The fly wheel is an important mechanical element in machines that is used to increase the moment of inertia of the machine and store rotational kinetic energy. A fly wheel of an engine starts rotating about a fixed axis from rest to 102 rad/s in 6.3 seconds. The wheel has a radius of 0.5 meters and mass of 88 kg. Calculate the net torque acting on the fly wheel during the time interval t=0; 6.3 s measured in Nm (Enter the number only).
The fly wheel is an important mechanical element in machines that is used to increase the moment of inertia of the machine and store rotational kinetic energy. A fly wheel of an engine starts rotating about a fixed axis from rest to 102 rad/s in 6.3 seconds. The wheel has a radius of 0.5 meters and mass of 88 kg. Calculate the net torque acting on the fly wheel during the time interval t=0; 6.3 s measured in Nm (Enter the number only).
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Chapter1: Units, Trigonometry. And Vectors
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![### Understanding Flywheels in Mechanical Systems
The flywheel is an essential mechanical element in machines designed to increase the moment of inertia and store rotational kinetic energy. These components are crucial for various machines, providing stability and energy storage that can be used to maintain a constant speed despite varying loads.
#### Example Calculation
Consider a flywheel integrated into an engine. This flywheel begins rotating from a stationary position and reaches an angular velocity of 102 rad/s within a time interval of 6.3 seconds.
**Specifications of the Flywheel:**
- **Radius:** 0.5 meters
- **Mass:** 88 kilograms
To find the net torque (\( \tau \)) acting on this flywheel during the specified time interval, we can use the following steps:
1. **Calculate Angular Acceleration (\( \alpha \))**
Angular acceleration is given by the change in angular velocity (\( \Delta \omega \)) over time (\( \Delta t \)):
\[ \alpha = \frac{\Delta \omega}{\Delta t} = \frac{102 \, \text{rad/s}}{6.3 \, \text{s}} \approx 16.19 \, \text{rad/s}^2 \]
2. **Moment of Inertia ( \( I \) )**
For a solid flywheel (solid disc), the moment of inertia about its central axis is:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass and \( r \) is the radius.
Substituting the given values:
\[ I = \frac{1}{2} \times 88 \, \text{kg} \times (0.5 \, \text{m})^2 \]
\[ I = \frac{1}{2} \times 88 \times 0.25 \]
\[ I = 11 \, \text{kg} \cdot \text{m}^2 \]
3. **Calculate Net Torque (\( \tau \))**
The net torque is given by the product of the moment of inertia and the angular acceleration:
\[ \tau = I \alpha \]
Substituting the known values:
\[ \tau = 11 \, \text{kg}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcde4937d-e883-446c-8941-8d2ade4d99b0%2Fa0bd6ec0-8730-4da3-9a95-f791456e72b3%2Fpx0xc4p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Flywheels in Mechanical Systems
The flywheel is an essential mechanical element in machines designed to increase the moment of inertia and store rotational kinetic energy. These components are crucial for various machines, providing stability and energy storage that can be used to maintain a constant speed despite varying loads.
#### Example Calculation
Consider a flywheel integrated into an engine. This flywheel begins rotating from a stationary position and reaches an angular velocity of 102 rad/s within a time interval of 6.3 seconds.
**Specifications of the Flywheel:**
- **Radius:** 0.5 meters
- **Mass:** 88 kilograms
To find the net torque (\( \tau \)) acting on this flywheel during the specified time interval, we can use the following steps:
1. **Calculate Angular Acceleration (\( \alpha \))**
Angular acceleration is given by the change in angular velocity (\( \Delta \omega \)) over time (\( \Delta t \)):
\[ \alpha = \frac{\Delta \omega}{\Delta t} = \frac{102 \, \text{rad/s}}{6.3 \, \text{s}} \approx 16.19 \, \text{rad/s}^2 \]
2. **Moment of Inertia ( \( I \) )**
For a solid flywheel (solid disc), the moment of inertia about its central axis is:
\[ I = \frac{1}{2} m r^2 \]
where \( m \) is the mass and \( r \) is the radius.
Substituting the given values:
\[ I = \frac{1}{2} \times 88 \, \text{kg} \times (0.5 \, \text{m})^2 \]
\[ I = \frac{1}{2} \times 88 \times 0.25 \]
\[ I = 11 \, \text{kg} \cdot \text{m}^2 \]
3. **Calculate Net Torque (\( \tau \))**
The net torque is given by the product of the moment of inertia and the angular acceleration:
\[ \tau = I \alpha \]
Substituting the known values:
\[ \tau = 11 \, \text{kg}
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---
**Explanation of Graphs or Diagrams:**
- The image provided shows the flywheel in question. It is a mechanical device with a large, solid, circular design typically used to store rotational kinetic energy. The flywheel pictured has distinctive red spokes and is mounted on an engine, as described in the problem prompt. This image helps to visually identify the type of component being discussed.
To solve the problem, use the following formula for angular acceleration (α):
\[ \alpha = \frac{\Delta \omega}{\Delta t} \]
where:
- \( \Delta \omega \) is the change in angular velocity,
- \( \Delta t \) is the change in time.
Given:
- Initial angular velocity, \( \omega_i = 0 \) rad/s (starts from rest),
- Final angular velocity, \( \omega_f = 102 \) rad/s,
- Time taken, \( \Delta t = 6.3 \) seconds,
Substitute these values into the formula:
\[ \alpha = \frac{\omega_f - \omega_i}{\Delta t} \]
\[ \alpha = \frac{102 \text{ rad/s} - 0 \text{ rad/s}}{6.3 \text{ s}} \]
\[ \alpha = \frac{102}{6.3} \]
\[ \alpha \approx 16.19 \text{ rad/s}^2 \]
Therefore, the magnitude of the average angular acceleration is approximately **16.19 rad/s²**.](https://content.bartleby.com/qna-images/question/5d107344-fc6e-4fa9-af9b-857bb71a0099/10c81fe1-1155-43ef-8371-02f1e609f634/kb1ijs_processed.jpeg)
Transcribed Image Text:### Question 2-a:
The flywheel is an important mechanical element in many machines used to increase the moment of inertia of the machine and to store rotational kinetic energy. A flywheel of an engine starts rotating about a fixed axis from rest to 102 rad/s in 6.3 seconds. The wheel has a radius of 0.5 meters and a mass of 88 kg. Calculate the magnitude of the average angular acceleration measured in rad/s². (Enter the number only).
---
**Explanation of Graphs or Diagrams:**
- The image provided shows the flywheel in question. It is a mechanical device with a large, solid, circular design typically used to store rotational kinetic energy. The flywheel pictured has distinctive red spokes and is mounted on an engine, as described in the problem prompt. This image helps to visually identify the type of component being discussed.
To solve the problem, use the following formula for angular acceleration (α):
\[ \alpha = \frac{\Delta \omega}{\Delta t} \]
where:
- \( \Delta \omega \) is the change in angular velocity,
- \( \Delta t \) is the change in time.
Given:
- Initial angular velocity, \( \omega_i = 0 \) rad/s (starts from rest),
- Final angular velocity, \( \omega_f = 102 \) rad/s,
- Time taken, \( \Delta t = 6.3 \) seconds,
Substitute these values into the formula:
\[ \alpha = \frac{\omega_f - \omega_i}{\Delta t} \]
\[ \alpha = \frac{102 \text{ rad/s} - 0 \text{ rad/s}}{6.3 \text{ s}} \]
\[ \alpha = \frac{102}{6.3} \]
\[ \alpha \approx 16.19 \text{ rad/s}^2 \]
Therefore, the magnitude of the average angular acceleration is approximately **16.19 rad/s²**.
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