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Moment of inertia of a stiff wire bent into a wheel with spokes

  • Thread starter mcphyssics
  • Start date
  • #1

Homework Statement



A stiff uniform wire of mass M0 and length L0 is cut, bent, and the parts soldered together so that it forms a circular wheel having four identical spokes coming out from the center. None of the wire is wasted, and you can neglect the mass of the solder.

What is the moment of inertia of this wheel about an axle through its center perpendicular to the plane of the wheel? Express your answer in terms of the given quantities.

Homework Equations



Irim = MR2
Ispoke = 1/3ML2
Itotal = Irim + 4*Ispoke

The Attempt at a Solution



The length of one spoke, which should also be the radius of the rim, should be L0/(2[itex]\pi[/itex]+4)

I am unsure of the mass of one spoke, as well as the mass of the rim. Would the mass of one spoke be the same ratio? (Mspoke = M0/(2[itex]\pi[/itex]+4)) ?
If so, then what would be the mass of the rim? I thought Mrim = M0 - 4*Mspoke
therefore Mrim = (M0*(1-4/(2[itex]\pi[/itex]+4))
but I am not getting it right when I submit it.
 

Answers and Replies

  • #2
lewando
Homework Helper
Gold Member
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What, exactly, are you submitting?
 
  • #3
This is what I have: I = (1-4/(2[itex]\pi[/itex]+4))*M0*(L0/(2[itex]\pi[/itex]+4))2+4*M0/(2[itex]\pi[/itex]+4)*(L0/(2[itex]\pi[/itex]+4))2

It seems very messy/cumbersome. Am I going about this wrong?
 

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