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Chapter 10: Problem 9

The flywheel of an engine has moment of inertia 2.50 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 \(\mathrm{rev} / \mathrm{min}\) in 8.00 \(\mathrm{s}\) starting from rest?

Short Answer

Expert verified
A torque of 13.1 N路m is required.

Step by step solution

01

Convert Angular Speed Units

First, we need to convert the final angular speed from revolutions per minute to radians per second. Since there are \(2\pi\) radians in one revolution and 60 seconds in a minute, we use the conversion \(\omega = 400 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{400 \times 2\pi}{60} \, \text{rad/s}\). This simplifies to \(\omega = \frac{400 \times 2\pi}{60} \, \text{rad/s} \approx 41.89\,\text{rad/s}\).
02

Calculate Angular Acceleration

Since the object starts from rest, the initial angular speed \(\omega_0 = 0\). The angular acceleration \(\alpha\) can be found using \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular speed and \(\Delta t\) is the time interval. Thus, \(\alpha = \frac{41.89 \, \text{rad/s} - 0 \, \text{rad/s}}{8 \, \text{s}} = \frac{41.89}{8} \, \text{rad/s}^2 \approx 5.24\,\text{rad/s}^2\).
03

Use Torque Formula

Torque \(\tau\) is related to angular acceleration by the equation \(\tau = I \cdot \alpha\), where \(I\) is the moment of inertia. Here, \(I = 2.50 \, \text{kg} \cdot \text{m}^2\). Using the previously calculated \(\alpha = 5.24 \, \text{rad/s}^2\), we find \(\tau = 2.50 \, \text{kg} \cdot \text{m}^2 \times 5.24 \, \text{rad/s}^2 = 13.1 \, \text{N} \cdot \text{m}\).

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Key Concepts

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

Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. Simply put, it tells us how quickly an object is speeding up or slowing down as it rotates. The formula for angular acceleration \(\alpha\) is given by:\[\alpha = \frac{\Delta \omega}{\Delta t}\]where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time it takes for this change.
In the case of our exercise, since the flywheel starts from rest, the initial angular speed is zero. After 8 seconds, it reaches a certain speed. By inserting these values into the formula, you can find the angular acceleration.
Remember that if an object speeds up, the angular acceleration is positive. Conversely, if the object slows down, the angular acceleration is negative. Understanding this concept helps us predict how fast a rotating object will spin at any given moment.
Moment of Inertia
Moment of inertia is a measure of an object's resistance to changes in its rotation. Think of it as the rotational "mass" of the object. Just as more massive objects are harder to move, objects with a high moment of inertia are harder to spin or stop spinning.
Mathematically, the moment of inertia \(I\) depends on the mass distribution of the object and is expressed in \(\text{kg} \cdot \text{m}^2\). In simple systems, you often see formulas like \(I = mr^2\) for point masses at a distance \(r\) from the rotation axis.
In this exercise, the flywheel has a moment of inertia of \(2.50 \, \text{kg} \cdot \text{m}^2\). This tells us how much torque is needed to achieve a certain angular acceleration. Combining moment of inertia with angular acceleration, you can calculate the torque required to make the flywheel spin.
Angular Speed Conversion
Angular speed or angular velocity is typically expressed in revolutions per minute (RPM) or radians per second (rad/s). Converting between these units is essential for solving many physics problems involving rotation.
Each revolution represents a full circle, equivalent to \(2\pi\) radians. Therefore, to convert angular speed from RPM to rad/s, you multiply by \(2\pi\) and divide by 60 (seconds per minute).
For the flywheel in the exercise, its speed was originally given as \(400\, \text{rev/min}\). By applying the conversion method, you end up with \(\approx 41.89\, \text{rad/s}\). This conversion step is crucial to ensure all units are consistent when calculating further properties like angular acceleration and torque.
Understanding the conversions helps simplify complex rotational problems by standardizing measurements into common units.

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Most popular questions from this chapter

A block with mass \(m=5.00 \mathrm{kg}\) slides down a a surface inclined \(36.9^{\circ}\) to the horizontal (Fig. P10.70). The coefficient of kinetic friction is \(0.25 .\) A string attached to the block is wrapped around a flywheel on a fixed axis at \(O .\) The flywheel has mass 25.0 \(\mathrm{kg}\) and moment of inertia 0.500 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200 \(\mathrm{m}\) from that axis. (a) What is the acceleration of the block down the plane? (b) What is the tension in the string?

A target in a shooting gallery consists of a vertical square wooden board, 0.250 \(\mathrm{m}\) on a side and with mass 0.750 \(\mathrm{kg}\) , that pivots on a horizontal axis along its top edge. The board is struck face-on at its center by a bullet with mass 1.90 \(\mathrm{g}\) that is traveling at 360 \(\mathrm{m} / \mathrm{s}\) and that remains embedded in the board. (a) What is the angular speed of the board just after the bullet's impact? (b) What maximum height above the equilibrium position does the center of the board reach before starting to swing down again? (c) What minimum bullet speed would be required for the board to swing all the way over after impact?

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of 7.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; (b) the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

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

A small block with mass 0.250 \(\mathrm{kg}\) is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.42). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 \(\mathrm{N} .\) What is the radius of the circle when the string breaks?
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