/*! 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 6 The initial angular velocity and... [FREE SOLUTION] | 91Ó°ÊÓ

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The initial angular velocity and the angular acceleration of four rotating objects at the same instant in time are listed in the table that follows. For each of the objects (a), (b), (c), and (d), determine the fi nal angular speed after an elapsed time of 2.0 s. $$ \begin{array}{lcc} & \begin{array}{c} \text { Initial angular } \\ \text { velocity } \omega_{0} \end{array} & \begin{array}{c} \text { Angular } \\ \text { acceleration } \alpha \end{array} \\ \text { (a) } & +12 \mathrm{rad} / \mathrm{s} & +3.0 \mathrm{rad} / \mathrm{s}^{2} \\ \hline \text { (b) } & +12 \mathrm{rad} / \mathrm{s} & -3.0 \mathrm{rad} / \mathrm{s}^{2} \\ \hline \text { (c) } & -12 \mathrm{rad} / \mathrm{s} & +3.0 \mathrm{rad} / \mathrm{s}^{2} \\ \hline \text { (d) } & -12 \mathrm{rad} / \mathrm{s} & -3.0 \mathrm{rad} / \mathrm{s}^{2} \\ \hline \end{array} $$

Short Answer

Expert verified
Final angular speeds: (a) 18 rad/s, (b) 6 rad/s, (c) -6 rad/s, (d) -18 rad/s.

Step by step solution

01

Understand the Angular Motion Equation

The final angular velocity \( \omega \) can be found using the equation: \( \omega = \omega_0 + \alpha t \), where \( \omega_0 \) is the initial angular velocity, \( \alpha \) is the angular acceleration, and \( t \) is the elapsed time. Here, \( t = 2.0 \) s.
02

Solve for Object (a)

For object (a), \( \omega_0 = +12 \) rad/s and \( \alpha = +3.0 \) rad/s². Substitute these values into the equation: \( \omega = 12 + 3.0 \times 2.0 = 12 + 6 = 18 \text{ rad/s} \).
03

Solve for Object (b)

For object (b), \( \omega_0 = +12 \) rad/s and \( \alpha = -3.0 \) rad/s². Substitute these values into the equation: \( \omega = 12 - 3.0 \times 2.0 = 12 - 6 = 6 \text{ rad/s} \).
04

Solve for Object (c)

For object (c), \( \omega_0 = -12 \) rad/s and \( \alpha = +3.0 \) rad/s². Substitute these values into the equation: \( \omega = -12 + 3.0 \times 2.0 = -12 + 6 = -6 \text{ rad/s} \).
05

Solve for Object (d)

For object (d), \( \omega_0 = -12 \) rad/s and \( \alpha = -3.0 \) rad/s². Substitute these values into the equation: \( \omega = -12 - 3.0 \times 2.0 = -12 - 6 = -18 \text{ rad/s} \).

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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 like speed but for objects that are spinning. It's a way to say how fast something is rotating. Imagine a bike wheel spinning—angular velocity tells us how quickly it’s turning. Just like how velocity measures how fast you're moving in a straight line, angular velocity does the same but for angles around a point. It's calculated in radians per second (rad/s), which might sound fancy but is just a way of measuring rotation. In the context of our problem, we’re starting with an initial angular velocity, which is how fast the object is already spinning at the very beginning.
Angular Acceleration
Angular acceleration is all about how the speed of rotation, or angular velocity, is changing. Imagine you're on a merry-go-round, and it starts speeding up. That increase in speed is due to angular acceleration. It can either increase or decrease the speed of something spinning, depending on whether it’s positive or negative. In simple terms, if angular acceleration is positive, the object speeds up. If it's negative, the object slows down. In our exercise, each spinning object has a specific angular acceleration, measured in radians per second squared (rad/s²), which factors into determining how their angular velocities evolve over time.
Equation for Final Angular Velocity
This handy equation helps us find out what the angular velocity will be after some time has passed if we know the initial angular velocity and the angular acceleration. The formula is \( \omega = \omega_0 + \alpha t \). Here, \( \omega_0 \) is the initial angular velocity, \( \alpha \) is the angular acceleration, and \( t \) is the time duration we're looking at. You can think of \( \omega_0 \) as your starting speed, \( \alpha \) as the rate of speed change, and \( t \) as how long that change is happening. By using this equation, we can calculate the final angular velocity after 2 seconds for each object in our exercise. It’s like a recipe where you plug in the ingredients \( \omega_0 \), \( \alpha \), and \( t \) to get the outcome \( \omega \).
Rotational Kinematics
Rotational kinematics is like the circle-dance version of regular kinematics. It deals with the motion of rotating objects, using angular versions of speed, acceleration, and displacement. Just as linear kinematics involves studying the velocity and acceleration of objects moving in a straight line, rotational kinematics applies similar principles but to objects that are spinning or rotating. This branch of physics allows us to understand the motion and predict how rotating objects behave over time. It's essential for calculating the paths and speeds of any rotating body, from planets to wheels. In problems like ours, rotational kinematics is the framework that ties together angular velocity and acceleration to predict future motion.

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Most popular questions from this chapter

A ball of radius 0.200 m rolls with a constant linear speed of \(3.60 \mathrm{m} / \mathrm{s}\) along a horizontal table. The ball rolls off the edge and falls a vertical distance of \(2.10 \mathrm{m}\) before hitting the floor. What is the angular displacement of the ball while the ball is in the air?

The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 8 blades and rotates at an angular speed of 1.25 rad/s. The opening between successive blades is equal to the width of a blade. A golf ball (diameter \(\left.4.50 \times 10^{-2} \mathrm{m}\right)\) has just reached the edge of one of the rotating blades (see the drawing). Ignoring the thickness of the blades, fi nd the minimum linear speed with which the ball moves along the ground, such that the ball will not be hit by the next blade.

A stroboscope is a light that flashes on and off at a constant rate. It can be used to illuminate a rotating object, and if the flashing rate is adjusted properly, the object can be made to appear stationary. (a) What is the shortest time between flashes of light that will make a three-bladed propeller appear stationary when it is rotating with an angular speed of \(16.7 \mathrm{rev} / \mathrm{s} ?\) (b) What is the next shortest time?

A rider on a mountain bike is traveling to the left in the figure. Each wheel has an angular velocity of \(+21.7 \mathrm{rad} / \mathrm{s},\) where, as usual, the plus sign indicates that the wheel is rotating in the counterclockwise direction. (a) To pass another cyclist, the rider pumps harder, and the angular velocity of the wheels increases from \(+21.7 \mathrm{to}+28.5 \mathrm{rad} / \mathrm{s}\) in a time of \(3.50 \mathrm{s}\) (b) After passing the cyclist, the rider begins to coast, and the angular velocity of the wheels decreases from \(+28.5 \mathrm{to}+15.3 \mathrm{rad} / \mathrm{s}\) in a time of \(10.7 \mathrm{s}\) Concepts: (i) Is the angular acceleration positive or negative when the rider is passing the cyclist and the angular speed of the wheels is increasing? (ii) Is the angular acceleration positive or negative when the rider is coasting and the angular speed of the wheels is decreasing? Calculations: In both instances, (a) and (b), determine the magnitude and direction of the angular acceleration (assumed constant) of the wheels.

?wo people start at the same place and walk around a circular lake in opposite directions. One walks with an angular speed of \(1.7 \times 10^{-3}\) \(\mathrm{rad} / \mathrm{s},\) while the other has an angular speed of \(3.4 \times 10^{-3} \mathrm{rad} / \mathrm{s} .\) How long will it be before they meet?

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