Winch drum

Originally an empty winch drum with a diameter of 20 cm and a width of 30 cm on the rescue car, he started winding a rope with a thickness of 1 cm beautifully from edge to edge. The winch stopped after 80 turns. It remains to spin 3.54m of rope (without hook). How long is the whole rope?

Result

x =  22.39 m

Solution:

D=20 cm→m=20/100 m=0.2 m s=30 cm→m=30/100 m=0.3 m s0=1 cm→m=1/100 m=0.01 m  n=80 x0=3.54 m n0=s/s0=0.3/0.01=30  r0=D/2=0.2/2=110=0.1 m r1=r0+s0=0.1+0.01=11100=0.11 m r2=r1+s0=0.11+0.01=325=0.12 m  x1=2π⋅ r0⋅ n0=2⋅ 3.1416⋅ 0.1⋅ 30≐18.8496 m x2=2π⋅ r1⋅ n0=2⋅ 3.1416⋅ 0.11⋅ 30≐20.7345 m x3=2π⋅ r2⋅ (n−2⋅ n0)=2⋅ 3.1416⋅ 0.12⋅ (80−2⋅ 30)≐15.0796 m  x=x0+x1+s2+s3=3.54+18.8496+s2+s3≐22.3896≐22.39 mD=20 \ cm \rightarrow m=20 / 100 \ m=0.2 \ m \ \\ s=30 \ cm \rightarrow m=30 / 100 \ m=0.3 \ m \ \\ s_{0}=1 \ cm \rightarrow m=1 / 100 \ m=0.01 \ m \ \\ \ \\ n=80 \ \\ x_{0}=3.54 \ \text{m} \ \\ n_{0}=s/s_{0}=0.3/0.01=30 \ \\ \ \\ r_{0}=D/2=0.2/2=\dfrac{ 1 }{ 10 }=0.1 \ \text{m} \ \\ r_{1}=r_{0}+s_{0}=0.1+0.01=\dfrac{ 11 }{ 100 }=0.11 \ \text{m} \ \\ r_{2}=r_{1}+s_{0}=0.11+0.01=\dfrac{ 3 }{ 25 }=0.12 \ \text{m} \ \\ \ \\ x_{1}=2 \pi \cdot \ r_{0} \cdot \ n_{0}=2 \cdot \ 3.1416 \cdot \ 0.1 \cdot \ 30 \doteq 18.8496 \ \text{m} \ \\ x_{2}=2 \pi \cdot \ r_{1} \cdot \ n_{0}=2 \cdot \ 3.1416 \cdot \ 0.11 \cdot \ 30 \doteq 20.7345 \ \text{m} \ \\ x_{3}=2 \pi \cdot \ r_{2} \cdot \ (n-2 \cdot \ n_{0})=2 \cdot \ 3.1416 \cdot \ 0.12 \cdot \ (80-2 \cdot \ 30) \doteq 15.0796 \ \text{m} \ \\ \ \\ x=x_{0}+x_{1}+s_{2}+s_{3}=3.54+18.8496+s_{2}+s_{3} \doteq 22.3896 \doteq 22.39 \ \text{m}

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