parallel Axis Theorem:

I_{arm} = I_{rod} + md^2

Substitute $I_{rod}$ and $d$ :

I_{arm} = \frac{1}{12}mL^2 + m\left(R + \frac{L}{2}\right)^2

There are two identical arms, so the total initial moment of inertia is:

I_i = I_{cyl} + 2I_{arm}

I_i = \frac{1}{2}MR^2 + 2\left[\frac{1}{12}mL^2 + m\left(R + \frac{L}{2}\right)^2\right]

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