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

[puipui_queen]

Homework Statement

Figure shows the cylindrical skater, http://hk.geocities.com/puipui_queen/cylinder_skater.jpg
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.

 

[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

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

 

[m2003]

I would approach this problem by first identifying the relevant equations and principles in Newtonian mechanics. In this case, we are dealing with rotational motion and therefore the equations for moment of inertia would be applicable. The moment of inertia is defined as the resistance of an object to rotational motion and is dependent on the mass distribution and shape of the object.

For the cylindrical skater, we can calculate the moment of inertia about the axis of rotation (I_body) using the equation I = 1/2MR^2, where M is the mass of the cylinder and R is its radius. From the given information, we can calculate the mass of the cylinder as M = p*pi*R^2*h = 4608p*pi*r. Therefore, the moment of inertia about the axis of rotation is 4608p*pi*r.

For the outstretched arms, we can use the parallel axis theorem to find the moment of inertia when they are perpendicular to the axis of rotation (I_out). This theorem states that the moment of inertia about an axis parallel to the axis of rotation is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the two axes. In this case, the distance between the two axes is r+R and the moment of inertia about the center of mass is equal to the moment of inertia of the cylinder (I_body). Therefore, I_out = I_body + Ms^2 = 4608p*pi*r + 16pr^2.

For the arms when they are parallel to the axis of rotation, we can use the same equation as for the body, I = 1/2MR^2. However, since the arms are now aligned with the axis of rotation, the moment of inertia (I_down) will be lower than when they are perpendicular. Using the given information, we can calculate the mass of each arm as M = p*pi*r^2*s = 256p*pi*r^3. Therefore, I_down = 1/2(256p*pi*r^3)*R^2 = 8192p*pi*r^5.

As for the other parts of the question, it would be best to open a new thread to avoid confusion and to allow others to provide their own solutions and insights.