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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 rad/s².

Step by step solution

01

Understanding the Formula

To find the angular acceleration, we can use the formula for torque: \( \tau = I \cdot \alpha \), where \( \tau \) is the torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration.
02

Rearranging the Formula

We need to solve for \( \alpha \). Rearrange the formula to get \( \alpha = \frac{\tau}{I} \).
03

Substituting the Values

Substitute the given values into the formula: \( \tau = 1.8 \, \mathrm{N} \cdot \mathrm{m} \) and \( I = 0.22 \, \mathrm{kg} \cdot \mathrm{m}^{2} \). Thus, \( \alpha = \frac{1.8}{0.22} \).
04

Calculating the Angular Acceleration

Perform the division: \( \alpha = \frac{1.8}{0.22} \approx 8.18 \, \mathrm{rad/s^2} \). The angular acceleration of the blades is approximately \( 8.18 \, \mathrm{rad/s^2} \).

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Key Concepts

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

Torque
Torque is a measure of how much a force acting on an object causes it to rotate. Think of it like the rotational equivalent of linear force. When we apply force to twist something, like turning a doorknob or opening a jar, we are applying torque.
Torque depends on two things: the size of the force applied and how far the force is from the pivot point, which is where the object rotates. The formula for torque, \( \tau \), is given by:\[ \tau = r \cdot F \cdot \sin(\theta) \]Where:
  • \( r \) is the distance from the pivot point to the point where the force is applied.
  • \( F \) is the magnitude of the force.
  • \( \theta \) is the angle between the force and the lever arm.
In our exercise, a net torque of \(1.8 \text{ N} \cdot \text{m}\) is applied to the ceiling fan's blades. This torque makes the fan blades spin about their axis.
Moment of Inertia
The moment of inertia determines how difficult it is to change the rotational state of an object, either starting to spin or stopping. It is the rotational analog of mass in linear motion. The larger the moment of inertia, the harder it is to change the rotational state.

The formula for the moment of inertia, \( I \), varies depending on the shape of the object and the axis it rotates around. For simple geometric objects rotating about known axes, such as fans or wheels, standard formulas can be used. The general idea is:\[ I = \sum m_i r_i^2 \]Where:
  • \( m_i \) is the mass of a point in the object.
  • \( r_i \) is the distance of the point from the axis of rotation.
In this problem, we know the moment of inertia of the fan blades is given as \(0.22 \text{ kg} \cdot \text{m}^{2}\). This value shows the resistance of the blades against rotational motion.
Rotational Motion
Rotational motion refers to the motion of an object around a center point or axis. Examples include wheels turning, gears rotating, and even planets orbiting the sun. In our exercise, the ceiling fan's blades experience rotational motion as they spin around the central motor axis.

Key quantities that describe rotational motion include:
  • Angular velocity: How fast an object spins. It's the rotational equivalent of linear velocity.
  • Angular acceleration: The rate of change of angular velocity with respect to time. This is directly related to torque and moment of inertia.
To find the angular acceleration, \( \alpha \), we use the formula:\[ \tau = I \cdot \alpha \]Rearrange it to solve for \( \alpha \):\[ \alpha = \frac{\tau}{I} \]By substituting the known values from the exercise, \( \tau = 1.8 \text{ N} \cdot \text{m} \) and \( I = 0.22 \text{ kg} \cdot \text{m}^{2} \), we find:\[ \alpha = \frac{1.8}{0.22} \approx 8.18 \text{ rad/s}^2 \]This is the angular acceleration experienced by the fan blades.

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Most popular questions from this chapter

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 4.0R. Assume that all of the star’s original mass is contained in the shell.

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