The turbine and associated rotating parts of a jet engine have a total moment of inertia of 23 kg m2. The turbine is accelerated uniformly from rest to an angular speed of 140 rad/s in a time of 27 s.

(a) Find the angular acceleration.


(b) Find the net torque required.


(c) Find the angle turned through in 27 s.


(d) Find the work done by the net torque.


(e) Find the kinetic energy of the turbine at the end of the 27 s.

(a) Well, to find the angular acceleration, we can use the formula:

angular acceleration = (final angular speed - initial angular speed) / time

Substituting in the given values, we get:

angular acceleration = (140 rad/s - 0 rad/s) / 27 s

Wow, that's some serious acceleration for a turbine! The angular acceleration comes out to be approximately 5.185 rad/s². Remember to wear your seatbelt!

(b) Now, to find the net torque required, we can use the formula:

net torque = moment of inertia * angular acceleration

Substituting the given values, we have:

net torque = 23 kg m² * 5.185 rad/s²

This torque is powerful enough to make your head spin, with a value of approximately 119.055 Nm. Hang on tight!

(c) To find the angle turned through in 27 s, we can use the formula:

angle = initial angular speed * time + (1/2) * angular acceleration * time²

Substituting the given values, we get:

angle = 0 rad/s * 27 s + (1/2) * 5.185 rad/s² * (27 s)²

Now, get ready for the twist! The angle turned through in 27 seconds is approximately 1984.5 radians. That's enough spin to impress any dance floor!

(d) To find the work done by the net torque, we use the formula:

work = (1/2) * moment of inertia * (final angular speed)²

Substituting the given values, we have:

work = (1/2) * 23 kg m² * (140 rad/s)²

The net torque really puts in some work, with a value of approximately 644,400 Joules. That's enough energy to power a small circus!

(e) Finally, to find the kinetic energy of the turbine at the end of the 27 seconds, we can use the formula:

kinetic energy = (1/2) * moment of inertia * (final angular speed)²

Substituting the given values, we get:

kinetic energy = (1/2) * 23 kg m² * (140 rad/s)²

The turbine is spinning with a kinetic energy of approximately 575,400 Joules. That's enough energy to make even the happiest clown jealous!

To find the angular acceleration of the turbine:

(a) Angular acceleration is given by the formula:
α = (ωf - ωi) / t
where α is the angular acceleration, ωf is the final angular speed, ωi is the initial angular speed, and t is the time taken.

Plugging in the given values:
α = (140 rad/s - 0 rad/s) / 27 s
α = 5.185 rad/s^2

So, the angular acceleration of the turbine is 5.185 rad/s^2.

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To find the net torque required:

(b) Net torque is given by the formula:
τ = I * α
where τ is the net torque, I is the moment of inertia, and α is the angular acceleration.

Plugging in the given values:
τ = 23 kg m^2 * 5.185 rad/s^2
τ = 119.055 Nm

So, the net torque required is 119.055 Nm.

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To find the angle turned through in 27 s:

(c) Angle turned is given by the formula:
θ = ωi * t + (1/2) * α * t^2
where θ is the angle turned, ωi is the initial angular speed, α is the angular acceleration, and t is the time taken.

Plugging in the given values:
θ = 0 rad/s * 27 s + (0.5) * 5.185 rad/s^2 * (27 s)^2
θ = 2467.745 rad

So, the turbine turns through an angle of 2467.745 rad in 27 s.

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To find the work done by the net torque:

(d) Work done is given by the formula:
W = (1/2) * I * ω^2
where W is the work done, I is the moment of inertia, and ω is the final angular speed.

Plugging in the given values:
W = (0.5) * 23 kg m^2 * (140 rad/s)^2
W = 226960 J

So, the work done by the net torque is 226960 J.

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To find the kinetic energy of the turbine at the end of the 27 s:

(e) Kinetic energy is given by the formula:
KE = (1/2) * I * ω^2
where KE is the kinetic energy, I is the moment of inertia, and ω is the final angular speed.

Plugging in the given values:
KE = (0.5) * 23 kg m^2 * (140 rad/s)^2
KE = 226960 J

So, the kinetic energy of the turbine at the end of the 27 s is 226960 J.

To find the answers to these questions, we will use the basic equations of rotational motion. These equations describe the relationship between angular velocity (ω), angular acceleration (α), moment of inertia (I), torque (τ), time (t), and angle turned (θ). Here are the equations we will use:

1. ω = α * t (Equation 1)
This equation relates angular velocity (ω), angular acceleration (α), and time (t).

2. τ = I * α (Equation 2)
This equation relates torque (τ), moment of inertia (I), and angular acceleration (α).

3. θ = 0.5 * α * t^2 (Equation 3)
This equation relates angle turned (θ), angular acceleration (α), and time (t).

4. Work (W) = τ * θ (Equation 4)
This equation relates work (W), torque (τ), and angle turned (θ).

5. Kinetic energy (KE) = 0.5 * I * ω^2 (Equation 5)
This equation relates kinetic energy (KE), moment of inertia (I), and angular velocity (ω).

Now let's solve each part of the question step by step:

(a) Find the angular acceleration:
Using Equation 1, we have:
140 rad/s = α * 27 s
Solving for α, we get:
α = 140 rad/s / 27 s ≈ 5.185 rad/s^2

(b) Find the net torque required:
Using Equation 2, we have:
τ = I * α
τ = 23 kg m^2 * 5.185 rad/s^2 ≈ 119.255 Nm

(c) Find the angle turned through in 27 s:
Using Equation 3, we have:
θ = 0.5 * α * t^2
θ = 0.5 * 5.185 rad/s^2 * (27 s)^2 ≈ 1839.352 rad

(d) Find the work done by the net torque:
Using Equation 4, we have:
W = τ * θ
W = 119.255 Nm * 1839.352 rad ≈ 219,250 J

(e) Find the kinetic energy of the turbine at the end of the 27 s:
Using Equation 5, we have:
KE = 0.5 * I * ω^2
KE = 0.5 * 23 kg m^2 * (140 rad/s)^2 ≈ 225,400 J

Note: In these calculations, we assumed that the angular velocity was at the end of the 27 s time interval.

You should not have to ask this question. I am sure exactly how to do it is in your text book.

(a) alpha = change in speed (140)/change in time (27)

(b) Torque = I alpha = 23 * alpha

(c) A = 0 + 0 t + (1/2) alpha t^2

(d)and (e) work done = change in kinetic energy
= (1/2) I w^2
where w = alpha t = 27 alpha