Question

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

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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

Key concepts

Angular Acceleration

Angular acceleration symbolizes the rate of change of angular velocity over time. When you think of straight-line motion, acceleration indicates how quickly an object speeds up or slows down. Similarly, angular acceleration ($\alpha$) explains how swiftly something spinning, like a wheel or turbine, gains or loses speed. To determine this, use the formula: $\alpha =\frac{\omega -{\omega }_0}{\mathrm{\Delta }t}$, where $\omega$ is the final angular velocity, ${\omega }_0$ is the initial angular velocity, and $\mathrm{\Delta }t$ is the time duration.

• In our exercise, the turbine started from rest (${\omega }_0=0$
• It reached an angular speed of $150\text{ rad/s}$ in $25\text{ seconds}$.
• Using the formula, you get: $\alpha =\frac{150}{25}=6{\text{ rad/s}}^2$.

This means the turbine's speed was increasing by 6 radians every second, squared.

Moment of Inertia

The moment of inertia, often denoted as $I$, plays a role similar to mass in linear motion but for rotational dynamics. It quantifies how spread out an object's mass is, which influences how easily it spins. Think of it as rotational mass. An object with a high moment of inertia requires more effort to start or stop spinning.

• A formula for moment of inertia is $I=\sum m_i{r}_i^2$, which adds up each piece of mass $m_i$ times its distance from the rotation axis squared $r_i^2$.
• In our example problem, the total $I$ for the engine parts was $25\text{ kg}\cdot {\text{m}}^2$.

This tells us how the mass of the turbine is distributed in relation to its spin.

Torque

Torque can be likened to the "twist" applied to make an object rotate. It occurs when you use force at some distance from the rotation axis, much like pushing a door open. The formula for torque ($\tau$) is:$\tau =I\cdot \alpha$.

• With a moment of inertia $I=25\text{ kg}\cdot {\text{m}}^2$ and angular acceleration $\alpha =6{\text{ rad/s}}^2$, the net torque required is $150\text{ N}\cdot \text{m}$.

This means you need 150 Newton-meters of rotational force to get the turbine to speed up at the given rate.

Kinetic Energy

Kinetic energy in the context of rotational motion represents the energy an object has due to its spinning. For a rotating system like a turbine, the kinetic energy ($K$) is calculated with:$K=\frac{1}{2}I{\omega }^2$.

• Here, $I=25\text{ kg}\cdot {\text{m}}^2$ and $\omega =150\text{ rad/s}$.
• The calculation gives $K=281250\text{ J}$, meaning the turbine has 281,250 Joules of energy at the end of the spin-up period.

This energy could then be used to do work, like powering part of the jet engine's operation.

Work Done

Work done due to torque in rotational systems is comparable to work done by force in linear systems. It's about how the applied torque moves the object over a certain angular distance. The formula to find the work done ($W$) is:$W=\tau \cdot \theta$.

• In the exercise, $\theta =1875\text{ rad}$ (angle turned in 25 seconds).
• With torque $\tau =150\text{ N}\cdot \text{m}$, the work done equaled $281250\text{ J}$.

This amount of energy, or work, was necessary to get the turbine rotating to its target speed within the given time.