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Chapter 9: Problem 34

A ceiling fan is turned on and a net torque of \(1.8 \mathrm{N} \cdot \mathrm{m}\) is applied to the blades. The blades have a total moment of inertia of \(0.22 \mathrm{kg} \cdot \mathrm{m}^{2} .\) What is the angular acceleration of the blades?

Short Answer

Expert verified
The angular acceleration is approximately \( 8.18 \mathrm{rad/s^2} \).

Step by step solution

01

Understand the Formula

The formula to calculate angular acceleration is given by \( \alpha = \frac{\tau}{I} \), where \( \alpha \) represents angular acceleration, \( \tau \) represents the torque applied, and \( I \) is the moment of inertia.
02

Identify Given Values

From the problem, we know that the net torque \( \tau \) is \( 1.8 \mathrm{N} \cdot \mathrm{m} \) and the moment of inertia \( I \) is \( 0.22 \mathrm{kg} \cdot \mathrm{m}^{2} \).
03

Substitute Values into Formula

Substitute \( \tau = 1.8 \) and \( I = 0.22 \) into the formula: \( \alpha = \frac{1.8}{0.22} \).
04

Calculate the Angular Acceleration

Perform the division to find \( \alpha = \frac{1.8}{0.22} \approx 8.18 \).
05

Interpret the Result

The angular acceleration of the blades is approximately \( 8.18 \mathrm{rad/s^2} \), which indicates how quickly the blades will speed up over time.

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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 fundamental concept in rotational dynamics. It's similar to force in linear motion but specifically relates to how effectively a force can cause an object to rotate around an axis. To visualize this, imagine trying to turn a door. Applying force close to the hinge creates minimal rotational effect, but pushing at the door handle causes the door to swing open more easily.

In mathematical terms, torque (\( au \)) is defined as:
  • \( au = r imes F imes ext{sin}( heta) \)
  • \( r \) is the distance from the rotation axis to where the force is applied.
  • \( F \) is the magnitude of the force.
  • \( \theta \) is the angle between the force vector and the lever arm.
The net torque results from summing all these effects. In this exercise, a net torque of \( 1.8 \, \mathrm{N \cdot m} \) is applied to the fan blades, influencing their rotational motion.
Defining Moment of Inertia
The moment of inertia, often symbolized by \( I \), is a measure of an object's resistance to changes in its rotational motion. Just like mass in linear motion determines how much an object resists acceleration, the moment of inertia impacts how an object behaves when torque is applied.

Think of it as rotational inertia; it's harder to spin a merry-go-round with more mass at the edge compared to one where the mass is concentrated at the center. The moment of inertia depends on:
  • The object's mass.
  • How this mass is distributed in relation to the rotational axis.
For the ceiling fan blades, an inertia of \( 0.22 \, \mathrm{kg \cdot m}^2 \) indicates how they resist change in speed once they start rotating.
Grasping Rotational Motion
Rotational motion is all about objects spinning or rotating around a central point or axis. The concept is similar to linear motion, but instead of moving in a straight line, all points in the rotating object travel in circular paths.

Key aspects of rotational motion include:
  • Angular displacement: Change in the angle as an object rotates.
  • Angular velocity: The rate of change of angular displacement over time.
  • Angular acceleration: The rate of change of angular velocity, which describes how quickly the object speeds up or slows down.
In our scenario, when the ceiling fan is turned on, the applied torque results in an angular acceleration of approximately \( 8.18 \, \mathrm{rad/s}^2 \). This indicates how quickly the fan blades will gain speed, transforming the net torque and moment of inertia into a tangible spinning motion. Understanding these elements makes it clear why the fan responds in this manner when force is applied.

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Most popular questions from this chapter

A rod is lying on the top of a table. One end of the rod is hinged to the table so that the rod can rotate freely on the tabletop. Two forces, both parallel to the tabletop, act on the rod at the same place. One force is directed perpendicular to the rod and has a magnitude of \(38.0 \mathrm{N}\). The second force has a magnitude of \(55.0 \mathrm{N}\) and is directed at an angle \(\theta\) with respect to the rod. If the sum of the torques due to the two forces is zero, what must be the angle \(\theta ?\)

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of the star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius \(R\) that is initially rotating at 2.0 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is \(4.0 R\). Assume that all of the star's original mass is contained in the shell.

A clay vase on a potter's wheel experiences an angular acceleration of \(8.00 \mathrm{rad} / \mathrm{s}^{2}\) due to the application of a \(10.0-\mathrm{N} \cdot \mathrm{m}\) net torque. Find the total moment of inertia of the vase and potter's wheel.

A 9.75-m ladder with a mass of \(23.2 \mathrm{kg}\) lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of \(245 \mathrm{N}\). At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of \(1.80 \mathrm{rad} / \mathrm{s}^{2}\) about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia?

A uniform board is leaning against a smooth vertical wall. The board is at an angle \(\theta\) above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is \(0.650 .\) Find the smallest value for the angle \(\theta\), such that the lower end of the board does not slide along the ground.
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