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Chapter 10: Problem 26

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in \(2.2 \mathrm{~h}\). (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 75 revimin, what is the tangential component of the linear acceleration of a flywheel particle that is \(50 \mathrm{~cm}\) from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Short Answer

Expert verified
(a) -1.136 rev/min虏; (b) 9915.76 revolutions; (c) -0.568 m/min虏; (d) 30.9 m/s虏.

Step by step solution

01

Convert Angular Velocity to Appropriate Units

The initial angular velocity is given as 150 revolutions per minute. To find the constant angular acceleration, we need the final angular velocity, which is 0 revolutions per minute. The stopping time is 2.2 hours. First, convert 2.2 hours to minutes:\[ 2.2 \text{ hours} \times 60 \frac{\text{minutes}}{\text{hour}} = 132 \text{ minutes} \]Now we have:- Initial angular velocity, \( \omega_i = 150 \text{ rev/min} \)- Final angular velocity, \( \omega_f = 0 \text{ rev/min} \)- Time, \( t = 132 \text{ min} \)
02

Calculate Angular Acceleration

Angular acceleration \( \alpha \) can be calculated using the equation:\[ \omega_f = \omega_i + \alpha t \]Rearrange this for \( \alpha \):\[ \alpha = \frac{\omega_f - \omega_i}{t} \]Substitute the values:\[ \alpha = \frac{0 - 150}{132} = \frac{-150}{132} \approx -1.136 \text{ rev/min}^2 \]
03

Calculate Total Revolutions before Stopping

The formula to find the total revolutions before stopping when angular acceleration is constant is:\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]Substitute the known values from previous steps:\[ \theta = 150 \times 132 + \frac{1}{2} \times (-1.136) \times (132)^2 \]Simplifying,\[ \theta = 19800 - \frac{1}{2} \times 1.136 \times 17424 \]\[ \theta = 19800 - 9894.24 \approx 9915.76 \text{ revolutions} \]
04

Find Tangential Component of Linear Acceleration

Tangential acceleration \( a_t \) is given by:\[ a_t = r \cdot \alpha \]At the instant the velocity is 75 rev/min, the angular acceleration remains \( -1.136 \text{ rev/min}^2 \) (since it's constant). Convert 50 cm to meters: 0.5 m.Substituting the values:\[ a_t = 0.5 \times (-1.136) \approx -0.568 \text{ m/min}^2 \]
05

Calculate Net Linear Acceleration Magnitude

The net linear acceleration \( a \) is the combination of tangential \( a_t \) and radial \( a_r \) components. When the wheel is at 75 rev/min, convert this to rad/s:\[ \omega = 75 \times \frac{2\pi}{60} = 7.854 \text{ rad/s} \]Radial acceleration \( a_r \) is given by:\[ a_r = \omega^2 \cdot r \]Substitute the known values:\[ a_r = (7.854)^2 \times 0.5 \approx 30.9 \text{ m/s}^2 \]Now compute the net acceleration using:\[ a = \sqrt{a_t^2 + a_r^2} \]\[ a = \sqrt{(-0.568)^2 + (30.9)^2} \approx 30.9 \text{ m/s}^2 \]

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Key Concepts

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

Tangential Acceleration
In rotational motion, tangential acceleration refers to the linear acceleration that is tangent to the circular path of an object. This type of acceleration occurs because of changes in the speed of the object moving along the circular path. For a particle on a rotating object, such as a flywheel, tangential acceleration can be calculated using the formula:\[a_t = r \cdot \alpha\]where:
  • \( a_t \) is the tangential acceleration,
  • \( r \) is the radius of the circular path (distance from the axis of rotation), and
  • \( \alpha \) is the angular acceleration.
In the context of the flywheel, as it decelerates, the angular acceleration is constant and negative due to friction, making the tangential acceleration also constant but negative. This negative sign indicates that the velocity in the circular path is decreasing.
Radial Acceleration
Radial acceleration (also known as centripetal acceleration) is the component of acceleration directed towards the center of the circular path of motion. It is responsible for changing the direction of the velocity vector of an object moving in a circle, keeping it on that path without changing its speed along the circular direction. The formula for radial acceleration is given by:\[a_r = \omega^2 \cdot r\]where:
  • \( a_r \) is the radial acceleration,
  • \( \omega \) is the angular velocity in radians per second, and
  • \( r \) is the radius of the circular path.
In our exercise, this acceleration ensures that the particles of the flywheel continue to follow the circular path as long as the wheel is still in motion. Given the rotational velocity of 75 revolutions per minute, converting this to radians per second allows for the precise calculation of radial acceleration, emphasizing the forces acting perpendicular to the path.
Revolutions
Revolutions in rotational motion refer to the number of complete turns an object makes around a fixed axis. This idea is central to understanding how far or how much an object has rotated over time. For the exercise, determining how many revolutions the flywheel underwent before coming to a stop involves knowing the initial rotational speed, the time over which it slows down, and the constant angular acceleration.The formula to calculate the total number of revolutions is:\[\theta = \omega_i t + \frac{1}{2} \alpha t^2\]where:
  • \( \theta \) is the total number of revolutions,
  • \( \omega_i \) is the initial angular velocity, and
  • \( \alpha \) is the angular acceleration, which is constant.
In this case, the flywheel starts at 150 revolutions per minute and slows down over 132 minutes due to negative angular acceleration, resulting in a final calculation of revolutions before the wheel finally stops.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves relative to another point, typically the center of the circular path. It is expressed in units such as revolutions per minute (rev/min) or radians per second (rad/s).In the problem, the flywheel's initial angular velocity is 150 rev/min, a critical factor in determining how long the wheel will take to stop and how many revolutions it will make before halting.Angular velocity is related to tangential speed through the formula for circular objects:\[v = \omega \cdot r\]where:
  • \( v \) is the tangential speed,
  • \( \omega \) is the angular velocity, and
  • \( r \) is the radius of the path.
By converting angular velocity to radians per second, one can more easily find related quantities like radial acceleration, further illuminating the dynamics of rotating systems like the flywheel in our scenario.

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Most popular questions from this chapter

Starting from rest at \(t=0\), a wheel undergoes a constant angular acceleration. When \(t=2.0 \mathrm{~s}\), the angular velocity of the wheel is \(5.0 \mathrm{rad} / \mathrm{s}\). The acceleration continues until \(t=20 \mathrm{~s}\), when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=40 \mathrm{~s}\) ?

A uniform cylinder of radius \(10 \mathrm{~cm}\) and mass \(20 \mathrm{~kg}\) is mounted so as to rotate freely about a horizontal axis that is parallel to and \(5.0 \mathrm{~cm}\) from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

An automobile crankshaft transfers energy from the engine to the axle at the rate of \(100 \mathrm{hp}(=74.6 \mathrm{~kW})\) when rotating at \(\mathrm{a}\) speed of 1800 rev/min. What torque (in newton-meters) does the crankshaft deliver?

A vinyl record on a turntable rotates at \(33 \frac{1}{3}\) revimin. (a) What is its angular speed in radians per second? What is the linear speed of a point on the record (b) \(15 \mathrm{~cm}\) and (c) \(7.4 \mathrm{~cm}\) from the turntable axis?

The angular position of a point on the rim of a rotating wheel is given by \(\theta=4.0 t-3.0 t^{2}+t^{3}\), where \(\theta\) is in radians and \(t\) is in seconds. What are the angular velocities at (a) \(t=2.0 \mathrm{~s}\) and \((\mathrm{b}) t=4.0 \mathrm{~s} ?\) (c) What is the average angular acceleration for the time interval that begins at \(t=2.0 \mathrm{~s}\) and ends at \(t=4.0 \mathrm{~s}\) ? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?
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