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2 years ago

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A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.

0cm . The power is off for 39.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 250 complete revolutions. At what rate is the flywheel spinning when the power comes back on? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

1 answer:

aleksley [76]2 years ago

5 0

Answer:

w = 25.05 rad / s , α = 0.7807 rad / s² , θ = 1972.75

Explanation:

This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off

θ = w₀ t - ½ α t²

α = (w₀t - θ) 2/t²

let's reduce the magnitudes to the SI system

w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s

θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad

let's calculate the angular acceleration

α = (55.5 39 - 1570.8) 2/39²

α = 0.7807 rad / s²

having the acceleration we can calculate the final speed

w = w₀ - ∝ t

w = 55.5 - 0.7807 39

w = 25.05 rad / s

the time to stop w = 0

0 = wo - alpha t

t = wo / alpha

t = 55.5 / 0.7807

t = 71.09 s

the angle traveled

w² = w₀⁹ - 2 α θ

w = 0

θ = w₀² / 2α

let's calculate

θ = 55.5 2 / (2 0.7807)

θ = 1972.75

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Answer:

Part a)

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Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

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so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

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Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

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$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

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Answer:

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Let's review the statements of the exercise

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Answer with Explanation:

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