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A record turntable rotating at \(33 \frac{1}{3}\) rev \(/\) min slows down and stops in \(30 \mathrm{~s}\) after the motor is turned off. (a) Find its (constant) angular acceleration in revolutions per minute-squared. (b) How many revolutions does it make in this time?

Short Answer

Expert verified
Angular acceleration is \(-66\frac{2}{3}\) rpm-squared, and it makes 8.33 revolutions.

Step by step solution

01

List Known Variables

First, identify the known variables from the problem. The initial angular velocity \(\omega_0\) is \(33 \frac{1}{3}\) revolutions per minute (rpm). The final angular velocity \(\omega\) is 0 rpm (since it stops). The time \(t\) is 30 seconds.
02

Convert Time to Minutes

Since the time is given in seconds but the angular velocity is in revolutions per minute (rpm), we need to convert the time from seconds to minutes. Divide 30 seconds by 60 to get \(t = 0.5\) minutes.
03

Calculate Angular Acceleration

The formula for angular acceleration \(\alpha\) is \(\alpha = \frac{\omega - \omega_0}{t}\). Substitute the known values: \(\alpha = \frac{0 - 33\frac{1}{3}}{0.5} = -66\frac{2}{3}\) revolutions per minute-squared. The negative sign indicates deceleration.
04

Calculate Total Revolutions

Use the formula for the angular displacement \(\theta = \omega_0 \cdot t + \frac{1}{2} \alpha \cdot t^2\). Substitute the known values: \(\theta = (33\frac{1}{3}) \times 0.5 + \frac{1}{2} \times (-66\frac{2}{3}) \times (0.5)^2 = 8.33 - 8.33 = 8.33\) revolutions.

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Key Concepts

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

Understanding Angular Velocity
In the world of rotation, angular velocity is like the rotational equivalent of linear velocity. It's a measure of how fast something spins around a central point. Imagine a record on a turntable spinning at a certain speed. That spinning speed is its angular velocity, which is often measured in revolutions per minute (rpm).
When we talk about angular velocity, there are a few key points to keep in mind:
  • Angular velocity, \(\omega\), is usually expressed in radians per second or revolutions per minute.
  • It describes the angle rotated per unit of time around a specific axis.
  • Calculating the change in angular velocity is vital to finding acceleration or deceleration rates.
The initial angular velocity in our turntable problem is \(33\frac{1}{3}\) rpm. This tells us how fast the record is turning before the motor is switched off. Its final angular velocity is 0 rpm, indicating it eventually stopped.
Exploring Angular Displacement
Angular displacement is a measure of the angle through which an object moves on a circular path. In other words, it tells us how much an object has rotated or turned.
For example, in the turntable exercise, when we talk about the number of revolutions, we are referring to angular displacement:
  • It is denoted by \(\theta\) and measured in degrees, radians, or revolutions.
  • To find angular displacement, you can use the formula: \(\theta = \omega_0 \cdot t + \frac{1}{2} \alpha \cdot t^2\), where \(\omega_0\) is the initial angular velocity, \(\alpha\) is the angular acceleration, and \(t\) is the time.
  • It gives a full account of how much rotation has occurred over a period.
From the given exercise, during the time it took for the record to stop, it made a total of 8.33 revolutions. This represents its total angular displacement.
Explaining Deceleration
Deceleration is the term used when an object's speed decreases—it is negative acceleration. In rotational contexts, this is known as negative angular acceleration. When something spins slower over time, it is decelerating.
Here’s how deceleration fits into our scenario:
  • Angular acceleration, \(\alpha\), measures how quickly angular velocity changes.
  • If the angular acceleration is negative, it means the object is slowing down.
  • The formulas for linear and angular acceleration are similar, with angular acceleration defined as \(\alpha = \frac{\omega - \omega_0}{t}\).
In the turntable problem, the initial angular velocity started at \(33\frac{1}{3}\) rpm and decreased to 0 rpm over 0.5 minutes. This means it decelerated with a constant angular acceleration of \(-66\frac{2}{3}\) revolutions per minute squared. This negative sign confirms it was slowing down until it came to a stop. This deceleration or negative acceleration is central to understanding how the completion of the rotation progress halts.

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