/*! 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 23 A flywheel with a diameter of \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A flywheel with a diameter of \(1.20 \mathrm{~m}\) is rotating at an angular speed of 200 rev \(/\) min. (a) What is the angular speed of the flywheel in radians per second? (b) What is the linear speed of a point on the rim of the flywheel? (c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel's angular speed to 1000 rev/min in 60.0 s? (d) How many revolutions does the wheel make during that \(60.0 \mathrm{~s} ?\)

Short Answer

Expert verified
(a) 20.94 rad/s (b) 12.56 m/s (c) 13.33 rev/min² (d) 1000 revolutions

Step by step solution

01

Convert angular speed to rad/s

To convert the flywheel's angular speed from revolutions per minute (rev/min) to radians per second (rad/s), use the conversion factors: 1 revolution = 2Ï€ radians and 1 minute = 60 seconds. The given speed is 200 rev/min. Therefore:\[\omega = 200 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{\text{rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{400\pi}{60} \text{ rad/s} \approx 20.94 \text{ rad/s}.\]
02

Calculate linear speed of a point on the rim

The linear speed \(v\) of a point on the rim can be calculated using the formula \(v = \omega r\), where \(r\) is the radius of the flywheel. The diameter is 1.20 m, so the radius \(r = \frac{1.20}{2} = 0.60\) m. Using \(\omega = 20.94 \text{ rad/s}\) from Step 1:\[v = 20.94 \text{ rad/s} \times 0.60 \text{ m} = 12.56 \text{ m/s}.\]
03

Determine angular acceleration

We need to find angular acceleration \(\alpha\) in rev/min² to increase the angular speed from 200 rev/min to 1000 rev/min in 60 seconds. Using the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega = 1000 - 200 = 800\) rev/min and \(\Delta t = 60\) s:\[\alpha = \frac{800 \text{ rev/min}}{60 \text{ s}} \approx 13.33 \text{ rev/min}^2.\]
04

Calculate number of revolutions

The number of revolutions \(N\) during acceleration can be found using:\[N = \omega_i t + \frac{1}{2} \alpha t^2,\]where \(\omega_i = 200 \text{ rev/min}\), converted to rev/s is \(\frac{200}{60}\), and \(t = 60\) s. \(\alpha = 13.33 \text{ rev/min}^2\), converted to rev/s² is \(\frac{13.33}{60}\).Substitute values:\[N = \left(\frac{200}{60} \text{ rev/s}\right) \times 60 \text{ s} + \frac{1}{2} \times \left(\frac{13.33}{60} \text{ rev/s}^2\right) \times (60 \text{ s})^2 = 600 + 400 = 1000 \text{ revolutions}.\]
05

Conclusion

The calculations show the necessary conversions and mathematical operations to solve each part of the exercise. The final answer to each part is extracted clearly.

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Key Concepts

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

Angular Speed
Angular speed is a measure of how quickly an object rotates or revolves relative to another point. It is typically expressed in units such as revolutions per minute (rev/min) or radians per second (rad/s). To convert angular speed from rev/min to rad/s, use the conversion factors: 1 revolution equals \(2\pi\) radians, and 1 minute equals 60 seconds. For instance, a flywheel rotating at 200 rev/min would have an angular speed of \(\omega = \frac{400\pi}{60} \approx 20.94\) rad/s. This conversion is essential because radians per second is the standard unit in physics for angular speed. Converting to this unit allows for consistent mathematical operations across different aspects of rotational motion.
Linear Speed
Linear speed refers to how fast a point travels along a path. In rotational motion, the linear speed of a point on the edge of a rotating object can be derived from its angular speed and the radius of rotation. This relationship is expressed by the formula \(v = \omega r\). The linear speed \(v\) of a point on the rim of a 1.20 m diameter flywheel, rotating at \(\omega = 20.94 \, \text{rad/s}\), is calculated as \(v = 20.94 \, \text{rad/s} \times 0.60 \, \text{m} = 12.56 \, \text{m/s}\). This linear speed signifies how fast the point moves along its circular path as a result of the flywheel's rotation.
Angular Acceleration
Angular acceleration is the rate of change of angular speed over time. It describes how quickly an object's rotation is speeding up or slowing down. To find angular acceleration when increasing a flywheel's speed from 200 rev/min to 1000 rev/min over 60 seconds, use the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\). Here, \(\Delta \omega = 800 \, \text{rev/min}\) and \(\Delta t = 60 \, \text{s}\). This yields \(\alpha \approx 13.33 \, \text{rev/min}^2\). This angular acceleration indicates how much the speed changes per minute squared, which can be used to predict future rotational speeds at specific times.
Rotational Kinematics
Rotational kinematics involve the motion of rotating objects without considering the forces causing the rotation. It is similar to linear kinematics but deals with angular variables. The kinematic equation \(N = \omega_i t + \frac{1}{2} \alpha t^2\) describes the number of revolutions \(N\) a rotating object makes over a time period \(t\). When using this equation, initial angular speed \(\omega_i\) and angular acceleration \(\alpha\) must be in consistent units such as revolutions or radians per second. For example, with \(\omega_i = \frac{200}{60} \, \text{rev/s}\) and \(\alpha = \frac{13.33}{60} \, \text{rev/s}^2\), the flywheel makes \(1000\) revolutions in 60 seconds. This comprehensive approach gives a clear understanding of how the object behaves as a whole.

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Most popular questions from this chapter

Two uniform solid cylinders, each rotating about its central (longitudinal) axis at \(235 \mathrm{rad} / \mathrm{s},\) have the same mass of \(1.25 \mathrm{~kg}\) but differ in radius. What is the rotational kinetic energy of (a) the smaller cylinder, of radius \(0.25 \mathrm{~m},\) and \((\mathrm{b})\) the larger cylinder, of radius \(0.75 \mathrm{~m} ?\)

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