- #1

simphys

- 322

- 45

- Homework Statement
- In order to increase her moment of inertia about a vertical axis, a spinning figure skater stretches out her arms horizontally; in order to reduce her

moment of inertia, she brings her arms down vertically along her sides. Calculate the change of moment of inertia between these two configurations of the arms. Assume that each arm is a thin, uniform rod of length 0.60 m and mass

2.8 kg hinged at the shoulder at a distance of 0.20 m from the axis of rotation.

- Relevant Equations
- xD

so calculated, the moment of inertia for a rod about an axis at the end of the rod is I = 1/3 * M * L^2

here for case 1: arms to the side

I is calculated to be ##I = 0.224##

for case 2: arms stretched

## I = 1 / 3 * M * L^2 + M * d^2 ## with L = 0.6 m (length of rod) and d = 0.2 (dinstance from 'hinge' to axis of rotation)

## I = 0.896## is

Yet the actual answer is ##I = 1.568##. I don't really understand what I am doing wrong here so can someone help me please?

Reasoning:

In case 2 I basically have the moment of inertia at the end (the 'hinge'/shoulder so to say) and need to add a dinstance d to get the moment of inertia about the rotation axis via the parallel axis theorem yet this isn't correct.

Thanks in advance

here for case 1: arms to the side

I is calculated to be ##I = 0.224##

for case 2: arms stretched

## I = 1 / 3 * M * L^2 + M * d^2 ## with L = 0.6 m (length of rod) and d = 0.2 (dinstance from 'hinge' to axis of rotation)

## I = 0.896## is

**my answer.**Yet the actual answer is ##I = 1.568##. I don't really understand what I am doing wrong here so can someone help me please?

Reasoning:

In case 2 I basically have the moment of inertia at the end (the 'hinge'/shoulder so to say) and need to add a dinstance d to get the moment of inertia about the rotation axis via the parallel axis theorem yet this isn't correct.

Thanks in advance