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Chapter 10: Problem 52

Each propeller on a King Air twin-engine airplane consists of three blades, each of mass \(10 \mathrm{kg}\) and length \(125 \mathrm{cm} .\) The blades may be treated approximately as uniform, thin rods, (a) What's the propeller's rotational inertia? (b) If the plane's engine develops a torque of \(2.7 \mathrm{kN} \cdot \mathrm{m}\), how long will it take to spin up the propeller from 1400 rpm to 1900 rpm?

Short Answer

Expert verified
The rotational inertia of the propeller is calculated first. Using the given torque, the angular acceleration is computed. Finally, after converting the initial and final speeds from rpm to rad/s, the time it takes for the propeller to reach the desired speed is calculated.

Step by step solution

01

Calculate the Moment of Inertia of the Propeller

The moment of inertia, \(I\), of a rod of mass, \(m\), and length, \(l\), about an axis through one end perpendicular to the length is given as \(I = \frac{1}{3}ml^2\). Given mass of the blade (\(m\)) is 10 kg and length of the blade (\(l\)) is 125 cm = 1.25 m, the moment of inertia for one blade would be calculated as \(I = \frac{1}{3}(10)(1.25)^2\). The propeller includes three blades, so the total moment of inertia will be 3 times this amount.
02

Calculate the Angular Acceleration

Torque (\(\tau\)) is equal to the Moment of inertia (\(I\)) times angular acceleration (\(\alpha\)), represented as \(\tau = I\alpha\). Given Torque (\(\tau\)) is 2.7 kN.m = 2700 N.m, we can calculate the angular acceleration from this formula as \(\alpha = \frac{\tau}{I}\).
03

Calculate the Time to Spin up the Propeller

Angular speed is measured in rad/s, so we first convert the given rpm to rad/s. The initial angular speed (\(\omega_{i}\)) is 1400 rev/min and the final angular speed (\(\omega_{f}\)) is 1900 rev/min. To convert them to rad/s, use the relation 1 rev = \(2\pi\) rad and 1 min = 60 s. Then use the angular form of the second equation of motion (\(\omega_{f} = \omega_{i} + \alpha t\)) to find the time (\(t\)).

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Most popular questions from this chapter

A 25 -cm-diameter circular saw blade spins at 3500 rpm. How fast would you have to push a straight hand saw to have the teeth move through the wood at the same rate as the circular saw teeth?

A circular saw spins at 5800 rpm, and its electronic brake is supposed to stop it in less than 2 s. As a quality-control specialist. you're testing saws with a device that counts the number of blade revolutions. A particular saw turns 75 revolutions while stopping. Does it meet its specs?

A solid 2.4 -kg sphere is rolling at \(5.0 \mathrm{m} / \mathrm{s}\). Find (a) its translational kinetic energy and (b) its rotational kinetic energy.

A solid sphere and a solid cube have the same mass, and the side of the cube is equal to the diameter of the sphere. The cube's rotation axis is perpendicular to two of its faces. Which has greater rotational inertia about an axis through the center of mass?

A \(320-\mathrm{N}\) frictional force acts on the rim of a 1.0 -m-diameter wheel to oppose its rotational motion. Find the torque about the wheel's central axis.
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