The initial angular speed when turned off and the angular deceleration.

Question

A Turbine is rotating at constant rate, when turned off it deccelerates at a constant rate.

The Turbine makes 5 complete rotations in the first second after it is switched off, it then takes another 4 seconds to make another 5 complete rotations.

1) What is the initial angular speed, 𝜔₀ of the drill when turned off.

2) What is the angular deceleration of the Turbine.

Answer 1

I) First I will rephrase the question, then

II) I will solved the rephrased version, then

III) I will explain why it is necessary to rephrase the question.

I)

At an "initial time" a turbine starts slowing down with a constant deceleration.

After one second the position of the turbine is different from its initial position by 5 rotations.

After another 4 seconds the position of the turbine is different from its initial position by 5 more rotations.

1) What is the initial angular speed, ω₀, of the turbine.

2) What is the angular deceleration of the turbine?

II)

Let

α be the constant angular acceleration (it will be negative, since the turbine is decelerating),

ω(t) be the angular velocity after t seconds since the initial time,

θ(t) be the number of rotations by which the position of the turbine differs from its initial position after t seconds.

We are given

θ(0) = 0, ω(0) = ω₀ to be calculated,

θ(1) = 5,

θ(5) = 10

I assume you know the following identity

θ(t) = θ(0) + ω₀t + αt²/2

If you need to know where this comes from, let me know.

Into that equation substitute given data for t = 1 and t = 5

5 = ω₀ + α/2

10 = 5ω₀ + 25α/2

The solution is

ω₀ = 5.75 revolutions per second

α = -1.5 revolutions per second square

Here comes the weird part.

The turbine will make 5 rotations in 1 second, i.e., θ(1) = 5,

then another 5 rotations in 1.66 seconds, i.e., θ(2.66) = 10,

then it will stop at time 3.833, i.e., ω(3.833) = 0, at which time θ(3.833) = 11

then it will start to rotate in the opposite direction, and come to θ(5) = 10.

That is, at time 5 it will differ from the initial time by the same number of rotations as at time 2.66.

III)

Here are the issues with the original wording.

"A turbine is rotating at a constant rate"

It is irrelevant how fast the turbine was rotating before the initial time.

"when turned off it decelerates at a constant rate"

This gives the impressing that the deceleration is due to drag alone.

But drag would not result in constant deceleration.

Constant deceleration can result only from a constant torque acting in direction opposite the initial rotation.

And this is key to solving the problem; drag alone would not reverse the direction of

rotation and there would be no solution.

While torque acting in the opposite direction will stop the turbine and then turn it in the opposite direction.

"then takes another 4 seconds to make another 5 complete rotations."

This wording gives the impression that there is no shorter time for the additional 5 rotations.

If we added this constraint then the problem would be over-constrained and there

would be no solution.

"ω0 of the drill when turned off"

I assume this is a typo: "drill" should be "turbine"