SOLUTION: Q.4
Ten turns of a wire are helically wrapped around a cylindrical tube with outside
circumference 4 inches and length 9 inches. The ends of the wire coincide with
ends of the
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-> SOLUTION: Q.4
Ten turns of a wire are helically wrapped around a cylindrical tube with outside
circumference 4 inches and length 9 inches. The ends of the wire coincide with
ends of the
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Question 389025: Q.4
Ten turns of a wire are helically wrapped around a cylindrical tube with outside
circumference 4 inches and length 9 inches. The ends of the wire coincide with
ends of the same cylindrical element. Find the length of the wire. Answer by Jk22(389) (Show Source):
You can put this solution on YOUR website! a description of the wire's curve can be done using r the radius, and L the length of the cylinder, n the number of turns :
x=r*cos(t)
y=r*sin(t)
z=a*t,
the latter a comes from : a* 2*Pi*n = L
the length is : l=int( sqrt(dx^2+dy^2+dz^2))=sqrt(r^2+L^2/(4*Pi^2n^2))2*Pi*n
here : l=sqrt(16/(4*Pi^2)+9^2/(4*Pi^2*100))*2*Pi*10
=sqrt(16+81/100)*10 = sqrt(1600+81) = 41 inches
