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Chapter 10: Problem 55

The turbine and associated rotating parts of a jet engine have a total moment of inertia of \(25 \mathrm{~kg} \mathrm{~m}^{2}\). The turbine is accelerated uniformly from rest to an angular speed of \(150 \mathrm{rad} / \mathrm{s}\) in a time of \(25 \mathrm{~s}\). Find a) the angular acceleration, b) the net torque required, c) the angle turned through in \(25 \mathrm{~s}\) d) the work done by the net torque, and e) the kinetic energy of the turbine at the end of the \(25 \mathrm{~s}\).

Short Answer

Expert verified
Question: Calculate the angular acceleration, net torque, angle turned in 25 seconds, work done by the net torque, and the kinetic energy of the turbine at the end of the 25 seconds for a rotating turbine given the moment of inertia of 25 kg · m² and final angular speed of 150 rad/s. Assume the turbine starts from rest. Answer: The angular acceleration is 6 rad/s², the net torque required is 150 N·m, the angle turned through in 25 seconds is 1875 rad, the work done by the net torque is 281,250 J, and the kinetic energy of the turbine at the end of the 25 seconds is 281,250 J.

Step by step solution

01

Find the angular acceleration (α)

We are given the initial angular speed (ω₀) which is 0, the final angular speed (ω), and the time interval (∆t). We can use the formula: ω = ω₀ + α∆t Rearranging to find α, we get: α = (ω - ω₀) / ∆t Now, plug in the given values: α = (150 rad/s - 0 rad/s) / 25 s
02

Calculate the angular acceleration.

By plugging the values into the formula, we get: α = 150 / 25 = 6 rad/s² The angular acceleration is 6 radians per second squared.
03

Find the net torque required (τ).

We can use the formula: τ = I * α Where τ is the torque and I is the moment of inertia. We're given I = 25 kg · m² and α = 6 rad/s².
04

Calculate the net torque required.

By plugging the values into the formula, we get: τ = 25 kg · m² · 6 rad/s² = 150 N·m The net torque required is 150 N·m.
05

Find the angle turned through in 25 seconds (θ).

Using the formula for angular displacement: θ = ω₀∆t + 0.5α(∆t)² Since the turbine starts from rest (ω₀ = 0), the formula simplifies to: θ = 0.5α(∆t)² We have α = 6 rad/s² and ∆t = 25 s.
06

Calculate the angle turned through in 25 seconds.

By inserting the values into the formula, we get: θ = 0.5 * 6 * (25)² = 0.5 * 6 * 625 = 1875 rad The angle turned in 25 seconds is 1875 radians.
07

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

We can use the formula for work done by torque: W = τ * θ Where τ is the torque and θ is the angle turned. We know τ = 150 N·m, and θ = 1875 rad.
08

Calculate the work done by the net torque.

By inserting the values into the formula, we get: W = 150 N·m * 1875 rad = 281250 J The work done by the net torque is 281,250 Joules.
09

Find the kinetic energy of the turbine at the end of the 25 seconds (K).

We can use the formula for kinetic energy of a rotating object: K = 0.5 * I * ω² Where I is the moment of inertia and ω is the final angular speed. We know I = 25 kg · m², and ω = 150 rad/s.
10

Calculate the kinetic energy of the turbine at the end of the 25 seconds.

By inserting the values into the formula, we get: K = 0.5 * 25 kg * m² * (150 rad/s)² = 0.5 * 25 * 22500 = 281250 J The kinetic energy of the turbine at the end of the 25 seconds is 281,250 Joules. From the steps above, we have found the following quantities for the rotating turbine: a) Angular acceleration: 6 rad/s² b) Net torque required: 150 N·m c) Angle turned through in 25 seconds: 1875 rad d) Work done by the net torque: 281,250 J e) Kinetic energy of the turbine at the end of the 25 seconds: 281,250 J

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