/*! 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 29 The drive propeller of a ship st... [FREE SOLUTION] | 91Ó°ÊÓ

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The drive propeller of a ship starts from rest and accelerates at \(2.90 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2}\) for \(2.10 \times 10^{3} \mathrm{s} .\) For the next \(1.40 \times 10^{3} \mathrm{s}\) the propeller rotates at a constant angular speed. Then it decelerates at \(2.30 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2}\) until it slows (without reversing direction) to an angular speed of \(4.00 \mathrm{rad} / \mathrm{s}\). Find the total angular displacement of the propeller.

Short Answer

Expert verified
The total angular displacement of the propeller is \(2.04 \times 10^4\ \mathrm{rad}\).

Step by step solution

01

Calculating Angular Velocity After Acceleration

To determine the angular velocity after the initial acceleration phase, we use the formula for angular velocity: \( \omega = \omega_0 + \alpha \cdot t \), where \( \omega_0 = 0 \), \( \alpha = 2.90 \times 10^{-3} \ \mathrm{rad/s^2} \), and \( t = 2.10 \times 10^3 \ \mathrm{s} \). This gives: \[ \omega = 0 + (2.90 \times 10^{-3})(2.10 \times 10^3) = 6.09 \ \mathrm{rad/s} \].
02

Calculating Angular Displacement During Acceleration

The angular displacement during the acceleration is given by \( \theta = \omega_0 \cdot t + \frac{1}{2} \alpha \cdot t^2 \). Since \( \omega_0 = 0 \), this simplifies to \( \theta = \frac{1}{2} \alpha \cdot t^2 \). Substituting \( \alpha = 2.90 \times 10^{-3} \ \mathrm{rad/s^2} \) and \( t = 2.10 \times 10^3 \ \mathrm{s} \): \[ \theta = \frac{1}{2} \times 2.90 \times 10^{-3} \times (2.10 \times 10^3)^2 = 6.38 \times 10^3 \ \mathrm{rad} \].
03

Calculating Angular Displacement During Constant Speed

During the phase of constant angular velocity, the angular displacement \( \theta \) is calculated as \( \theta = \omega \cdot t \) with \( \omega = 6.09 \ \mathrm{rad/s} \) and \( t = 1.40 \times 10^3 \ \mathrm{s} \): \[ \theta = 6.09 \times 1.40 \times 10^3 = 8.53 \times 10^3 \ \mathrm{rad} \].
04

Calculating Time for Deceleration

Using the equation \( \omega_f = \omega_i - \alpha \cdot t \) for deceleration, solve for \( t \), where \( \omega_f = 4.00 \ \mathrm{rad/s} \), \( \omega_i = 6.09 \ \mathrm{rad/s} \), and \( \alpha = 2.30 \times 10^{-3} \ \mathrm{rad/s^2} \): \[ 4.00 = 6.09 - 2.30 \times 10^{-3} \times t \]. Solving for \( t \) gives: \[ t = \frac{6.09 - 4.00}{2.30 \times 10^{-3}} = 907.83 \ \mathrm{s} \].
05

Calculating Angular Displacement During Deceleration

For the deceleration phase, use \( \theta = \omega_i \cdot t - \frac{1}{2} \alpha \cdot t^2 \), substituting \( \omega_i = 6.09 \ \mathrm{rad/s} \), \( t = 907.83 \ \mathrm{s} \), and \( \alpha = 2.30 \times 10^{-3} \ \mathrm{rad/s^2} \): \[ \theta = (6.09 \times 907.83) - \frac{1}{2} \times 2.30 \times 10^{-3} \times (907.83)^2 = 5.53 \times 10^3 \ \mathrm{rad} \].
06

Calculating Total Angular Displacement

Add the angular displacements from all phases: the acceleration, constant speed, and deceleration phases give \( 6.38 \times 10^3 \), \( 8.53 \times 10^3 \), and \( 5.53 \times 10^3 \) radians, respectively. The total angular displacement is: \[ 6.38 \times 10^3 + 8.53 \times 10^3 + 5.53 \times 10^3 = 2.04 \times 10^4 \ \mathrm{rad} \].

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Key Concepts

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

Angular Velocity
Angular velocity is an important concept when dealing with rotating objects. It refers to how fast an object is rotating around a point or axis. Think of it like how fast a clock's hands move around its center.
The unit for angular velocity is radians per second (\(\mathrm{rad/s}\)), which tells you how much area the spinning object sweeps through per second. In the scenario of the ship's propeller, we start calculating angular velocity from rest over a phase of acceleration. This requires us to utilize the equation:
  • \(\omega = \omega_0 + \alpha \cdot t\)
where \(\omega\) is the final angular velocity, \(\omega_0\) is the initial angular velocity (in this case, zero since it starts from rest), \(\alpha\) is the angular acceleration, and \(t\) is the elapsed time.
This helps determine how fast the propeller is spinning after a period of speeding up, providing us with a crucial figure for understanding the behavior of the rotating system.
Angular Acceleration
Angular acceleration refers to the rate at which angular velocity changes over time. If you imagine a merry-go-round, angular acceleration describes how quickly it speeds up or slows down. It is measured in radians per second squared (\(\mathrm{rad/s^2}\)). This unit shows how much the angular velocity changes every second.
In exercises like ours with the propeller, the object experiences two types of angular acceleration: speeding up initially and slowing down later. Each phase can be calculated using specific sets of data and equations.
  • During acceleration, use: \(\theta = \omega_0 \cdot t + \frac{1}{2} \alpha \cdot t^2\).
  • During deceleration, use a similar concept but adjust for the change in velocity.
Understanding these calculations helps highlight changes in speed over time and distance covered during these periods.
Constant Angular Speed
Constant angular speed refers to situations where an object rotates at the same speed continuously, without speeding up or slowing down. This concept is comparable to the steady ticking of a metronome.
When a rotating object, like our ship's propeller, reaches a phase where its angular velocity remains unchanged, we say it maintains constant angular speed. This phase is crucial since it tells us that no additional energy inputs or removals are affecting its motion.
To calculate the angular displacement for this period, employ the equation:
  • \(\theta = \omega \cdot t\)
where \(\omega\) is the angular velocity and \(t\) is the time duration. This direct calculation shows how much the object has rotated while maintaining a consistent speed, crucial for understanding total motion.

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Most popular questions from this chapter

Multiple-Concept Example 7 explores the approach taken in problems such as this one. The blades of a ceiling fan have a radius of \(0.380 \mathrm{m}\) and are rotating about a fixed axis with an angular velocity of \(+1.50 \mathrm{rad} / \mathrm{s}\) When the switch on the fan is turned to a higher speed, the blades acquire an angular acceleration of \(+2.00 \mathrm{rad} / \mathrm{s}^{2} .\) After \(0.500 \mathrm{s}\) has elapsed since the switch was reset, what is (a) the total acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) of a point on the tip of a blade and (b) the angle \(\phi\) between the total acceleration \(\overrightarrow{\mathbf{a}}\) and the centripetal acceleration \(\overrightarrow{\mathbf{a}}_{\mathbf{c}} ?\)

The earth spins on its axis once a day and orbits the sun once a year \(\left(365 \frac{1}{4}\right.\) days \()\) Determine the average angular velocity (in rad/s) of the earth as it (a) spins on its axis and (b) orbits the sun. In each case, take the positive direction for the angular displacement to be the direction of the earth’s motion.

A compact disc (CD) contains music on a spiral track. Music is put onto a CD with the assumption that, during playback, the music will be detected at a constant tangential speed at any point. since \(v_{\mathrm{T}}=r \omega, \mathrm{a} \mathrm{CD}\) rotates at a smaller angular speed for music near the outer edge and a larger angular speed for music near the inner part of the disc. For music at the outer edge \((r=0.0568 \mathrm{m}),\) the angular speed is 3.50 rev/s. Find \((\mathrm{a})\) the constant tangential speed at which music is detected and (b) the angular speed (in rev/s) for music at a distance of \(0.0249 \mathrm{m}\) from the center of a CD.

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.00 \(\mathrm{rad} / \mathrm{s}^{2} .\) Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of \(-25.0 \mathrm{rad} / \mathrm{s} .\) While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is \(8.3 \mathrm{m}\) above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 1.6 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

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