Question
Asked Oct 27, 2019
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Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of diameter ?d=0.273 m.Dd=0.273 m. The mass of the hoop part is ?h=0.120 kgmh=0.120 kg and the mass of the disk part is ?d=0.050 kg.md=0.050 kg. Imagine making a boomerang that has the same total moment of inertia around its center as the sport disk. The boomerang is to be constructed in the shape of an "X," which can be approximated as two thin, uniform rods joined at their midpoints.

If the total mass of the boomerang is to be ?b=0.255 kg,mb=0.255 kg, what must be the length ?bLb of the boomerang?

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D d Cross-sectional view

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Given Diameter of thin hoop, Dh 0.273m Diameter of Disk, Dd0- 273 m 1 Mars of hoop, mh= 0120 kg MaM of clisk, md = 0.05o kg Mass of boomerang, m = 0.255ks.

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Amume length of one thin od of boomerans as Since, beomerang fa madle by ining two thfn od ten each od has mam MR

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Moment of inertia formula for Aome obrects at center of mas is os fol ouw Cm cm Disk Thin hoop Thin rod ( Moment of inerHa of dlak, u = mD (i) Moment of inertia of thin hoopP, I = MD ( of fnextia of red, IR= 2m (ni) Moment 0

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