/*! 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 73 A rotating door is made from fou... [FREE SOLUTION] | 91Ó°ÊÓ

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A rotating door is made from four rectangular sections, as indicated in the drawing. The mass of each section is \(85 \mathrm{kg} .\) A person pushes on the outer edge of one section with a force of \(F=68 \mathrm{N}\) that is directed perpendicular to the section. Determine the magnitude of the door's angular acceleration.

Short Answer

Expert verified
Angular acceleration is approximately \( 2 \ \text{rad/s}^2 \).

Step by step solution

01

Understand the Problem

We need to find the angular acceleration of a rotating door when a force is applied at the outer edge. This requires understanding that the force creates a torque about the center of the door, causing rotational motion.
02

Calculate the Torque

Torque (Ï„) is calculated using the formula \( \tau = F \times r \), where \( F \) is the force applied and \( r \) is the distance from the pivot point (center of rotation). For a door section, \( r \) is the radius of the door. We assume a typical door radius for calculation if it's not given, say \( r = 1.2 \) meters.
03

Apply the Torque Formula

Using the assumed radius \( r = 1.2 \ \text{m} \):\[\tau = 68 \ \text{N} \times 1.2 \ \text{m} = 81.6 \ \text{Nm}\]
04

Determine Moment of Inertia

For a rectangular section (regarded here as a thin rod rotated about one end):\[I = \frac{1}{3}mr^2\]Let's calculate using \( m = 85 \ \text{kg} \) and \( r = 1.2 \ \text{m} \):\[I = \frac{1}{3} \times 85 \ \text{kg} \times (1.2 \ \text{m})^2 = 40.8 \ \text{kg} \cdot \text{m}^2\]
05

Calculate Angular Acceleration

Use the relation \( \tau = I \cdot \alpha \) to solve for angular acceleration \( \alpha \):\[\alpha = \frac{\tau}{I} = \frac{81.6 \ \text{Nm}}{40.8 \ \text{kg} \cdot \text{m}^2} \approx 2 \ \text{rad/s}^2\]
06

Conclusion

After the calculations, we find that the angular acceleration of the door is approximately \( 2 \ \text{rad/s}^2 \).

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Key Concepts

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

Angular Acceleration
Angular acceleration refers to how quickly an object speeds up or slows down its rotation. It is a crucial concept when dealing with objects in circular motion, such as a door on a hinge or a spinning wheel. The notation for angular acceleration is \( \alpha \), and it is measured in radians per second squared \( \text{rad/s}^2 \). Angular acceleration provides insight into the change in rotational speed over time. It's analogous to linear acceleration, which measures how fast an object speeds up or slows down in a straight line.

Calculating Angular Acceleration

To find angular acceleration, we use the equation \( \tau = I \cdot \alpha \), where \( \tau \) is the torque applied to the object, and \( I \) is the moment of inertia. By rearranging, the formula becomes:
  • \( \alpha = \frac{\tau}{I} \)
For instance, in the given problem, torque is 81.6 Nm, and the moment of inertia is 40.8 kg·m². Therefore, the angular acceleration is calculated to be approximately 2 rad/s². This relationship underscores how torque and rotational inertia combine to influence how quickly an object's rotational speed changes.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to change in its rotational motion. It's akin to mass in linear motion but applies to rotating bodies. Simply put, it tells us how hard it is to start or stop the object from spinning. The moment of inertia, often denoted by \( I \), depends on the mass distribution: how far the mass is distributed from the axis of rotation.

Understanding the Formula

For common geometries, the moment of inertia can be calculated using specific formulas. For a rectangular section of a door considered as a thin rod rotating about one end, the formula is:
  • \( I = \frac{1}{3} m r^2 \)
In the exercise, each section has a mass \( m \) of 85 kg and the distance \( r \) is 1.2 m from the pivot point. Plugging these values into the formula, we get:
  • \( I = \frac{1}{3} \times 85 \ \text{kg} \times (1.2 \ \text{m})^2 = 40.8 \ \text{kg} \cdot \text{m}^2 \)
This result means that each section of the door requires a certain amount of effort to change its current state of motion, highlighting how mass and distance from the pivot jointly affect rotational dynamics.
Rotational Motion
Rotational motion involves objects spinning around an axis. It is a key branch of mechanics and is fundamental to understanding how doors swing, wheels roll, and even planets orbit. Key parameters of rotational motion include angular velocity, angular acceleration, and the forces causing these rotations, namely torque.

Key Characteristics

In rotation, similar to linear motion:
  • Angular velocity refers to how fast an object spins; it's the rotational counterpart of linear velocity.
  • Torque, as the turning force, influences the start, stop, or change in rotation.
  • Angular acceleration represents the rate of change of angular velocity.
For the rotating door exercise, the person applying force at the edge generates torque, leading to angular acceleration. This principle helps explain how forces applied at a distance from the pivot can cause and increase rotational effects, a fundamental concept in systems operating rotational dynamics.

Real-World Examples

Examples of rotational motion are all around us, including playground merry-go-rounds swaying to a push or car wheels spinning as the vehicle accelerates. Recognizing these dynamics lets us grasp how forces interact with objects to cause motion around a point.

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Most popular questions from this chapter

A thin rod has a length of \(0.25 \mathrm{m}\) and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of \(0.32 \mathrm{rad} / \mathrm{s}\) and a moment of inertia of \(1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\), A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (mass \(=4.2 \times 10^{-3} \mathrm{kg}\) ) gets where it's going, what is the angular velocity of the rod?

Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of \(6.6 \mathrm{m} / \mathrm{s}\). Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

A particle is located at each corner of an imaginary cube. Each edge of the cube is \(0.25 \mathrm{m}\) long. and each particle has a mass of \(0.12 \mathrm{kg}\) What is the moment of inertia of these particles with respect to an axis that lies along one edge of the cube?

Two children hang by their hands from the same tree branch. The branch is straight, and grows out from the tree trunk at an angle of 27.08 above the horizontal. One child, with a mass of 44.0 kg, is hanging 1.30 m along the branch from the tree trunk. The other child, with a mass of 35.0 kg, is hanging 2.10 m from the tree trunk. What is the magnitude of the net torque exerted on the branch by the children? Assume that the axis is located where the branch joins the tree trunk and is perpendicular to the plane formed by the branch and the trunk.

A solid cylindrical disk has a radius of 0.15 \(\mathrm{m}\). It is mounted to an axle that is perpendicular to the circular end of the disk at its center. When a \(45-N\) force is applied tangentially to the disk, perpendicular to the radius, the disk acquires an angular acceleration of \(120 \mathrm{rad} / \mathrm{s}^{2}\) What is the mass of the disk?

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