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

Short Answer

Expert verified
(a) 195.84 m/s; (b) Approximately 1151.66 times gravity.

Step by step solution

01

Convert RPM to Radians Per Second

First, we need to convert the rotational speed from revolutions per minute (RPM) to radians per second (rad/s). Given that 1 revolution is equal to \(2\pi\) radians, and there are 60 seconds in a minute:\[\omega = 550\, \text{rev/min} \times \frac{2\pi\, \text{rad}}{1\, \text{rev}} \times \frac{1\, \text{min}}{60\, \text{s}} \]Calculating this gives:\[\omega = \frac{550 \times 2\pi}{60} \approx 57.60\, \text{rad/s}\]
02

Calculate the Linear Speed of the Blade Tip

The linear speed \(v\) of the blade tip can be calculated using the formula:\[ v = r \cdot \omega \]where \(r = 3.40\, \text{m}\) is the radius.\[ v = 3.40\, \text{m} \times 57.60\, \text{rad/s} \approx 195.84\, \text{m/s} \]
03

Determine the Radial (Centripetal) Acceleration

The radial acceleration \(a_r\) is given by the formula:\[ a_r = \omega^2 \cdot r \]Substituting the known values:\[ a_r = (57.60\, \text{rad/s})^2 \times 3.40\, \text{m} \approx 11292.16\, \text{m/s}^2 \]
04

Express Radial Acceleration as Multiple of g

Since the acceleration due to gravity \(g\) is approximately \(9.81\, \text{m/s}^2\), we find \(a_r\) as a multiple of \(g\) by:\[ \frac{a_r}{g} = \frac{11292.16\, \text{m/s}^2}{9.81\, \text{m/s}^2} \approx 1151.66 \]Thus, the radial acceleration is approximately 1151.66 \(g\).

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Key Concepts

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

Linear Speed Calculation
Understanding linear speed in the context of rotational motion is key to grasping how fast an object moves along its path. In rotational scenarios like our helicopter rotor, linear speed refers to how swiftly the tip of the rotor blade travels through space.
To compute linear speed, the formula we use is:
  • \( v = r \cdot \omega \)
Here, \( v \) is the linear speed, \( r \) is the radius (or the length from the central shaft to the blade tip), and \( \omega \) is the angular speed in radians per second (rad/s).

By multiplying the radius of 3.40 meters by the angular speed of the rotor, which is approximately 57.60 rad/s, we find that the linear speed of the blade tip is roughly 195.84 meters per second. This calculation allows us to understand just how quickly the tip of each rotor blade is moving.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. This acceleration keeps the object moving in a curved trajectory instead of flying off in a straight line.
The formula used for radial acceleration is:
  • \( a_r = \omega^2 \cdot r \)
Here, \( a_r \) is the radial acceleration, \( \omega \) is the angular speed in rad/s, and \( r \) is the radius.

For the helicopter rotors, when substituting \( \omega = 57.60 \) rad/s and \( r = 3.40 \) meters, we calculate the radial acceleration as approximately 11292.16 meters per second squared.
To express this as a multiple of gravity, we divide this value by the acceleration due to gravity \( g = 9.81 \) m/s². Therefore, the radial acceleration is spectacularly high at about 1151.66 times that of gravity.
Revolutions Per Minute to Radians Per Second Conversion
Understanding how to convert rotational speed from revolutions per minute (RPM) to radians per second is an essential skill in physics, especially when working with calculations involving angular motion.

To convert RPM to rad/s, we use the relationship that one complete revolution of a circle is equivalent to \( 2\pi \) radians. Additionally, since there are 60 seconds in a minute, the conversion formula becomes:
  • \( \omega = \, \text{RPM} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \)
For a rotor spinning at 550 RPM, this results in an angular speed of \( 550 \times \frac{2\pi}{60} \approx 57.60 \) rad/s.

This conversion is vital because angular velocity in radians per second is required for further calculations, including linear speed and radial acceleration in rotational motion scenarios.

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Most popular questions from this chapter

A physics book slides off a horizontal tabletop with a speed of 1.10 \(\mathrm{m} / \mathrm{s} .\) It strikes the floor in 0.350 s. Ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor. (d) Draw \(x-t, y-t, v_{x^{-}},\) and \(v_{y}-t\) graphs for the motion.

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