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8 days ago

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When switched on, the grinding machine accelerates from rest to its operating speed of 3550 rev/min in 10 seconds. When switched

off, it coasts to rest in 31 seconds. Determine the number of revolutions turned during both the startup and shutdown periods. Also determine the number of revolutions turned during the first half of each period. Assume uniform angular acceleration in both cases.

1 answer:

serg [3.2K]8 days ago

5 0

Response:

At startup, the total revolutions

$N_1 = 296 turn$

After the first half of the duration, total revolutions

$N_3 = \frac{29.6 + 0}{2}(5) = 74 turns$

Total revolutions until it halts

$N_2 = \frac{59.17 + 0}{2}(31) = 917 turn$

Following the first half of the duration, the total revolutions are

$N_4 = 688 turns$

Clarification:

At the beginning, the machine is stationary and then accelerates to a speed of 3550 rev/min

At this point we calculate

$f = \frac{3550}{60} = 59.17 rev/s$

Since it took 10 seconds to achieve the speed

the angular acceleration can be expressed as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{59.17- 0}{10} = 5.917 rev/s^2$

It comes to a stop after 31 seconds, leading to the angular deceleration being defined as

$\alpha_2 = -\frac{59.17}{31} = -1.91 rev/s^2$

Initially, the revolutions taken to reach the specified speed

$N_1 = \frac{59.17 + 0}{2}(10) = 296 turn$

the number of revolutions during the stopping phase is

$N_2 = \frac{59.17 + 0}{2}(31) = 917 turn$

Now, considering the increase in speed after t = 5 seconds is

$\omega_1 = (5.917)(5) = 29.6 rev/s$

$N_3 = \frac{29.6 + 0}{2}(5) = 74 turns$

During the stopping phase, the speed after half of the duration is calculated as

$\omega_2 = 59.17 - (1.91)(15.5) = 29.56 rev/s$

Lastly, the total revolutions are calculated as

$N_4 = \frac{59.17 + 29.56}{2}(15.5)$

$N_4 = 688 turns$

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