/*! 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 8 A wheel is rotating about an axi... [FREE SOLUTION] | 91Ó°ÊÓ

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A wheel is rotating about an axis that is in the \(z\) -direction. The angular velocity \(\omega_{z}\) is \(-6.00 \mathrm{rad} / \mathrm{s}\) at \(t=0,\) increases linearly with time, and is \(+4.00 \mathrm{rad} / \mathrm{s}\) at \(t=7.00 \mathrm{~s}\). We have taken counterclockwise rotation to be positive. (a) Is the angular acceleration during this time interval positive or negative? (b) During what time interval is the speed of the wheel increasing? Decreasing? (c) What is the angular displacement of the wheel at \(t=7.00 \mathrm{~s}\) ?

Short Answer

Expert verified
Find the numerical value of \(\alpha\) in step 2, and use this value to determine its sign, and identify the intervals when the speed of the wheel increases or decreases in step 3. Lastly, calculate the value of angular displacement in step 4.

Step by step solution

01

Finding angular acceleration

First, calculate the angular acceleration \(\alpha\). Angular acceleration is defined as the change in angular velocity divided by the change in time. Denoted mathematically as \(\alpha = \frac{\Delta \omega_z}{\Delta t}\). Substituting given values gives \(\alpha = \frac{+4.00 \mathrm{ rad/s } - (-6.00 \mathrm{ rad/s })}{7.00 \mathrm{s} - 0 \mathrm{s}}\)
02

Determine the sign of angular acceleration

Solve the expression obtained in step 1 to find the value of \(\alpha\). The sign of \(\alpha\) will indicate whether the angular acceleration is positive or negative.
03

Identify time when speed increases and decreases

The speed of the wheel increases when its angular velocity and angular acceleration have the same sign, and decreases when they have opposite signs. Since the wheel’s angular velocity increases from -6.00 rad/s to +4.00 rad/s, and \(\alpha\) is found in step 2, its sign can be compared with the sign of the angular velocity at different times to identify when the speed is increasing or decreasing.
04

Calculate angular displacement

The angular displacement (\(\Delta \theta\)) can be calculated with the equation of motion for uniform acceleration \(\Delta \theta = \omega_{z_0}t + 0.5\alpha t^2\). Substitute the given values \(\omega_{z_0} = -6.00 \mathrm{ rad/s }, \alpha =\) the value found in step 1, and \(t = 7.00 \mathrm{s}\) into the equation to find the angular displacement at \(t = 7.00 \mathrm{s}\).

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Key Concepts

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

Angular Velocity
Understanding angular velocity is crucial when dealing with rotating objects. Angular velocity, often denoted as \( \omega \) in physics, represents the rate of change of angular displacement over time. It measures how fast an object is rotating and is typically measured in radians per second (rad/s). In the context of a rotating wheel, if it turns faster, its angular velocity increases, similar to how a car accelerates when it speeds up.

For example, if a wheel starts with an angular velocity of -6.00 rad/s and changes to +4.00 rad/s over 7.00 seconds, there's an evident change in how quickly it's spinning. The negative sign indicates a counterclockwise rotation, which is considered the positive direction in many physics problems. Therefore, an increase from a negative to a positive value suggests that the wheel's rotation direction has reversed.

In many cases, you can envision angular velocity as being analogous to linear velocity but applied to rotational motion. It allows us to describe how rotations 'speed up' or 'slow down' over time.
Angular Displacement
Angular displacement plays a pivotal role in understanding circular motion. It represents the angle through which a point or line has been rotated in a specified sense about a specified axis. Angular displacement is denoted by \( \Delta \theta \) and is measured in radians. Unlike angular velocity, displacement gives us information about the rotation itself, rather than its rate.

Interpreting Angular Displacement

Consider a wheel that has rotated from a certain initial position. The measurement of that rotation, in terms of angular distance, is the wheel's angular displacement. Think of it almost as you would the distance traveled by a car, but for an object rotating in place, the 'route' it takes is along the circumference of its circular path. In the exercise provided, calculating the angular displacement at \( t = 7.00 \mathrm{s} \) allows us to understand how far, in angular terms, the wheel has rotated from the starting point. This quantity provides valuable insight into the wheel's movement over the given period.
Equation of Motion for Uniform Acceleration
The equation of motion for uniform acceleration is an essential tool for predicting the future state of moving objects. When acceleration is constant—which is termed 'uniform acceleration'—specific equations can be applied to describe motion.

The fundamental equation \( \Delta \theta = \omega_{z_0}t + 0.5\alpha t^2 \) relates several key quantities in rotational motion: angular displacement (\(\Delta \theta\)), initial angular velocity (\(\omega_{z_0}\)), angular acceleration (\(\alpha\)), and time (\(t\)).

Uniform Acceleration in Practice

Applying this equation enables us to solve problems involving spinning objects. For instance, by knowing the initial angular velocity, the constant angular acceleration, and the time period over which the object is rotating, one can predict the object's angular displacement. The inclusion of the '0.5' factors in this equation comes from integrating the constant acceleration over time—a fundamental concept from calculus applied to physics. In the exercise, we used these principles to determine the wheel's angular displacement at the end of 7.00 seconds.

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Most popular questions from this chapter

A pulley on a frictionless axle has the shape of a uniform solid disk of mass \(2.50 \mathrm{~kg}\) and radius \(20.0 \mathrm{~cm}\). A \(1.50 \mathrm{~kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has \(4.50 \mathrm{~J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

A uniform sphere made of modeling clay has radius \(R\) and moment of inertia \(I_{1}\) for rotation about a diameter. It is flattened to a disk with the same radius \(R .\) In terms of \(I_{1},\) what is the moment of inertia of the disk for rotation about an axis that is at the center of the disk and perpendicular to its flat surface?

At \(t=3.00 \mathrm{~s}\) a point on the rim of a \(0.200-\mathrm{m}\) -radius wheel has a tangential speed of \(50.0 \mathrm{~m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude \(10.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00 \mathrm{~s}\) and \(t=0 .\) (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00 \mathrm{~s} ?\) (d) At what time will the radial acceleration equal \(g ?\)

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A uniform bar has two small balls glued to its ends. The bar is \(2.00 \mathrm{~m}\) long and has mass \(4.00 \mathrm{~kg},\) while the balls each have mass \(0.300 \mathrm{~kg}\) and can be treated as point masses. Find the moment of inertia of this combination about an axis (a) perpendicular to the bar through its center; (b) perpendicular to the bar through one of the balls; (c) parallel to the bar through both balls; and (d) parallel to the bar and \(0.500 \mathrm{~m}\) from it.

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