Engine pulley

The engine has a 1460 rev / min (RPM). Disc diameter is 350 mm. What will be the disc peripheral speed in RPM? Pulleys on the engine has diameter 80mm, on a disc has diameter 160mm.

Correct result:

n2 =  730 1/min
n3 =  13.378 m/s

Solution:

D1=80 mm D2=160 mm  n2=1460⋅ D1/D2=1460⋅ 80/160=730 1/minD_{1}=80 \ \text{mm} \ \\ D_{2}=160 \ \text{mm} \ \\ \ \\ n_{2}=1460 \cdot \ D_{1}/D_{2}=1460 \cdot \ 80/160=730 \ 1\text{/min}
D3=350 mm→m=350/1000 m=0.35 m n3=π⋅ D3⋅ n2/60=3.1416⋅ 0.35⋅ 730/60=13.378 m/sD_{3}=350 \ mm \rightarrow m=350 / 1000 \ m=0.35 \ m \ \\ n_{3}=\pi \cdot \ D_{3} \cdot \ n_{2}/60=3.1416 \cdot \ 0.35 \cdot \ 730/60=13.378 \ \text{m/s}

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