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Zielflug [23.3K]
2 years ago
4

A flywheel rotates at 7200 rpm when the power is suddenly cut off. The flywheel decelerates at a constant rate of 2.1 rad/s2 and

comes to rest 5 min later. How many revolutions does the flywheel make before coming to rest?
Physics
2 answers:
Mnenie [13.5K]2 years ago
7 0

Answer:

20940.3 revolution

Explanation:

Using,

θ = ω₀t+1/2αt² ................... Equation 1

Where θ = number of revolution made by the flywheel, t = time, ω₀ = angular initial velocity, α = angular acceleration.

Given: ω₀ = 7200 rpm = 7200(0.1047) = 754 rad/s, t = 5 min = 300 s, α = -2.1 rad/s²( decelerating)

Substitute into equation 1

θ = 754(300)+1/2(-2.1)(300²)

θ = 226200-94500

θ = 131700 rad.

If 1 rad = 0.159 rev,

Then 131700 rad = 0.159×131700 = 20940.3 rev.

Hence the flywheel makes 20940.3 revolution before coming to rest

dmitriy555 [2]2 years ago
6 0

Answer:

20960.7 revolutions

Explanation:

Convert from revolutions-per-minute to angular speed (rad/s) knowing that each revolution has a 2π angle and each minute has 60 seconds

7200 rpm = 7200 (rev/min) * 2π (rad/rev) * (1/60) (min/sec) = 754 rad/s

We can use the following equation of motion to calculate the angle swept by the flywheel during the deceleration of t = 5 minute = 300 s

\theta = \omega_0 t + \alpha t^2/2

\theta = 754 * 300 - 2.1 * 300^2/2 = 131700 rad

Again we can convert this to number of revolution

131700 rad * (1/2π) (rev/rad) = 20960.7 rev

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49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

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<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
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The formula we follow for this problem are as follows:

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Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

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Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

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Answer:

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Substituting the values we get

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