/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 The flywheel of an engine has mo... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
A constant torque of approximately 8.38 N·m is required.

Step by step solution

01

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 \(\omega = \frac{400 \times 2\pi}{60} \approx 41.89 \text{ rad/s}\).
02

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, \(\alpha = \frac{41.89}{8.00} \approx 5.24 \text{ rad/s}^2\).
03

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 \(\tau \approx 8.38 \text{ N} \cdot \text{m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • Cylindrical objects: Moment of inertia is calculated using the formula for cylinders.
  • Rectangular beams: Utilize the corresponding formula for beams.
Understanding the moment of inertia makes it easier to predict how an object will behave when acted upon by a torque.
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.
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.

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Most popular questions from this chapter

You are designing a slide for a water park. In a sitting position, park guests slide a vertical distance \(h\) down the waterslide, which has negligible friction. When they reach the bottom of the slide, they grab a handle at the bottom end of a 6.00-m-long uniform pole. The pole hangs vertically, initially at rest. The upper end of the pole is pivoted about a stationary, frictionless axle. The pole with a person hanging on the end swings up through an angle of 72.0\(^\circ\), and then the person lets go of the pole and drops into a pool of water. Treat the person as a point mass. The pole's moment of inertia is given by \(I = \frac{1}{3} ML^2\), where \(L =\) 6.00 m is the length of the pole and \(M =\) 24.0 kg is its mass. For a person of mass 70.0 kg, what must be the height h in order for the pole to have a maximum angle of swing of 72.0\(^\circ\) after the collision?

One force acting on a machine part is \(\overrightarrow{F} = (-5.00 N)\hat{\imath} + (4.00 N)\hat{\jmath}\). The vector from the origin to the point where the force is applied is \(\overrightarrow{r} = (-0.450 m)\hat{\imath} + (0.150 m)\hat{\jmath}\). (a) In a sketch,show \(\overrightarrow{r}, \overrightarrow{F},\) and the origin. (b) Use the right- hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

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 rest, and then a constant force \(F =\) 30.0 N is applied tangent to the rim of the disk. (a) What is the magnitude \(v\) of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution? (b) What is the magnitude \(a\) of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolution?

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 kg, traveling perpendicular to the door at 12.0 m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

(a) Compute the torque developed by an industrial motor whose output is 150 kW at an angular speed of 4000 rev/min. (b) A drum with negligible mass, 0.400 m in diameter, is attached to the motor shaft, and the power output of the motor is used to raise a weight hanging from a rope wrapped around the drum. How heavy a weight can the motor lift at constant speed? (c) At what constant speed will the weight rise?

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