Rotational Motion and moments of inertia

[FluffCorgi]

Homework Statement:

A 65kg student is ice skating as shown. She is initially rotating at with an angular speed of 2.5rad/s with her arms straight out. At this time she can be modeled as a vertical and horizontal cylinder as shown.

She then brings her arms in, at which point she can be modeled as only a vertical cylinder with the properties shown.

The moment of inertia for the vertical cylinders spinning around their axes is I = 1/2mr2.
The moment of inertia for the horizontal cylinder spinning around it's center, perpendicular to it's axis is I = (1/12)md^2.

Determine all of the following moments of inertia, angular speeds and angular momenta relative to the axis of rotation.

Relevant Equations:

arms
m = 6kg
d = 1.7m
core
m = 59kg
r = 0.18m
full body
m = 65kg
r = 0.18m

I have the moment of inertia for the core(initial) and full body(final) but my answer for the moment of inertia for the arms(initial) was incorrect.
Arms(initial) moment of inertia:(1/12)(6)(1.7^2)=1.445 this is incorrect for some reason
Core(initial) moment of inertia: .9558
Full Body(final): 1.053
I also have to find the angular velocity and I have done this correctly for the arms and core but I could not figure out how to find the angular velocity for the full body(final)
Arms: 2.5 rad/s
Core: 2.5 rad/s
Full Body(final): I don't think I can solve this without the arms initial moment of inertia
Angular Momentum
Arms: my moment of inertia was incorrect so this was as well
Core: 2.3895
Full Body(final): My moment of inertia for the arms was incorrect so this was as well but this was my work to attempt to find it.
(1.445+.9558)2.5 = (1.053)w
w = 5.699 this was wrong but i tried to solve it anyway
I need help with the initial moment of inertia for the arms.

 

[tnich]

The moment of inertia of the outstretched arms is proportional $r^2$. You have put in 1.7m for the radius. Do you think that is a reasonable length for a skater's arm?

 

[FluffCorgi]

Why is d defined as 1.7 if they don't want you to use it in the formula?

 

[kuruman]

The moment of inertia of the outstretched arms is proportional $r^2$. You have put in 1.7m for the radius. Do you think that is a reasonable length for a skater's arm?

Not quite. The moment of inertia of a rod of length $L$ about its midpoint is $I=\frac{1}{12}mL^2$. Here $L=1.7 ~\mathrm{m}$ and I get the same value as OP for the moment of inertia of just the arms. I also think that 1.7 m is a reasonable length for the skater's wingspan which should be approximately equal to her height at under 6 ft. See here for details.

 

[FluffCorgi]

I think the website is incorrect.

 

[kuruman]

I think the website is incorrect.

This wouldn't be the first (or last) time that a website is incorrect. Talk to the person who assigned this question. He/she needs to know.

 

[haruspex]

Full Body(final): 1.053

I would take the retracted arms as point masses at the full radius of the torso. Similarly, the extended arms should not be taken as a uniform rod through the body. Take them as starting at the periphery of the torso.

You have put in 1.7m for the radius.

No, as the fingertip to fingertip span, as @kuruman notes.

Not quite. The moment of inertia of a rod of length L about its midpoint is

@tnich only said it was proportional to r² (but maybe didn't notice the 1/12).

 

[kuruman]

Similarly, the extended arms should not be taken as a uniform rod through the body. Take them as starting at the periphery of the torso.

Yes, but the figure accompanying the post says "At this time she can be modeled as a vertical and horizontal cylinder as shown." And further down , "The moment of inertia for the horizontal cylinder spinning about it's (sic) center, perpendicular to it's axis is $\frac{1}{12}md^2$." Parameter $d$ is the length of the horizontal cylinder in the figure.

 

[haruspex]

Yes, but the figure accompanying the post says "At this time she can be modeled as a vertical and horizontal cylinder as shown." And further down , "The moment of inertia for the horizontal cylinder spinning about it's (sic) center, perpendicular to it's axis is $\frac{1}{12}md^2$." Parameter $d$ is the length of the horizontal cylinder in the figure.

Ok, I hadn't noticed the attachment.
In effect it treats the shoulders and the space between as part of the arms.
I note also it says to treat the arms as a cylinder but provides the formula for a thin rod and no thickness for the arms.

 

[kuruman]

Ok, I hadn't noticed the attachment.
In effect it treats the shoulders and the space between as part of the arms.
I note also it says to treat the arms as a cylinder but provides the formula for a thin rod and no thickness for the arms.

Right. Out of curiosity I looked up the moment of inertia about the midpoint of a cylinder of length $L$ and radius $r$; it is $$I=\frac{1}{12}m(3r^2+L^2)=\frac{1}{12}mL^2\left(1+3\frac{r^2}{L^2}\right).$$Using the given value L = 170 cm and assuming a reasonable arm radius of 2-3 cm, the added term $3\frac{r^2}{L^2}$ is of order 10^-4. It seems that the thin rod approximation for the extended arms is good.