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

• puipui_queen
In summary, the cylindrical skater shown in the figure can be modeled as a homogeneous cylinder with outstretched arms, all with lengths that are simple multiples of the radius. The moment of inertia about the axis of rotation for the cylindrical body can be calculated using the equation 1/2MR^2. For the outstretched arms, the rotational inertia can be calculated using the equation 1/12ML^2, with an offset of (s/2 + R) from the center of mass of the skater. The moment of inertia of each arm when it is parallel to the axis of rotation can be found using the parallel axis theorem, where I = Icm +

## Homework Statement

Figure shows the cylindrical skater, http://hk.geocities.com/puipui_queen/cylinder_skater.jpg [Broken]
as she spins,she may be modeled 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. Homework 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:
... 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

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