Respuesta :
Answer:
During start total turns
[tex]N_1 = 296 turn[/tex]
After half of the time total turns
[tex]N_3 = \frac{29.6 + 0}{2}(5) = 74 turns[/tex]
Total number of turns during it stop
[tex]N_2 = \frac{59.17 + 0}{2}(31) = 917 turn[/tex]
After half of the time total turns
[tex]N_4 = 688 turns[/tex]
Explanation:
Initially the machine is at rest and then starts rotating with speed 3550 rev/min
now we will have
[tex]f = \frac{3550}{60} = 59.17 rev/s[/tex]
now we know that it took 10 s to reach the speed
so angular acceleration is given as
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
[tex]\alpha = \frac{59.17- 0}{10} = 5.917 rev/s^2[/tex]
now it stops in 31 s so the angular deceleration is given as
[tex]\alpha_2 = -\frac{59.17}{31} = -1.91 rev/s^2[/tex]
now initially number of turn to reach the given speed
[tex]N_1 = \frac{59.17 + 0}{2}(10) = 296 turn[/tex]
number of turns during it stop
[tex]N_2 = \frac{59.17 + 0}{2}(31) = 917 turn[/tex]
Now during startup speed after t = 5 s is given as
[tex]\omega_1 = (5.917)(5) = 29.6 rev/s[/tex]
[tex]N_3 = \frac{29.6 + 0}{2}(5) = 74 turns[/tex]
now during it stop the speed after half the time is given as
[tex]\omega_2 = 59.17 - (1.91)(15.5) = 29.56 rev/s[/tex]
now the number of turns is given as
[tex]N_4 = \frac{59.17 + 29.56}{2}(15.5)[/tex]
[tex]N_4 = 688 turns[/tex]
