Problem 9
Question
The flywheel of an engine has moment of inertia 1.60 kg \(\cdot\) m\(^2\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8.00 s, starting from rest?
Step-by-Step Solution
Verified Answer
A constant torque of approximately 8.38 N·m is required.
1Step 1: Convert Angular Speed to Radians per Second
Start by converting the final angular speed from revolutions per minute (rev/min) to radians per second (rad/s). Recall that 1 revolution is equal to \(2\pi\) radians and 1 minute is 60 seconds. So, \(400 \text{ rev/min} = \frac{400 \times 2\pi}{60} \text{ rad/s}\). This simplifies to approximately \(\omega = \frac{400 \times 2\pi}{60} \approx 41.89 \text{ rad/s}\).
2Step 2: Calculate Angular Acceleration
Next, find the angular acceleration (\(\alpha\)). Since the flywheel starts from rest, its initial angular speed is 0. Use the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega = 41.89 \text{ rad/s}\) and \(\Delta t = 8.00 \text{ s}\). Thus, \(\alpha = \frac{41.89}{8.00} \approx 5.24 \text{ rad/s}^2\).
3Step 3: Use Torque Equation
Torque (\(\tau\)) is related to angular acceleration and moment of inertia (\(I\)) by the equation \(\tau = I\alpha\). Given that \(I = 1.60 \text{ kg} \cdot \text{m}^2\), substitute \(\alpha = 5.24 \text{ rad/s}^2\) into the equation to find \(\tau = 1.60 \times 5.24\). Calculate this to get \(\tau \approx 8.38 \text{ N} \cdot \text{m}\).
Key Concepts
Moment of InertiaAngular AccelerationTorque Calculation
Moment of Inertia
Moment of Inertia is a crucial concept when dealing with rotational motion, similar to mass in linear motion. It essentially measures how much torque is required for a desired angular acceleration about a rotational axis. Think of it as the rotational equivalent of mass, reflecting how much "heaviness" the object feels while rotating.
For instance, a heavier flywheel or one with mass concentrated further from the axis of rotation will have a higher moment of inertia. In the exercise above, the flywheel's moment of inertia is given as 1.60 kg·m². This indicates how resistant it is to changes in its rotational speed given a torque is applied.
Calculators for moment of inertia often depend on both the shape of the object and the distribution of mass across the object. Examples include:
For instance, a heavier flywheel or one with mass concentrated further from the axis of rotation will have a higher moment of inertia. In the exercise above, the flywheel's moment of inertia is given as 1.60 kg·m². This indicates how resistant it is to changes in its rotational speed given a torque is applied.
Calculators for moment of inertia often depend on both the shape of the object and the distribution of mass across the object. Examples include:
- Cylindrical objects: Moment of inertia is calculated using the formula for cylinders.
- Rectangular beams: Utilize the corresponding formula for beams.
Angular Acceleration
Angular Acceleration describes how the angular velocity of a rotating object changes with time. It's the rotational counterpart to linear acceleration and is expressed in radians per second squared (rad/s²).
In the context of the exercise, the flywheel starts from rest, so the initial angular velocity is zero. To find angular acceleration, we use the formula: \\(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time taken for this change.
This is crucial because, in real-world applications like engine flywheels, rapid changes need to be efficiently managed. The provided solution calculates \(\alpha\) as approximately 5.24 rad/s², indicating how rapidly the flywheel's speed increases, highlighting the link between velocity and time under constant conditions.
In the context of the exercise, the flywheel starts from rest, so the initial angular velocity is zero. To find angular acceleration, we use the formula: \\(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time taken for this change.
This is crucial because, in real-world applications like engine flywheels, rapid changes need to be efficiently managed. The provided solution calculates \(\alpha\) as approximately 5.24 rad/s², indicating how rapidly the flywheel's speed increases, highlighting the link between velocity and time under constant conditions.
Torque Calculation
Torque is the force that causes an object to rotate around an axis. It's akin to a twist or a turn that you apply to get an object spinning. In rotational motion, torque is related to both angular acceleration and moment of inertia.
Mathematically, torque (\(\tau\)) is calculated using the equation: \(\tau = I\alpha\), depicting the direct relationship between the applied torque, moment of inertia, and resulting angular acceleration. In the exercise, the torque required was approximately 8.38 N·m. This value indicates the strength of the rotational force needed to accelerate the flywheel from rest to the desired angular speed.
Torque is measured in newton-meters (N·m) and plays an essential role in designing mechanical systems, where controlled rotational acceleration is necessary for efficiency. Understanding how to calculate and apply torque allows for more effective design and operation of machinery involving rotational elements.
Mathematically, torque (\(\tau\)) is calculated using the equation: \(\tau = I\alpha\), depicting the direct relationship between the applied torque, moment of inertia, and resulting angular acceleration. In the exercise, the torque required was approximately 8.38 N·m. This value indicates the strength of the rotational force needed to accelerate the flywheel from rest to the desired angular speed.
Torque is measured in newton-meters (N·m) and plays an essential role in designing mechanical systems, where controlled rotational acceleration is necessary for efficiency. Understanding how to calculate and apply torque allows for more effective design and operation of machinery involving rotational elements.
Other exercises in this chapter
Problem 6
A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \ma
View solution Problem 8
A uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at
View solution Problem 10
A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on t
View solution Problem 11
A machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3.00 cm. It is spinning about a frictionless axle through its center, but at o
View solution