/*! 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 12 A propeller is rotating about an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A propeller is rotating about an axis perpendicular to its center, as the drawing shows. The axis is parallel to the ground. An arrow is fired at the propeller, travels parallel to the axis, and passes through one of the open spaces between the propeller blades. The angular open spaces between the three propeller blades are each \(\pi / 3\) rad \(\left(60.0^{\circ}\right) .\) The vertical drop of the arrow may be ignored. There is a maximum value \(\omega\) for the angular speed of the propeller, beyond which the arrow cannot pass through an open space without being struck by one of the blades. Find this maximum value when the arrow has the lengths \(L\) and speeds \(v\) shown in the following table. $$ \begin{array}{lll} & L & v \\ (\mathrm{a}) & 0.71 \mathrm{m} & 75.0 \mathrm{m} / \mathrm{s} \\ \hline(\mathbf{b}) & 0.71 \mathrm{m} & 91.0 \mathrm{m} / \mathrm{s} \\ \hline(\mathbf{c}) & 0.81 \mathrm{m} & 91.0 \mathrm{m} / \mathrm{s} \\ \hline \end{array} $$

Short Answer

Expert verified
(a) 111.28 rad/s, (b) 134.95 rad/s, (c) 118.62 rad/s.

Step by step solution

01

Understand the Motion of the Arrow

When the arrow is fired through the propeller, it must pass completely through an open space of angular width \( \pi/3 \) radians without being hit by a blade. The time it takes for the arrow to pass through the propeller is critical to determining the maximum angular speed \( \omega \).
02

Find Time for Arrow to Pass

The time \( t \) it takes for the arrow to pass through the open space is determined by its length \( L \) and speed \( v \). Since the motion is linear, \( t = \frac{L}{v} \).
03

Calculate Angular Displacement

During the time \( t \), the propeller should not move an angular distance that exceeds the size of the open space. The angular displacement \( \theta \) the propeller moves in time \( t \) is \( \theta = \omega \cdot t \).
04

Set Up Equation for Angular Limit

Since the angular open space is \( \pi/3 \), set up the inequality: \( \omega \cdot t \leq \frac{\pi}{3} \). Substitute \( t = \frac{L}{v} \) to find \( \omega \).
05

Solve for Maximum Angular Speed \( \omega \)

Rearrange the inequality to solve for \( \omega \): \( \omega \leq \frac{\pi}{3} \times \frac{v}{L} \). Substitute possible values for \( L \) and \( v \) for each case. - For (a): \( \omega \leq \frac{\pi}{3} \times \frac{75.0}{0.71} \)- For (b): \( \omega \leq \frac{\pi}{3} \times \frac{91.0}{0.71} \)- For (c): \( \omega \leq \frac{\pi}{3} \times \frac{91.0}{0.81} \).
06

Calculate Specific Values for Each Case

Perform the calculations:- For (a): \( \omega \leq \frac{\pi}{3} \times \frac{75.0}{0.71} = \frac{\pi \times 75.0}{3 \times 0.71} \approx 111.28 \) rad/s- For (b): \( \omega \leq \frac{\pi}{3} \times \frac{91.0}{0.71} = \frac{\pi \times 91.0}{3 \times 0.71} \approx 134.95 \) rad/s- For (c): \( \omega \leq \frac{\pi}{3} \times \frac{91.0}{0.81} = \frac{\pi \times 91.0}{3 \times 0.81} \approx 118.62 \) rad/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Projectile Motion
Projectile motion is a fascinating aspect of physics that deals with objects moving through space affected only by gravitational forces. It's based on parameters like initial velocity and the angle of projection. In the case of a propeller and arrow, even though the physics of projectile is involved, we simplify the situation by ignoring the gravitational drop of the arrow given its high speed and short traversal distance. The key here is the arrow's speed (75.0 m/s or 91.0 m/s) combined with its length (0.71 m or 0.81 m). This allows us to calculate the time it takes to pass through an open space, which is fundamental when determining the interaction of projectiles with moving objects like a rotating propeller. Understanding projectile motion helps break down complex motions into simpler straight-line tasks for ease of problem-solving.
When applying the basic projectile equation, distance equals rate times time, we can easily figure the time taken by dividing the arrow’s length by its speed, simplifying the trajectory analysis.
Rotational Motion
Rotational motion involves objects rotating around a fixed axis, which is central to understanding the propeller's action in this problem. The key to solving this problem lies in considering the angular velocity (\(\omega\)), which describes how fast the propeller blades turn. This angular velocity must be low enough to allow the arrow to pass unimpeded through the open spaces. The equal angular spacing (\(\pi/3\) radians) between blades becomes crucial because it defines the rotational window the arrow must pass through.
Angular displacement (\(\theta\)) during the time the arrow traverses is given by the product of angular velocity and time (\(\theta = \omega \cdot t\)). In our scenario, the objective is to ensure angular displacement doesn’t exceed the space’s angular size as the arrow passes. This provides a tangible application of rotational motion principles to practical problem-solving.
Physics Problem Solving
Physics problem-solving tactics involve breaking complex scenarios into manageable parts through equations and logical reasoning. By identifying known quantities, like arrow speed (\(v\)) and length (\(L\)), and linking them to desired outcomes, such as maximum angular speed (\(\omega\)), we can systematically address the problem.
Start by understanding the relationships and substitutions that simplify the process. In this exercise, once we determine how long the arrow takes to traverse the gap, we can use inequalities to establish acceptable limits for the angular velocity. For a problem solving approach:
  • Define the conditions, like the angular spacing (\(\pi/3\) radians).
  • Translate this into problems solvable by basic physics equations.
  • Reorganize and use inequalities to ensure realistic solutions are found.
Ultimately, physics is about making sense of the complex motions and forces acting on objects, and problem solving is about using intuition and calculation to find solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An automatic dryer spins wet clothes at an angular speed of \(5.2 \mathrm{rad} / \mathrm{s} .\) Starting from rest, the dryer reaches its operating speed with an average angular acceleration of \(4.0 \mathrm{rad} / \mathrm{s}^{2} .\) How long does it take the dryer to come up to speed?

Suppose you are riding a stationary exercise bicycle, and the electronic meter indicates that the wheel is rotating at \(9.1 \mathrm{rad} / \mathrm{s}\). The wheel has a radius of \(0.45 \mathrm{m} .\) If you ride the bike for \(35 \mathrm{min},\) how far would you have gone if the bike could move?

A racing car, starting from rest, travels around a circular turn of radius \(23.5 \mathrm{m}\). At a certain instant, the car is still accelerating, and its angular speed is \(0.571 \mathrm{rad} / \mathrm{s}\). At this time, the total acceleration (centripetal plus tangential) makes an angle of \(35.0^{\circ}\) with respect to the radius. (The situation is similar to that in Interactive Figure \(8.12 b .\) ) What is the magnitude of the total acceleration?

A dragster starts from rest and accelerates down a track. Each tire has a radius of \(0.320 \mathrm{m}\) and rolls without slipping. At a distance of \(384 \mathrm{m},\) the angular speed of the wheels is 288 rad/s. Determine (a) the linear speed of the dragster and (b) the magnitude of the angular acceleration of its wheels.

A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from 3.00 to 5.00 rev/s while rotating through one-half of a revolution. How much time does this maneuver take?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.