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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In \(5.0 \mathrm{~s}\), it rotates 25 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the \(5.0 \mathrm{~s}\) ? (d) With the angular acceleration unchanged, through what additional angle will the disk turn during the next \(5.0 \mathrm{~s}\) ?

Short Answer

Expert verified
The angular acceleration is 2 rad/s², average angular velocity is 5 rad/s, instantaneous angular velocity at 5 s is 10 rad/s, and additional angle turned is 75 rad.

Step by step solution

01

Understand the given data and formulas

We need to find different angular measurements based on the information provided. The disk starts from rest, therefore initial angular velocity \( \omega_0 = 0 \). It rotates through an angle \( \theta = 25 \text{ rad} \) in \( t = 5 \text{ s} \). We will utilize the equations of motion for rotational kinematics: \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \) and \( \omega = \omega_0 + \alpha t \).
02

Calculate the angular acceleration \(\alpha\)

The equation for angular displacement is \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \). Given that \( \omega_0 = 0 \), the equation simplifies to \( \theta = \frac{1}{2} \alpha t^2 \). By substituting \( \theta = 25 \text{ rad} \) and \( t = 5 \text{ s} \), we have \( 25 = \frac{1}{2} \alpha (5)^2 \). Solving for \( \alpha \), we get \[ \alpha = \frac{2 \times 25}{5^2} = 2 \text{ rad/s}^2 \].
03

Calculate the average angular velocity \(\omega_{avg}\)

The average angular velocity \( \omega_{avg} \) can be calculated using the formula \( \omega_{avg} = \frac{\theta}{t} \). Substituting \( \theta = 25 \text{ rad} \) and \( t = 5 \text{ s} \), we find \( \omega_{avg} = \frac{25}{5} = 5 \text{ rad/s} \).
04

Determine the instantaneous angular velocity at \(5\text{ s}\)

To find the instantaneous angular velocity at the end of \(5\text{ s}\), use \( \omega = \omega_0 + \alpha t \). With \( \omega_0 = 0 \), \( \alpha = 2 \text{ rad/s}^2 \), and \( t = 5 \text{ s} \), we have \( \omega = 0 + 2 \times 5 = 10 \text{ rad/s} \).
05

Calculate additional angle turned in the next \(5\text{ s}\)

We use the kinematic equation again for the next interval: \( \theta' = \omega t + \frac{1}{2} \alpha t^2 \), where \( \omega = 10 \text{ rad/s}\) from Step 3, \( \alpha = 2 \text{ rad/s}^2 \), and \( t = 5 \text{ s} \). This gives \( \theta' = 10 \times 5 + \frac{1}{2} \times 2 \times 5^2 = 50 + 25 = 75 \text{ rad} \).

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Key Concepts

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

Angular Acceleration
Angular acceleration describes how quickly an object's angular velocity changes over time. For rotational motion, it's the equivalent of linear acceleration. Just like with any acceleration, this can be due to speeding up or slowing down. In rotational kinematics, angular acceleration is denoted by \( \alpha \), and is expressed in radians per second squared (rad/s²).
When dealing with problems of rotational motion, like our spinning disk, it's important to identify the constant acceleration affecting the system. Since the disk begins from rest, our initial angular velocity \( \omega_0 \) is zero. Using the equation \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), and knowing our disk rotates 25 radians in 5 seconds, we isolated \( \alpha \) and found it to be 2 rad/s². This indicates a steady increase in angular velocity, crucial to understanding the motion's development.
Through constant angular acceleration, we predict future motion, making this concept a vital tool in kinematics.
Instantaneous Angular Velocity
Instantaneous angular velocity is like taking a snapshot of an object's rotational speed at a specific moment. It gives you the exact rate at which an object rotates at that instant. It's different from average velocity because it accounts for changes happening over time.
For our disk example, we calculated the instantaneous angular velocity at the end of 5 seconds. By using the equation \( \omega = \omega_0 + \alpha t \), we applied our known values: initial angular velocity as 0, acceleration \( \alpha \) of 2 rad/s², and time \( t = 5 \) seconds. This resulted in an instantaneous angular velocity of \( 10 \text{ rad/s} \).
Understanding this helps predict how the object will perform in subsequent moments, grasping more than just the average experience.
Average Angular Velocity
Average angular velocity provides a simpler concept to understand how fast an object rotates over a period. It essentially smooths out the rotation to provide a single value indicating the average rotation rate.
The formula \( \omega_{avg} = \frac{\theta}{t} \) helps us find this average rate given the angular displacement \( \theta \), and the time \( t \). From our example, with the disk rotating 25 radians over 5 seconds, the average angular velocity is \( \frac{25}{5} = 5 \text{ rad/s} \).
Averaging is especially useful in providing a big picture view of motion, regardless of fluctuations in speed. It gives an overall sense of rotational speed over a significant duration.
Angular Displacement
Angular displacement represents the angle through which a point or line has been rotated in a given sense, around a specific axis, all starting from an initial position. Unlike linear displacement, which deals with point to point movement, angular displacement considers the rotation aspect.
In our situation with the rotating disk, the initial angular displacement is given as 25 radians, describing how much the disk rotated over the initial period. For the subsequent 5-second interval, after having gained a momentum, we calculated the further displacement using \( \theta' = \omega t + \frac{1}{2} \alpha t^2 \). With \( \omega = 10 \text{ rad/s} \), \( \alpha = 2 \text{ rad/s}^2 \), and \( t = 5 \text{ s} \), the additional angular displacement turned out to be 75 radians.
Understanding angular displacement helps us track how far an object has rotated over time and is key in predicting future rotational positions.

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Most popular questions from this chapter

A wheel, starting from rest, rotates with a constant angular acceleration of \(2.00 \mathrm{rad} / \mathrm{s}^{2}\). During a certain \(3.00 \mathrm{~s}\) interval, it turns through \(90.0\) rad. (a) What is the angular velocity of the wheel at the start of the \(3.00 \mathrm{~s}\) interval? (b) How long has the wheel been turning before the start of the \(3.00\) s interval?

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A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a 30 \(\mathrm{kg}\) box as shown in Fig. \(10-56\). The outer radius \(R\) of the device is \(0.50 \mathrm{~m}\), and the radius \(r\) of the hub is \(0.20 \mathrm{~m}\). When a constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(140 \mathrm{~N}\) is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude \(0.80 \mathrm{~m} / \mathrm{s}^{2} .\) What is the rotational inertia of the device about its axis of rotation?

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