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 =  13.3779 m/s

Solution:

$$\begin{gathered} D_1=80\text{ mm } \\ {D}_2=160\text{ mm } \\ {n}_2=1460\cdot D_1\mathrm{/}{D}_2=1460\cdot 80\mathrm{/}160=7301\text{/min} \end{gathered}$$

$$\begin{gathered} D_3=350\text{ mm}\to \text{ m}=350\mathrm{/}1000\text{ m}=0.35\text{ m } \\ {n}_3=\pi \cdot D_3\cdot n_2\mathrm{/}60=3.1416\cdot 0.35\cdot 730\mathrm{/}60=13.3779\text{ m/s} \end{gathered}$$

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