Answer:
Following are the solution to this question:
[tex]N_p= 1600 \ rpm\\\\N_G = 400 \ \frac{rev}{min}\\\\ p = 9.42 \ mm\\\\C = 112.5 \ mm\\[/tex]
Explanation:
[tex]t= \text{number of tooth in pinion}= 15 \\\\T= \text{number of tooth on Gear}= 60\\\\m= module = 3 \ mm\\\\N_p= 1600 \ rpm\\\\G= gear ratio= \frac{I}{t} = \frac{60}{15} =4 =\frac{N_p}{N_G} \\\\N_G =\frac{1600}{4} = 400 \ rpm \\\\Circular\ pitch = \frac{\pi D}{T} = \pi m= \pi (3) = 9.42 mm \\\\\\text{Center to center distance} = \frac{m}{2}(T+t)\\\\[/tex]
[tex]=\frac{3}{2} (15+60)\\\\=\frac{3}{2} \times 75\\\\=3 \times 37.5\\\\= 112.5 \ mm[/tex]
