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A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. (a) What is the linear speed of the blade tip, in m/s? (b) What is the radial acceleration of the blade tip expressed as a multiple of \(g\)?

Short Answer

Expert verified
(a) The blade tip's linear speed is approximately 195.67 m/s. (b) The radial acceleration is about 1150.34 times the acceleration due to gravity.

Step by step solution

01

Find the Rotation Speed in Revolutions per Second

The rotation speed given is 550 revolutions per minute. First, we need to convert this to revolutions per second by dividing 550 by 60. \[\text{Revolutions per second} = \frac{550}{60} \approx 9.1667\, \text{rev/s}\]
02

Calculate Angular Speed in Radians per Second

Angular speed \( \omega \) can be calculated by multiplying the revolutions per second by \( 2\pi \), since one revolution corresponds to an angle of \( 2\pi \) radians.\[\omega = 9.1667 \times 2\pi \approx 57.55\, \text{rad/s}\]
03

Compute Linear Speed of the Blade Tip

The linear speed \( v \) of the blade tip is given by the product of angular speed \( \omega \) and the radius (length of the blade) \( r \).\[ v = \omega \times r = 57.55 \times 3.40 \approx 195.67\, \text{m/s} \]
04

Determine Radial Acceleration of the Blade Tip

The radial (centripetal) acceleration \( a_r \) can be found using the formula \( a_r = \omega^2 \times r \).\[ a_r = 57.55^2 \times 3.40 \approx 11286.85\, \text{m/s}^2 \]
05

Express Radial Acceleration as a Multiple of \( g \)

The acceleration due to gravity \( g \) is approximately \( 9.81 \text{m/s}^2 \). To express the radial acceleration as a multiple of \( g \), divide \( a_r \) by \( g \).\[\frac{a_r}{g} = \frac{11286.85}{9.81} \approx 1150.34\]

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Key Concepts

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

Linear Speed Calculation
Linear speed is a crucial concept when discussing moving objects, like helicopter blades. It tells us how fast something moves along its path. For the blade tip of a rotor moving in a circular path, the linear speed can be found. This is done by multiplying its angular speed by the radius of the circle. Here, the radius is the length of the blade.

The formula used is:
  • Linear Speed (\( v \)) = Angular Speed (\( \omega \)) x Radius (\( r \))
In our example, after calculating the angular speed as 57.55 rad/s and knowing that the length of the blade is 3.40 meters, the linear speed turns out to be approximately 195.67 meters per second.

This means the tip of the blade travels 195.67 meters for every second it spins. Such high speeds emphasize the power and efficiency of helicopter rotors.
Angular Speed
Angular speed denotes how fast an object rotates or spins around a fixed point. For helicopter rotors, it indicates how quickly the blades revolve around the central shaft. Unlike linear speed, which is about path distance, angular speed is about rotation angle. Angular speed, expressed in radians per second, forms a key part of circular motion concepts.

To determine angular speed in this scenario, use:
  • Angular Speed (\( \omega \)) = Revolutions per second \( \times 2\pi \)
By converting 550 revolutions per minute to approximately 9.1667 revolutions per second, then multiplying by \( 2\pi \), we find the angular speed to be around 57.55 radians per second. This measurement helps us understand the rate at which the rotor blades complete one full circle.
Radial Acceleration
When an object moves in a circle, it constantly accelerates towards the center, ensuring it stays on its circular path. This kind of acceleration, called radial or centripetal acceleration, is crucial for understanding dynamics in circular motion. For our helicopter rotor, the radial acceleration can be worked out using the angular speed.

The formula is:
  • Radial Acceleration (\( a_r \)) = \( \omega^2 \times r \)
Here, \( \omega \) is the angular speed previously found (57.55 rad/s), and \( r \) is the blade length (3.40 m). Substituting these values, the radial acceleration calculates to about 11286.85 m/s\(^2\). This powerful acceleration illustrates how the blades are pulled inwards to keep them whirling in their path.
Revolutions per Minute
Revolutions per minute (RPM) is a unit measuring rotation rate, specifically how many full circles an object makes each minute. It's a more familiar form of expressing how fast an engine or rotor spins. We start with RPM because it's a common measurement in machinery like helicopter rotors.

Initially given in 550 RPM, this specifies the blades make 550 full turns every minute. To simplify further calculations, we convert this to revolutions per second by dividing by 60, giving us about 9.1667 revolutions per second. This conversion is necessary because many computations in physics require understanding speed per second, given their reliance on SI units like seconds and meters.
Centripetal Acceleration
Centripetal acceleration is a vital concept when analyzing circular motion. It describes how an object moving along a curve constantly changes direction. While its speed may stay the same, the continual direction change means there's acceleration directed toward the circle's center. This ensures the object doesn't fly off tangentially, allowing it to maintain its curved path.

Often, centripetal acceleration is directly related to the object's velocity and the curve's radius:
  • Centripetal Acceleration = \( \frac{v^2}{r} \)
In the helicopter rotor scenario, it is equivalent to the radial acceleration of 11286.85 m/s\(^2\). Without such centripetal force, the rotor blades would merely move in a straight line, losing the intricate control necessary for helicopter flight.

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Most popular questions from this chapter

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