# Homework Help: Newtonian mechanics - extended rigid body's rotation, moment of inertia

1. Feb 12, 2007

### puipui_queen

1. The problem statement, all variables and given/known data

Figure shows the cylindrical skater, http://hk.geocities.com/puipui_queen/cylinder_skater.jpg [Broken]
as she spins,she may be modelled as a homogeneous cylinder of raduis R, height h, and density p, with outstretched arms. The arms are cylinders as welll, also of density p, radius r and length s. All these lengths are simple multiples of r.

Calculate the moment of inertia about the axis of cylindical body I_body, the moment of inertia of each outstretched arm I_out, and the moment of inertia of each arm when it is parallel to the axis of rotation, I_down, as shown in the figure.
2. Relevant equations

R=4r
h=36r
s=16r

3. The attempt at a solution

I_body = 1/2MR^2 = 4608p*pi*r
I_out, I tried using the eq'n 1/3ML^2, where L= s+R, but someone told me this is not right since it is the eq'n for thin rod, and the cylinder is not thin.

I_down, I am thinking about parallel axis theorm, where I = Icm + Md^2, and Icm = I_body and d = r+R, but then i don't think it is very logical... so i'm stuck...

And just wondering, there's more parts to this question, can I post the question under the same thread, or should I open a new thread for next part of the question.

Last edited by a moderator: May 2, 2017
2. Feb 12, 2007

### andrevdh

.... I_body = 1/2MR^2 = 4608p*pi*r .....

it should be r^5

For an outstretched arm:
The rotational inertia of a rod about an axis perpendicular to its length through its middle is

$$I_{rod} = \frac{1}{12}ML^2$$

about the skater's axis this will be an offset of

$$\frac{s}{2} + R$$

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