/*! 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 gymnast is performing a floor ... [FREE SOLUTION] | 91Ó°ÊÓ

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A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from 3.00 to \(5.00 \mathrm{rev} / \mathrm{s}\) while rotating through one-half of a revolution. How much time does this maneuver take?

Short Answer

Expert verified
The maneuver takes approximately 0.785 seconds.

Step by step solution

01

Understand the Problem

We need to find the time it takes for the gymnast to increase her angular velocity from 3.00 revolutions per second to 5.00 revolutions per second, while she rotates through half a revolution.
02

Identify Known Variables

Initial angular velocity \( \omega_i = 3.00 \, \text{rev/s} \). Final angular velocity \( \omega_f = 5.00 \, \text{rev/s} \). Angular displacement \( \theta = 0.5 \, \text{rev} \).
03

Convert Angular Displacement to Radians

Convert the angular displacement from revolutions to radians using the conversion factor \(2\pi \text{ radians} = 1 \text{ revolution}\). So, \( \theta = 0.5 \, \text{rev} \times 2\pi \, \text{rad/rev} = \pi \, \text{rad} \).
04

Use the Angular Kinematics Equation

We need to use the equation \( \omega_f^2 = \omega_i^2 + 2\alpha\theta \) to solve for the angular acceleration \( \alpha \). First, rearrange the equation: \( \alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta} \).
05

Calculate Angular Acceleration

Substitute the values into the equation: \( \alpha = \frac{5.00^2 - 3.00^2}{2 \times \pi} \approx \frac{25 - 9}{2\pi} = \frac{16}{2\pi} = \frac{8}{\pi} \text{ rad/s}^2 \).
06

Find Time Using Kinematics Equation

Use the equation \( \omega_f = \omega_i + \alpha t \) to solve for time \( t \). Rearrange to get \( t = \frac{\omega_f - \omega_i}{\alpha} \).
07

Calculate the Time

Substitute the known values into the time equation: \( t = \frac{5.00 - 3.00}{8/\pi} = \frac{2 \pi}{8} = \frac{\pi}{4} \text{ seconds} \approx 0.785 \text{ seconds} \).

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins around a particular axis. Think about it like the speedometer of a spinning object. The unit commonly used is revolutions per second (rev/s) or radians per second (rad/s).
To imagine this, consider a CD lightly spinning in a player. If it completes several rotations in quick succession, it has high angular velocity. In the context of our gymnast, the change in angular velocity from 3.00 rev/s to 5.00 rev/s represents how she speeds up her spin mid-air.
The formula for angular velocity is often abbreviated as \( \omega \), hence an object's angular velocity is denoted as \( \omega_i \) for initial and \( \omega_f \) for final angular velocities.
Understanding how this velocity changes is crucial in determining other aspects of motion such as angular acceleration and total time of the maneuver.
Angular Acceleration
Angular acceleration refers to the rate of change of angular velocity over time. If the spinning speed of an object like our gymnast increases or decreases, it experiences angular acceleration. It's akin to pressing a gas pedal (positive acceleration) or a brake pedal (negative acceleration) in a car.
The unit for angular acceleration is rad/s² (radians per second squared), indicating how much the angular velocity changes in one second. For the gymnast, knowing the initial and final angular velocities helps to compute angular acceleration using the formula:
  • \( \alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta} \)

This calculation involves substituting the initial and final velocities and the angular displacement in radians to find out how rapidly her spinning speed adjusts during the maneuver. A positive value here shows an increase in angular speed.
Radians Conversion
Radians are a standard unit of angular measure used in many areas of mathematics and physics. They offer a direct relationship between the arc length of a circle and its radius. Since rotations are usually measured in revolutions, conversions to radians are essential in calculations.
To convert revolutions to radians, recall that one full revolution is \(2\pi\) radians. Therefore, to convert any given rotational measure to radians, multiply by \(2\pi\).
In our gymnast's case, she spins through half a revolution. Half a revolution, or 0.5 revs, converts to radians as:
  • \( \theta = 0.5 \times 2\pi = \pi \text{ radians} \)

This conversion is crucial for applying kinematic equations where distances and angles are used as input in radians, allowing for more straightforward mathematical manipulation in the formulas.
Kinematics Equation
Kinematics equations for rotational motion are used to describe the relationships among angular displacement, initial and final angular velocities, angular acceleration, and time.
These equations are similar in structure to the familiar linear kinematics equations but apply to rotational analogs:
  • \( \omega_f^2 = \omega_i^2 + 2\alpha\theta \)
  • \( \omega_f = \omega_i + \alpha t \)
These help in determining unknown variables such as angular acceleration or time.
In our problem, the equation \( \omega_f^2 = \omega_i^2 + 2\alpha\theta \) helps find the angular acceleration, \( \alpha \). The subsequent equation \( \omega_f = \omega_i + \alpha t \) solves for time, \( t \), showing how long it takes for the gymnast to complete her spin maneuver.

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Most popular questions from this chapter

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has a length of \(64 \mathrm{cm}\) and is wound around the top at a spot where its radius is \(2.0 \mathrm{cm}\). The thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of the string, thereby unwinding it and giving the top an angular acceleration of \(+12 \mathrm{rad} / \mathrm{s}^{2} .\) What is the final angular velocity of the top when the string is completely unwound?

The trap-jaw ant can snap its mandibles shut in as little as \(1.3 \times 10^{-4}\) s. In order to shut, each mandible rotates through a \(90^{\circ}\) angle. What is the average angular velocity of one of the mandibles of the trap-jaw ant when the mandibles snap shut?

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