/*! 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 36 Two thin rectangular sheets \((0... [FREE SOLUTION] | 91Ó°ÊÓ

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Two thin rectangular sheets \((0.20 \mathrm{~m} \times 0.40 \mathrm{~m})\) are identical. In the first sheet the axis of rotation lies along the \(0.20-\mathrm{m}\) side, and in the second it lies along the \(0.40-\mathrm{m}\) side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in \(8.0 \mathrm{~s}\). How long does it take for the second sheet, starting from rest, to reach the same angular velocity?

Short Answer

Expert verified
The second sheet takes 32 seconds to reach the same angular velocity.

Step by step solution

01

Understand the Moment of Inertia Concept

The moment of inertia of a rectangular sheet depends on the axis of rotation. For a sheet with mass \( m \) and dimensions \( a \times b \), rotating about an edge, it is given by \( I = \frac{1}{3} m b^2 \) (if axis is along side \( a \)) and \( I = \frac{1}{3} m a^2 \) (if axis is along side \( b \)). Since both sheets are identical, their masses are equal.
02

Apply Moment of Inertia for Each Sheet

For the first sheet, rotating about the \(0.20 \, \mathrm{m}\) side, the moment of inertia is \( I_1 = \frac{1}{3} m (0.20)^2 \). For the second sheet, rotating about the \(0.40 \, \mathrm{m}\) side, the moment of inertia is \( I_2 = \frac{1}{3} m (0.40)^2 \).
03

Write Torque Equation

The torque equation \( \tau = I \alpha \) is used, where \( \alpha \) is the angular acceleration. Since the same torque \( \tau \) is applied to both sheets, \( \tau = I_1 \alpha_1 = I_2 \alpha_2 \).
04

Solve for Angular Accelerations

Since \( I_1 = \frac{1}{3} m (0.20)^2 \) and \( I_2 = \frac{1}{3} m (0.40)^2 \), and \( \tau = I_1 \alpha_1 = I_2 \alpha_2 \), the ratio of angular accelerations is \( \alpha_2 / \alpha_1 = I_1 / I_2 = (0.20/0.40)^2 = 0.25 \). Thus, \( \alpha_2 = 0.25 \alpha_1 \).
05

Relate Angular Velocity and Time

For constant angular acceleration, \( \omega = \alpha t \). Since both reach the same final angular velocity \( \omega \), then \( \alpha_1 t_1 = \alpha_2 t_2 \). Using \( t_1 = 8 \, \mathrm{s} \) and \( \alpha_2 = 0.25 \alpha_1 \), you get \( 8 \alpha_1 = 0.25 \alpha_1 t_2 \).
06

Solve for Time \( t_2 \)

From \( 8 \alpha_1 = 0.25 \alpha_1 t_2 \), solve for \( t_2 \): divide both sides by \( 0.25 \alpha_1 \) to get \( t_2 = 8 / 0.25 = 32 \, \mathrm{s} \).

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Key Concepts

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

Angular Acceleration
Angular acceleration is a key concept in rotational dynamics and describes how quickly an object's rotational speed changes over time. It is analogous to linear acceleration, which describes changes in speed along a straight line. For a rotating object, angular acceleration is expressed in radians per second squared (\(\text{rad/s}^2\)). This tells us how much the angular velocity of an object changes every second.

In equations, angular acceleration is often represented by \(\alpha\). It is directly related to the applied torque and the object's moment of inertia by the equation \( \tau = I \alpha \). Here, \( \tau \) is the torque applied to the object, and \( I \) is the object's moment of inertia. When you apply a torque to a sheet, just like in the exercise, the angular acceleration is inversely proportional to the moment of inertia, assuming a constant torque is applied. This means that, for two objects of different moments of inertia, the one with the larger moment will have a smaller angular acceleration if the same torque is applied.

Understanding this relationship is crucial when solving problems where you're dealing with objects of different shapes, sizes, or rotation axes. Angular acceleration tells us how forceful or gentle the rotation starts, which is a critical part of analyzing rotational motion.
Torque
Torque is the rotational equivalent of force and is crucial for causing changes in an object's rotation. It measures how effectively a force causes an object to rotate around an axis and is calculated as the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force. The equation \( \tau = I \alpha \) encapsulates the relationship between torque, moment of inertia, and angular acceleration.

Here is how you consider torque in practical terms:
  • Imagine using a wrench to tighten a bolt: the longer the wrench, the easier it is to apply a greater torque because the distance from the axis is longer.
  • Torque is measured in newton-meters (Nm). A higher torque means a more powerful twisting force.
For example, in the exercise, the same torque is applied to two sheets of the same mass but different dimensions. As torque is constant, how these sheets rotate under different moments of inertia is what creates different angular accelerations.

Understanding torque helps us not only solve problems but also grasp the deeper mechanics of rotational movements in everyday life, such as engines, swings, and spinning tops.
Rectangular Sheet Rotation
When we rotate rectangular sheets, understanding their moment of inertia is vital. Moment of inertia depends heavily on the axis about which the object rotates. For rectangular sheets, it differs based on whether the rotation is along the shorter or longer side.

Consider these two identical sheets from the exercise, one rotating about its 0.20 m side and the other about its 0.40 m side. Although they share the same material and dimensions, their rotational behaviors differ due to their different moments of inertia.
  • For the sheet rotating about the shorter side, its moment of inertia is smaller as calculated by \( I = \frac{1}{3} m (a)^2 \).
  • On the longer side, it is larger, \( I = \frac{1}{3} m (b)^2 \).
These differences arise because more mass is distributed further from the axis of rotation along the longer side, making it harder for the same torque to rotate it. This results in a slower angular acceleration for the sheet with the larger moment of inertia.

Understanding these principles of rotation helps us predict how objects like doors, paddles, and even planets might behave when set in motion. It's all about the distribution of mass and the position of the rotational axis, key factors when designing anything from machinery to fun amusement park rides.

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Most popular questions from this chapter

The drawing shows a bicycle wheel resting against a small step whose height is \(h=0.120 \mathrm{~m}\). The weight and radius of the wheel are \(W=25.0 \mathrm{~N}\) and \(r=0.340 \mathrm{~m}\). A horizontal force \(\overrightarrow{\mathbf{F}}\) is applied to the axle of the wheel. As the magnitude of \(\overrightarrow{\mathbf{F}}\) increases, there comes a time when the wheel just begins to rise up and loses contact with the ground. What is the magnitude of the force when this happens?

Concept Questions As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. (a) What applies the force to the person to create the torque causing this acceleration? What is the direction of this force? (b) According to Newton's actionreaction law, what can you say about the direction of the force applied to the carousel by the person and about the nature (clockwise or counterclockwise) of the torque that it creates? (c) Does the torque identified in part (b) increase or decrease the angular speed of the carousel? Problem The carousel has a radius of \(1.50 \mathrm{~m}\), an initial angular speed of \(3.14 \mathrm{rad} / \mathrm{s}\), and a moment of inertia of \(125 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The mass of the person is \(40.0 \mathrm{~kg}\). Find the final angular speed of the carousel after the person climbs aboard. Verify that your answer is consistent with your answers to the Concept Questions.

A baggage carousel at an airport is rotating with an angular speed of \(0.20 \mathrm{rad} / \mathrm{s}\) when the baggage begins to be loaded onto it. The moment of inertia of the carousel is 1500 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Ten pieces of baggage with an average mass of \(15 \mathrm{~kg}\) each are dropped vertically onto the carousel and come to rest at a perpendicular distance of \(2.0 \mathrm{~m}\) from the axis of rotation. (a) Assuming that no net external torque acts on the system of carousel and baggage, find the final angular speed. (b) In reality, the angular speed of a baggage carousel does not change. Therefore, what can you say qualitatively about the net external torque acting on the system?

A cylindrically shaped space station is rotating about the axis of the cylinder to create artificial gravity. The radius of the cylinder is \(82.5 \mathrm{~m}\). The moment of inertia of the station without people is \(3.00 \times 10^{9} \mathrm{~kg} \cdot \mathrm{m}^{2}\). Suppose 500 people, with an average mass of \(70.0 \mathrm{~kg}\) each, live on this station. As they move radially from the outer surface of the cylinder toward the axis, the angular speed of the station changes. What is the maximum possible percentage change in the station's angular speed due to the radial movement of the people?

A helicopter has two blades (see Figure 8-12), each of which has a mass of \(240 \mathrm{~kg}\) and can be approximated as a thin rod of length \(6.7 \mathrm{~m}\). The blades are rotating at an angular speed of \(44 \mathrm{rad} / \mathrm{s}\). (a) What is the total moment of inertia of the two blades about the axis of rotation? (b) Determine the rotational kinetic energy of the spinning blades.

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