/*! 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 6 An airplane propeller has a mass... [FREE SOLUTION] | 91Ó°ÊÓ

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An airplane propeller has a mass of \(70 \mathrm{~kg}\) and a radius of gyration of \(75 \mathrm{~cm}\). Find its moment of inertia. How large a torque is needed to give it an angular acceleration of \(4.0 \mathrm{rev} / \mathrm{s}^{2} ?\) $$ I=M k^{2}=(70 \mathrm{~kg})(0.75 \mathrm{~m})^{2}=39 \mathrm{~kg} \cdot \mathrm{m}^{2} $$ To be able to use \(\tau=I \alpha\), we must have \(\alpha\) in \(\mathrm{rad} / \mathrm{s}^{2}\) : $$ \alpha=\left(4.0 \frac{\mathrm{rev}}{\mathrm{s}^{2}}\right)\left(2 \pi \frac{\mathrm{rad}}{\mathrm{rev}}\right)=8.0 \pi \mathrm{rad} / \mathrm{s}^{2} $$ Then \(\tau=I \alpha=\left(39 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(8.0 \pi \mathrm{rad} / \mathrm{s}^{2}\right)=0.99 \mathrm{kN} \cdot \mathrm{m}\)

Short Answer

Expert verified
The moment of inertia is 39 kg·m² and the required torque is approximately 980 N·m.

Step by step solution

01

Define Given Quantities

We have the mass of the propeller as \( M = 70 \text{ kg} \) and the radius of gyration as \( k = 75 \text{ cm} = 0.75 \text{ m} \). We need to find the moment of inertia \( I \) and the required torque \( \tau \).
02

Calculate Moment of Inertia

The moment of inertia \( I \) is calculated using the formula \( I = M k^2 \). Substitute the given values: \[ I = 70 \times (0.75)^2 = 39 \text{ kg} \cdot \text{m}^2 \].
03

Convert Angular Acceleration to Radians per Second Squared

The given angular acceleration is \( \alpha = 4.0 \text{ rev/s}^2 \). To convert this to radians per second squared, use the conversion factor \( 2\pi \text{ rad/rev} \): \[ \alpha = 4.0 \times 2\pi = 8.0\pi \text{ rad/s}^2 \].
04

Calculate Torque Required

Using the relation \( \tau = I \alpha \), substitute \( I = 39 \text{ kg} \cdot \text{m}^2 \) and \( \alpha = 8.0\pi \text{ rad/s}^2 \). Calculate \( \tau \): \[ \tau = 39 \times 8.0\pi = 312\pi \approx 980 \text{ N} \cdot \text{m} \] (rounded to three significant figures, expressed in kN).

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Key Concepts

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

Understanding Torque
Torque is a crucial concept in physics, especially in rotational dynamics. It is often referred to as a "rotational force." Think of torque as the push or pull needed to make an object rotate. This can be compared to the force we use to turn a wrench. The general formula for torque \( \tau \) is given by:
  • \( \tau = I \alpha \)
In this formula, \( I \) represents the moment of inertia, and \( \alpha \) is the angular acceleration. For our case with the airplane propeller, torque is needed to achieve a specific angular acceleration, which we're looking to calculate.
Calculating torque involves multiplying the calculated moment of inertia with the angular acceleration (converted appropriately into radians per second squared). This results in the amount of torque needed to achieve a desired level of rotation. Torque is measured in Newton-meters (Nm), indicating the force applied at a distance.
Angular Acceleration in Radians
Angular acceleration \( \alpha \) is how quickly the rotational speed of an object is changing. It highlights acceleration in a rotating system, such as airplane propeller motion. In physics, measurements in radians are preferred because they allow for simple integration with angular movement equations. One complete rotation is \( 2\pi \) radians.
In our scenario, the angular acceleration was initially provided in revolutions per second squared. Since revolutions aren't standard in many physics equations, we needed to convert this into radians per second squared. This transformation was done as follows:
  • \( \alpha = 4.0 \frac{\text{rev}}{\text{s}^2} \times 2\pi \frac{\text{rad}}{\text{rev}} = 8.0\pi \frac{\text{rad}}{\text{s}^2} \)
Understanding this conversion process is vital, ensuring that your results align correctly when plugged into torque equations.
Radius of Gyration and its Importance
The radius of gyration, denoted as \( k \), is a measure that helps simplify the calculation of an object's moment of inertia (\( I \)). Think of it as the distance from the axis of rotation at which the entire mass of the body can be thought to concentrate; this simplifies our complex calculation.
In mathematical terms, the moment of inertia can be calculated using the formula:
  • \( I = M k^2 \)
Here, \( M \) is the mass of the object. The radius of gyration provides a powerful abstraction, allowing us to calculate rotational dynamics without needing to understand the complete mass distribution of an object.
For the airplane propeller exercise, the radius of gyration (\( 0.75 \text{ m} \)) was used directly to find inertia: \( I = 70 \times (0.75)^2 = 39 \text{ kg} \cdot \text{m}^2 \). Knowing \( k \) simplifies many calculations in physics, especially in engineering, where complex shapes and mass distributions are often involved.

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Most popular questions from this chapter

An electric motor runs at \(900 \mathrm{rpm}\) and delivers \(2.0 \mathrm{hp}\). How much torque does it deliver?

An \(8.0-\mathrm{kg}\) wheel has a radius of gyration of \(25 \mathrm{~cm} .(a)\) What is its moment of inertia? \((b)\) How large a torque is required to give it an angular acceleration of \(3.0 \mathrm{rad} / \mathrm{s}^{2} ?\)

A \(4.0-\mathrm{kg}\) wheel of 20 -cm radius of gyration is rotating at \(360 \mathrm{rpm}\). The retarding frictional torque is \(0.12 \mathrm{~N} \cdot \mathrm{m}\). Compute the time it will take the wheel to coast to rest.

Suppose that a satellite goes around the Moon in an elliptical orbit. At its closest approach it has a speed \(v_{c}\) and a radius \(r_{c}\) from the center of the Moon. At its farthest distance, it has a speed \(v_{f}\) and a radius \(r_{f}\). Find the ratio \(v_{c} / v_{f}\). [Hint: Angular momentum is conserved, and, moreover, the satellite can be treated as a point mass.]

A \(0.75-\) hp motor acts for \(8.0 \mathrm{~s}\) on an initially nonrotating wheel having a moment of inertia \(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Find the angular speed developed in the wheel, assuming no losses. Work done by motor in \(8.0 \mathrm{~s}=\mathrm{KE}\) of wheel after \(8.0 \mathrm{~s}\) $$ \begin{aligned} (\text { Power) }(\text { Time })&=\frac{1}{2} I \omega^{2} \\ (0.75 \mathrm{hp})(746 \mathrm{~W} / \mathrm{hp})(8.0 \mathrm{~s}) &=\frac{1}{2}\left(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\right) \omega^{2} \end{aligned} $$ from which \(\omega=67 \mathrm{rad} / \mathrm{s}\).

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