Moment of Inertia of a butterfly-shaped yoyo

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SUMMARY

The discussion focuses on calculating the moment of inertia for a butterfly-shaped yoyo, which has a total mass of 61.5 grams and varying radii. The moment of inertia formula I = m(r^2) is initially applied, but integration across varying disk sizes is necessary due to the yoyo's unique shape. Participants emphasize the importance of careful drawings to determine the appropriate radii for integration. The conversation highlights the need for a solid understanding of mechanics and integration techniques to solve the problem accurately.

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waughcraft
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Homework Statement


As interpreted from butterflyyoyo1cm.png, assuming that they are the same yoyos, each "half-yoyos" has 1cm of length, also, the center gap consists of 0.2cm, making the yoyo widths 2.2cm across in diameter exteriorly(1.1cm in radius).

For the "interior" length, as shown in butterflyyoyo1.1cm.PNG, the yoyo has 1.1cm in radius interiorly.

As shown in image BEST3124BU.jpg, the yoyo has a radius of 2.5cm in length.

Ignoring the mass of the string, half of the yoyo weighs 30.75g(0.03075kg), which adds up to 61.5grams(0.0615kg) for the object as a whole.

Question: Assuming the yoyo has the same mass density in every part of the yoyo(except the center gap), what is the value of the moment of inertia of the yoyo?

Homework Equations


Moment of Inertia: I = m(r^2)

The Attempt at a Solution


Used the formula I = 0.5M(R^2) from a physics textbook, stating that solid cylinder about cylindrical axis, but it does not work for this case.
 

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waughcraft said:
Used the formula I = 0.5M(R^2) from a physics textbook, stating that solid cylinder about cylindrical axis, but it does not work for this case.

Welcome to PF.

Well it does ... but. In this case you have to then integrate across the varying size disks that are determined by the displacement along the axis of rotation. Symmetry will let you at least double it once you've found 1 side.

Looks like you have to do a careful drawing to determine the appropriate radii of all the incremental disks.

Too bad it's not simply a solid sphere cut in half and the halves swapped back to back. That you could get out of your book.
 
LowlyPion said:
Welcome to PF.
Well it does ... but. In this case you have to then integrate across the varying size disks that are determined by the displacement along the axis of rotation. Symmetry will let you at least double it once you've found 1 side.
Thanks for the reply first of all.
However, sorry that I feel kinda frustrated about it since I don't know how to "integreate across the varying size disks" due to my insufficient knowledge in mechanics for that I'm at the last year of high school.

May you, or someone, kindly explain how to integrate the various disk sizes in this case?

Also, I would like to kindly ask, for the insufficient data that you've just commented, what sets of data do I need to collect to make the data sets "sufficient"?

Thank you!
 
First of all you have sufficient data, if you make the assumption that the radius of curvature of the halves are spherical. So what you have then are missing parts, or parts of a whole which ever way you want to look at it.

Integration is really your best option and the most exact.

Like I said before you want to make a careful drawing. And then determine which parts you can integrate across. (i.e. from which radius to which radius these disks will be summed from)

The center post is a cylinder so that's easy. You have that portion, though its contribution will be slight.

As to the spherical sections here is the example of developing the moment of inertia for a sphere.

http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html#sph4

Your limits are not R to -R however. That you will need to determine from your careful drawing.
 

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