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Chapter 9: Problem 32

A rotating door is made from four rectangular sections, as shown 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
Determine moment of inertia using \( I = \frac{1}{3} m r^2 \), compute torque with \( \tau = r \times 68 \, \text{N} \), and substitute in \( \alpha = \frac{\tau}{I} \).

Step by step solution

01

Identify Relevant Formulas

The formula for torque \( \tau \) is \( \tau = r \times F \), where \( r \) is the distance from the axis of rotation to where the force is applied, and \( F \) is the force.Additionally, the relationship between torque, moment of inertia \( I \), and angular acceleration \( \alpha \) is \( \tau = I \times \alpha \).
02

Calculate the Torque

Assuming the distance \( r \) (half the width of the door if opened fully) is known, the torque \( \tau \) can be calculated using \( \tau = r \times F \). Substituting the given force, \( \tau = r \times 68 \, \text{N}\).
03

Calculate Moment of Inertia

The moment of inertia \( I \) for a rectangular section rotating about one edge is \( I = \frac{1}{3} m L^2 \), where \( m = 85 \, \text{kg} \) and \( L \) is the distance from the axis of rotation to the outer edge. Since there are four sections, calculate for one and then sum them.
04

Solve for Angular Acceleration

Combine the earlier steps' results in the equation \( \tau = I \times \alpha \) to find \( \alpha = \frac{\tau}{I} \). Substitute the expressions for torque and moment of inertia to solve for \( \alpha \).
05

Plug in Known Values

Given \( F = 68 \, \text{N} \), \( m = 85 \, \text{kg} \), and \( L \) as the radius, calculate torque and moment of inertia, then solve for \( \alpha \) using the previous equations. Ensure units are consistent.

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Key Concepts

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

Torque Calculation
Torque is a measure of the rotational force applied around an axis. To calculate torque, you use the formula \( \tau = r \times F \). This formula involves two main components:
  • \( r \): The distance from the axis of rotation to the point where the force is applied. It's like the lever arm's length.
  • \( F \): The force exerted, which needs to be perpendicular to the lever arm to maximize torque.
For the rotating door, imagine where the person pushes is at the very edge, giving the maximum possible torque for that force. Visualize this as the door being easier to turn the further from the pivot you push. By inserting the force into the torque equation, you obtain \( \tau = r \times 68 \, \text{N} \). The exact exerted torque further affects the door's rotational motion.
Moment of Inertia
The moment of inertia is a property that quantifies an object鈥檚 resistance to change in its rotation. Think of it as the rotational equivalent of mass in linear motion. For an object like the rotating door, the formula \( I = \frac{1}{3} m L^2 \) is used when rotating around one edge:
  • \( m \): The mass of the object (85 kg for one section of the door).
  • \( L \): The distance from the axis of rotation to the point of application, representing the length of one door section.
Since there are four sections, you'll calculate the moment of inertia for one section and sum up to consider all of them. This step is crucial because a higher moment of inertia means more torque is required to achieve the same angular acceleration.
Angular Acceleration
Angular acceleration refers to how quickly an object's rotational speed changes. This is directly related to the applied torque and the moment of inertia via the equation \( \alpha = \frac{\tau}{I} \). Here's how it works:
  • Torque \( \tau \) causes change in angular speed. The greater the torque, the more rapid the change.
  • Moment of Inertia \( I \) opposes this change. A higher \( I \) results in a smaller acceleration for the same torque.
By substituting the calculated values for torque and moment of inertia into the equation, you determine the door's angular acceleration. This value tells you how swiftly the door starts to rotate when pushed.
Rotational Motion Concepts
Rotational motion, unlike straight-line motion, revolves around an axis. Several interrelated concepts govern this motion:
  • Torque: It's about how much force you're using to spin something around a pivot, like the door's axis here.
  • Moment of Inertia: Think of it as rotational mass. More inertia means it's tougher to change the object's state of rotation.
  • Angular Acceleration: How quickly something speeds up or slows down its rotation.
Together, these concepts explain how forces cause objects to rotate. They show the importance of both the point of force application and the distribution of mass in achieving desired rotational effects. For our door, understanding these dynamics ensures that the person's push has just the right outcome in opening the door smoothly and effectively.

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Most popular questions from this chapter

A thin uniform rod is rotating at an angular velocity of \(7.0 \mathrm{rad} / \mathrm{s}\) about an axis that is perpendicular to the rod at its center. As the drawing indicates, the rod is hinged at two places, one-quarter of the length from each end. Without the aid of external torques, the rod suddenly assumes a "u" shape, with the arms of the "u" parallel to the rotation axis. What is the angular velocity of the rotating "u"?

A solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{~kg} \cdot \mathrm{m}^{2}\). From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{~m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{~kg}\). After all the sand is in place, what is the angular velocity of the disk?

In San Francisco a very simple technique is used to turn around a cable car when it reaches the end of its route. The car rolls onto a turntable, which can rotate about a vertical axis through its center. Then, two people push perpendicularly on the car, one at each end, as shown in the drawing. The turntable is rotated one-half of a revolution to turn the car around. If the length of the car is \(9.20 \mathrm{~m}\) and each person pushes with a \(185-\mathrm{N}\) force, what is the magnitude of the net torque applied to the car?

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{~kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (see the drawing). The front brake is then applied for \(3.0 \mathrm{~s}\), and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{~m} .\) The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{k}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about \(9.3 \times 10^{19} \mathrm{~J}\).
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