/*! 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 21 A wheel of diameter 40.0 \(\math... [FREE SOLUTION] | 91Ó°ÊÓ

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A wheel of diameter 40.0 \(\mathrm{cm}\) starts from rest and rotates with a constant angular acceleration of 3.00 \(\mathrm{rad} / \mathrm{s}^{2} .\) At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship \(a_{\text { rad }}=\omega^{2} r\) and \((b)\) from the relationship \(a_{\text { rad }}=v^{2} / r\) .

Short Answer

Expert verified
The radial acceleration is approximately 150 \( \text{m/s}^2 \) using both methods.

Step by step solution

01

Calculate the final angular velocity \( \omega \) after two revolutions

First, determine the angular distance for two revolutions, which is \( \theta = 4\pi \text{ radians} \). Use the kinematic equation for angular motion, \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), where \( \omega_0 = 0 \text{ rad/s} \) and \( \alpha = 3 \text{ rad/s}^2 \). Solving for \( t \), we have:\[ 4\pi = \frac{1}{2} \times 3 \times t^2 \]\[ t^2 = \frac{8\pi}{3} \]\[ t = \sqrt{\frac{8\pi}{3}} \text{ seconds} \]Then calculate the final angular velocity \( \omega \) using \( \omega = \omega_0 + \alpha t \):\[ \omega = 3 \times \sqrt{\frac{8\pi}{3}} \]
02

Calculate the radial acceleration using \( a_{\text{rad}} = \omega^2 r \)

First, find the radius \( r = 20 \text{ cm} = 0.2 \text{ m} \). Using \( a_{\text{rad}} = \omega^2 r \), substitute the values:1. \( \omega^2 = \left(3 \times \sqrt{\frac{8\pi}{3}} \right)^2 \).2. Calculate \( a_{\text{rad}} = \left(3 \times \sqrt{\frac{8\pi}{3}} \right)^2 \times 0.2 \).
03

Calculate the tangential velocity \( v \) at this instant

Using the relationship between angular velocity and tangential speed, \( v = \omega r \):1. Substitute the known values: \( v = \left(3 \times \sqrt{\frac{8\pi}{3}} \right) \times 0.2 \).2. Compute \( v \), the linear velocity at the edge of the wheel.
04

Compute the radial acceleration using \( a_{\text{rad}} = \frac{v^2}{r} \)

Now, using the velocity derived in the previous step, compute \( a_{\text{rad}} \):1. Compute \( v^2 \) from \( v \) obtained in step 3.2. Use \( a_{\text{rad}} = \frac{v^2}{0.2} \) to find the radial acceleration.

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Key Concepts

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

Angular Motion
Angular motion refers to the motion of a body around a fixed point or axis. It includes various parameters like angular displacement, angular velocity, and angular acceleration. Angular displacement (\( \theta \)) is the angle through which a point or line has been rotated in a specified sense about a specified axis. For example, in the given problem, the wheel undergoes an angular displacement of two revolutions, which is \( 4\pi \text{ radians} \).
Angular velocity (\( \omega \)) measures how quickly an object rotates or spins. It is defined as the change in angular displacement over time. In the case where the wheel starts from rest, we use the formula \( \omega = \omega_0 + \alpha t \) to find the final angular velocity, utilizing the given angular acceleration and time of rotation.
Finally, angular acceleration (\( \alpha \)) is the rate at which the angular velocity changes with time. A constant angular acceleration, as in this problem, simplifies calculations using the kinematic equations.
Tangential Velocity
Tangential velocity is the linear speed of something moving along a circular path. It is related to angular velocity by the equation \( v = \omega r \), where \( r \) is the radius of the circular path. In our exercise, we calculate the tangential velocity to determine how fast a point on the outer rim of the wheel is moving as it spins.
Given the wheel's radius of 0.2 meters, we compute the tangential velocity once the final angular velocity (\( \omega \)) is determined. This step highlights the relationship between linear and angular properties in rotational motion. Understanding tangential velocity is crucial for calculating radial acceleration using the formula \( a_{\text{rad}} = \frac{v^2}{r} \).
This keeps rotational dynamics grounded in familiar linear terms, making complex motion easier to understand.
Kinematic Equations
Kinematic equations describe the motion of points, bodies, and systems of bodies without consideration of the forces that cause them to move. They are especially useful in analyzing angular motion when rotation starts from rest, as demonstrated in this problem.
We apply the fundamental kinematic equation for angular motion:\[\theta = \omega_0 t + \frac{1}{2} \alpha t^2\]
This allows us to solve for the time (\( t \)) the wheel requires to complete two revolutions. Since the initial angular velocity \( \omega_0 \) is zero, calculations simplify, and we focus on finding the time and subsequent angular velocity by leveraging the known angular acceleration (\( \alpha \)).
These equations offer a systematic approach to solving rotational dynamics by connecting angular displacement, velocity, acceleration, and time — making them indispensable tools in physics and engineering.
Angular Acceleration
Angular acceleration is the change in angular velocity over time. It tells us how quickly a rotating object speeds up or slows down. For the wheel in our problem, the angular acceleration is given as \( 3.00 \text{ rad/s}^2 \), indicating that the wheel's spin rate is increasing steadily.
Angular acceleration plays a crucial role when using kinematic equations to calculate other parameters like time and velocity. Knowing that the wheel starts from rest simplifies these calculations significantly. The constant rate of \( 3.00 \text{ rad/s}^2 \) allows us to predict changes over time with straightforward formulas.
In practical terms, angular acceleration is akin to how we perceive cars speeding up on a highway. A car accelerating smoothly from a stop closely mirrors the wheel accelerating from rest, providing a tangible way to comprehend angular acceleration in everyday life.

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Most popular questions from this chapter

The spin cycles of a washing machine have two angular speeds, 423 rev/min and 640 rev/min. The internal diameter of the drum is 0.470 \(\mathrm{m}\) . (a) What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed? (b) What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed? (c) Find the laundry's maximum tangential speed and the maximum radial acceleration, in terms of \(g\) .

An airplane propeller is 2.08 \(\mathrm{m}\) in length (from tip to tip) with mass 117 \(\mathrm{kg}\) and is rotating at 2400 \(\mathrm{rpm}(\mathrm{rev} / \mathrm{min})\) about an axis through its center. You can model the propeller as a slender rod. (a) What is its rotational kinetic energy? (b) Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0\(\%\) of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

How \(I\) Scales. If we multiply all the design dimensions of an object by a scaling factor \(f,\) its volume and mass will be multiplied by \(f^{3}\) (a) By what factor will its moment of inertia be multiplied? (b) If a \(\frac{1}{48}-\) scale model has a rotational kinetic energy of \(2.5 \mathrm{J},\) what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

A cylinder with radius \(R\) and mass \(M\) has density that increases linearly with distance \(r\) from the cylinder axis, \(\rho=\alpha r\) where \(\alpha\) is a positive constant. (a) Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of \(M\) and \(R .\) (b) Is your answer greater or smaller than the moment of inertia of a cylinder of the same mass and radius but of uniform density? Explain why this result makes qualitative sense.

A safety device brings the blade of a power mower from an initial angular speed of \(\omega_{1}\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_{3}\) that was three times as great, \(\omega_{3}=3 \omega_{1} ?\)

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