/*! 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 7 The tub of a washer goes into it... [FREE SOLUTION] | 91Ó°ÊÓ

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The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for \(8.00 \mathrm{s}\), at which time it is turning at 5.00 rev/s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in \(12.0 \mathrm{s}\) Through how many revolutions does the tub turn while it is in motion?

Short Answer

Expert verified
The tub turns for a total of 50 revolutions while it is in motion.

Step by step solution

01

Determine the angular acceleration while speeding up

First, convert the speed to radians per second from revolutions per second by multiplying by \(2\pi\) (one revolution = \(2\pi\) rad). So, \(5 \, rev/s = 5 \times 2\pi = 10\pi \, rad/s\).\nThen, calculate the angular acceleration (\(\alpha_1\)) using the equation \(\alpha = \frac{\Delta \omega}{\Delta t}\). Substituting the given values, we have \(\alpha_1 = \frac{10\pi \, rad/s}{8 \, s} = 1.25\pi \, rad/s^2\).
02

Determine the angular acceleration while slowing down

Do a similar calculation for the deceleration phase. Note that since the tub is slowing down, the final velocity is 0, therefore \(\alpha_2 = \frac{-10\pi \, rad/s}{12 \, s} = -\frac{5}{6}\pi \, rad/s^2\). The negative sign indicates deceleration.
03

Determine the revolutions made in each phase

For each phase, calculate the angular displacement (\(\theta\)) using the equation \(\theta = \omega_0 t + 0.5 \alpha t^2\). In the first phase, \(\theta_1 = 0.5 \times 1.25\pi \times (8 s)^2 = 40\pi \, rad\). In the second phase, \(\theta_2 = 10\pi \times 12 s + 0.5 \times -\frac{5}{6}\pi \times (12 s)^2 = 60\pi \, rad\). Convert both to revolutions by dividing by \(2\pi\), resulting in \(20 \, rev\) and \(30 \, rev\) respectively.
04

Find the total revolutions

Add the revolutions in each phase to find the total revolutions, that is \(20 \, rev + 30 \, rev = 50 \, rev\).

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Key Concepts

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

Angular Acceleration
Imagine you are on a merry-go-round that is just starting to spin. As it picks up speed, you feel the change in rotational velocity. This change over time is what we call angular acceleration. It measures how quickly a rotating object speeds up or slows down. Mathematically, it's represented by the symbol \(\alpha\) and is calculated using the formula \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) stands for the change in angular velocity and \(\Delta t\) is the change in time.

In our washer tub example, the tub began at rest and increased its rotational speed until it reached 5 rev/s over a period of 8 seconds. By converting revolutions per second into radians per second (the standard unit for angular measurements), and employing the formula, we deduced that the angular acceleration during the spin up was \(1.25\pi \, rad/s^2\). Similarly, when the tub decelerates to a stop, the angular acceleration is negative, indicating it's slowing down, and was calculated as \(\text{-} \frac{5}{6}\pi \, rad/s^2\). This negative or positive sign in the angular acceleration denotes the direction of the change: speeding up or slowing down.
Angular Displacement
Angular displacement is the measure of the angle through which an object rotates, or in simpler terms, how many degrees, radians, or revolutions it has turned. It's quite like how you might measure the distance traveled by a car in miles or kilometers, only for rotation. For any rotational motion, angular displacement reflects the angle between the starting and ending point.

The equation used for calculating angular displacement \((\theta)\) is \(\theta = \omega_0 t + 0.5 \alpha t^2\), where \(\omega_0\) is the initial angular velocity, \((t\) is the time, and \(\alpha\) is the angular acceleration. In the washer's case, angular displacement allowed us to compute how many revolutions the tub completed before it stopped. By applying a separate calculation for both the acceleration and deceleration phases and then adding these two values together, we found that the tub turned through a total of 50 revolutions while it was in motion.
Angular Velocity
When you see an ice skater spinning or a wheel turning, you're witnessing angular velocity. It answers the question of 'how fast' an object is rotating. Specifically, angular velocity quantifies the rate of change of angular displacement over time and is often represented by the Greek letter \(\omega\). If you think of a car's speedometer showing the speed in kilometers per hour, angular velocity would be the equivalent for rotational motion, but instead measured in revolutions per minute (rpm) or, more scientifically, in radians per second.

In our example, the angular velocity of the washer's tub increased from 0 to 5 rev/s during its upswing. By leveraging the angular velocity, along with time and angular acceleration, we could accurately determine the angular displacement for both the periods of acceleration and deceleration. Knowing how to measure angular velocity allows you to understand rotational dynamics and helps to calculate other important values in physics, such as torque and moment of inertia.

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Most popular questions from this chapter

The tires of a \(1500-\mathrm{kg}\) car are \(0.600 \mathrm{m}\) in diameter, and the coefficients of friction with the road surface are \(\mu_{s}=0.800\) and \(\mu_{k}=0.600 .\) Assuming that the weight is evenly distributed on the four wheels, calculate the maximum torque that can be exerted by the engine on a driving wheel without spinning the wheel. If you wish, you may assume the car is at rest.

The combination of an applied force and a friction force produces a constant total torque of \(36.0 \mathrm{N} \cdot \mathrm{m}\) on a wheel rotating about a fixed axis. The applied force acts for \(6.00 \mathrm{s} .\) During this time the angular speed of the wheel increases from 0 to 10.0 rad/s. The applied force is then removed, and the wheel comes to rest in 60.0 s. Find (a) the moment of inertia of the wheel, (b) the magnitude of the frictional torque, and (c) the total number of revolutions of the wheel.

A horizontal \(800-\mathrm{N}\) merry-go-round is a solid disk of radius \(1.50 \mathrm{m},\) started from rest by a constant horizontal force of \(50.0 \mathrm{N}\) applied tangentially to the edge of the disk. Find the kinetic energy of the disk after \(3.00 \mathrm{s}\)

A \(4.00-\mathrm{m}\) length of light nylon cord is wound around a uniform cylindrical spool of radius \(0.500 \mathrm{m}\) and mass 1.00 kg. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude \(2.50 \mathrm{m} / \mathrm{s}^{2}\). (a) How much work has been done on the spool when it reaches an angular speed of \(8.00 \mathrm{rad} / \mathrm{s} ?\) (b) Assuming there is enough cord on the spool, how long does it take the spool to reach this angular speed? (c) Is there enough cord on the spool?

An electric motor rotating a grinding wheel at \(100 \mathrm{rev} / \mathrm{min}\) is switched off. With constant negative angular acceleration of magnitude \(2.00 \mathrm{rad} / \mathrm{s}^{2},\) (a) how long does it take the wheel to stop? (b) Through how many radians does it turn while it is slowing down?

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