Moment of inertia of a yo-yo

Question

Considering the yo-yo like two CDs with a hollow cylinder between them, what is the moment of inertia of that object?

The axis that I must choose can't pass through the CM and be parallel to rotational axis that passes through the center of mass. A force $vec{T}$ is pulling the object.

I can't use the Steiner theorem here, right? All because that cylinder cannot be ignored.

Another try that I do is calculating the moment of inertia of the three (well, two) objects and sum, but I think that this isn't correct.

David · Answer

Use Steiner theorem and ignore the cilinder between the two disc.

ezfzx · Answer

The inertia is calculated through the axis of rotation. for a yo-yo, I think we're safe to assume that is the yo'yo's center of mass.
So, the total inertial moment is just the sum of the parts.
We can assume each side of the yo-yo is identical, and maybe can be modeled as a solid cylinder.
What might be tricky is the mass. If the yo-yo is in hand, it could simply be weighed. If not, we can find the volume and density of the material and use that to estimate the mass.
Conceivably, we could add the tiny inertia of the tiny axle in between, but truthfully, it's virtually negligible.
The torque applied does not affect the inertia.
However, when torque IS applied, the inertia can be used to find the angular acceleration of the yo-yo. Keep in mind that the moment arm is the tiny radius of the axle, not the yo-yo body.
FYI, the typical yo-yo toy has a body radius of around 3.3 cm and an axle radius of about 0.21 cm (which is very small).
When "dropped", a yo-yo can take about 5 seconds to un-spool down the height of 1 meter.

Moment of inertia of a yo-yo

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