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3 years ago

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A motor running at 2600 rev/min is suddenly switched off and decelerates uniformly to rest after 10 s. Find the angular decelera

tion and the number of rotations to come to rest?

1 answer:

Mama L [17]3 years ago

6 0

Angular acceleration = (change in angular speed) / (time for the change)

change in angular speed = (zero - 2,600 RPM) = -2,600 RPM

time for the change = 10 sec

Angular acceleration = -2600 RPM / 10 sec = -260 rev / min-sec

(-260 rev/min-sec) x (1 min / 60 sec) = <em>-(4 1/3) rev / sec²</em>

Since the acceleration is negative, the motor is slowing down.
You might call that a 'deceleration' of (4 1/3) rev/sec² .

The average speed is 1/2(2,600 + 0) = 1,300 rev/min = (21 2/3) rev/sec.

Number of revs = (average speed) x (time) = (21 2/3) x (10sec) = <em>(216 2/3) revs</em>

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$W=\frac{P_2V_2-P_1V_1}{n-1}$

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$W=57.48 kJ$

From Energy balance

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Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

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$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

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