/*! 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 84 A ceiling fan has four blades, e... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 84

A ceiling fan has four blades, each with a mass of \(0.35 \mathrm{kg}\) and a length of \(60 \mathrm{cm} .\) Model each blade as a rod connected to the fan axle at one end. When the fan is turned on, it takes 4.35 s for the fan to reach its final angular speed of 1.8 rev/s. What torque was applied to the fan by the motor? Ignore torque due to the air.

Short Answer

Expert verified
Answer: The torque applied to the fan by the motor is approximately \(0.437\,\text{N m}\).

Step by step solution

01

Calculate the moment of inertia of one fan blade

Model each blade as a rod connected to the fan axle at one end. The moment of inertia of a rod with mass \(m\) and length \(L\) rotating about one end is given by the formula: \(I = \frac{1}{3}mL^2\). Given the mass \(m = 0.35\,\text{kg}\) and length \(L = 60\,\text{cm}\) (in meters, \(L = 0.6\,\text{m}\)), we can find the moment of inertia for one blade: \(I = \frac{1}{3}(0.35\,\text{kg})(0.6\,\text{m})^2 \approx 0.042\,\text{kg m}^2\)
02

Calculate the total moment of inertia of the fan

The fan has 4 blades, so the total moment of inertia \(I_{\text{total}}\) is given by the sum of the moments of inertia of each individual blade: \(I_{\text{total}} = 4I = 4(0.042\,\text{kg m}^2) = 0.168\,\text{kg m}^2\)
03

Find the angular acceleration of the fan

The fan starts at an angular speed of \(0\,\text{rev/s}\) and takes \(4.35\,\text{s}\) to reach its final angular speed of \(1.8\,\text{rev/s}\). Recall that in the SI unit, angular speed is typically expressed in radians/second. To convert from rev/s to rad/s, we use the conversion factor \(\text{1 rev} = 2\pi\,\text{rad}\): \(\omega = 1.8\,\text{rev/s} \times \frac{2\pi\,\text{rad}}{\text{1 rev}} \approx 11.31\,\text{rad/s}\) Now we can find the angular acceleration \(\alpha\) using the equation \(\alpha = \frac{\Delta\omega}{\Delta t}\). In this case, \(\Delta\omega=\omega\) and \(\Delta t = 4.35\,\text{s}\): \(\alpha = \frac{11.31\,\text{rad/s}}{4.35\,\text{s}} \approx 2.60\,\text{rad/s}^2\)
04

Determine the torque applied by the motor

Finally, we can determine the torque \(\tau\) applied by the motor using the relation between torque, moment of inertia, and angular acceleration: \(\tau = I_{\text{total}}\alpha\): \(\tau = (0.168\,\text{kg m}^2)(2.60\,\text{rad/s}^2) \approx 0.437\,\text{N m}\) Therefore, the torque applied to the fan by the motor is approximately \(0.437\,\text{N m}\).

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Most popular questions from this chapter

A bicycle has wheels of radius \(0.32 \mathrm{m}\). Each wheel has a rotational inertia of \(0.080 \mathrm{kg} \cdot \mathrm{m}^{2}\) about its axle. The total mass of the bicycle including the wheels and the rider is 79 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?
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