/*! 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 10 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-dry cycle, starting from rest and reaching an angular speed of \(5.0 \mathrm{rev} / \mathrm{s}\) in \(8.0 \mathrm{~s}\). At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in \(12.0 \mathrm{~s}\). Through how many revolutions

Short Answer

Expert verified
The tub spins through 50 revolutions during the spin-dry cycle

Step by step solution

01

Find the Angular Acceleration

One physical quantity that can be used in our analysis of the washing machine's spinning is angular acceleration. This is a measure of the rate of change in angular velocity over time. We can find the angular acceleration, denoted as \(\alpha\), using the formula \[ \alpha = \frac{\omega_2 - \omega_1}{t} \]where \(\omega_1\) and \(\omega_2\) are the initial and final angular velocities (in revolutions/second), and \(t\) is the time (in seconds). By substituting the given values into the formula, we have: \[ \alpha = \frac{5 rev/s - 0 rev/s}{8 s} = 0.625 rev/s^2 \]
02

Calculate Rotations During Acceleration Phase

The number of rotations during the acceleration phase can be calculated using the equation \[ \theta = \omega_1 t + \frac{1}{2} \alpha t^2 \] Here, \(\theta\) is the number of rotations, \(\omega_1\) is the initial angular velocity, \(t\) is the time, and \(\alpha\) is the angular acceleration. By using the calculated angular acceleration from step 1 and the given values, we get: \[ \theta = 0 rev/s * 8s + \frac{1}{2} * 0.625 rev/s^2 * (8s)^2 = 20 revolutions\]
03

Calculate the Angular Deceleration

The angular deceleration can be calculated using the same formula as for acceleration, but in this case with the final angular velocity being 0 rev/s (since it comes to rest) and the initial angular velocity as 5 rev/s. Hence, substituting these values into the formula, we get the angular deceleration as: \[ \alpha = \frac{0 rev/s - 5 rev/s}{12s} = -0.4167 rev/s^2 \] We take the magnitude as we are only interested in the magnitude of deceleration.
04

Calculate Rotations During Deceleration Phase

The number of rotations during the deceleration phase can then be calculated using the same equation as in Step 2, but with the angular deceleration from Step 3 replacing \(\alpha\) and the time it takes for the machine to come to rest replacing \(t\). This yields: \[ \theta = 5 rev/s * 12s + \frac{1}{2} * -0.4167 rev/s^2 * (12s)^2 = 30 revolutions\]
05

Total Number of Rotations

Finally, by adding the number of rotations during the acceleration phase (from Step 2) and the deceleration phase (from Step 4), we get the total number of rotations. Hence, the total number of rotations are: \[ 20 revolutions + 30 revolutions = 50 revolutions \] Thus, the tub spins through 50 revolutions during the spin-dry cycle.

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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 key concept in understanding any object that rotates or revolves. Just like velocity tells us how fast an object is moving along a straight path, angular velocity tells us how fast an object is rotating around a point or axis. The standard unit for angular velocity is radians per second (rad/s), but it can also be expressed in revolutions per second (rev/s),

as seen in the example with the washing machine's spin cycle. In our exercise, the tub reached an angular speed of 5.0 rev/s. This angular speed represents how many revolutions the tub completes every second. To visualize this, picture the second hand on a clock which completes one revolution per minute; if this were a washing machine tub at 5.0 rev/s, it would make five complete rotations in that one second!

When the magnitude of angular velocity changes, the object experiences angular acceleration or deceleration, depending on whether the speed increases or decreases. This brings us to the importance of calculating angular acceleration to determine how quickly the washer's tub reaches its operating speed or comes to a stop.
Revolutions per Second
When we talk about revolutions per second, we're discussing a unit of angular velocity that's particularly intuitive for many people, as it corresponds to the number of full rotations an object makes within one second. It's a simple yet illustrative way to measure and understand the speed of rotating objects. In everyday machinery, such as fans, car wheels, or even our washer tub from the exercise, knowing the revolutions per second helps us gauge the performance and effectiveness of the motion.

For instance, stating that a car's tires rotate at a certain rate per second can give us a sense of what speed the car is moving at. Similarly, in our exercise, the tub's performance is characterized by how quickly it reaches 5 rev/s during the spin-dry cycle. It's also valuable for calculating the total number of revolutions throughout different phases of motion, as demonstrated in the steps provided for the exercise, ultimately enabling us to comprehend the tub's motion from beginning to end.
Physical Quantity in Rotational Motion
Physical quantities are important in describing the characteristics of rotational motion, which include angular displacement, angular velocity, and angular acceleration. These quantities parallel linear displacement, velocity, and acceleration in linear motion. For rotational motion, angular displacement refers to the change in position of a rotating body, angular velocity refers to how fast an object rotates, and angular acceleration refers to any changes in the angular velocity.

In the context of the washer exercise, we utilized the concept of angular acceleration to find out how the tub's speed increases and decreases over time. The equation \( \alpha = \frac{\omega_2 - \omega_1}{t} \) used in the example illustrates how we determine the rate of change in the tub's angular velocity, represented by \(\alpha\), over a given time interval \(t\). Understanding these physical quantities helps us to predict and analyze rotational motion in a wide range of applications, from simple machines to complex engineering systems.

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Most popular questions from this chapter

ECP A machine part rotates at an angular speed of \(0.06 \mathrm{rad} / \mathrm{s}\); its speed is then increased to \(2.2 \mathrm{rad} / \mathrm{s}\) at an angular acceleration of \(0.70 \mathrm{rad} / \mathrm{s}^{2}\). (a) Find the angle through which the part rotates before reaching this final speed. (b) In general, if both the initial and final angular speed are doubled at the same angular acceleration, by what factor is the angular displacement changed? Why? Hint: Look at the form of Equation \(7.9 .\)

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of \(20.0^{\circ}\) with the horizontal. A \(30.0\) -kg piece of luggage is placed on the carousel, \(7.46 \mathrm{~m}\) from the axis of rotation. The travel bag goes around once in \(38.0 \mathrm{~s}\). Calculate the force of static friction between the bag and the carousel. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to a position \(7.94 \mathrm{~m}\) from the axis of rotation. The bag is on the verge of slipping as it goes around once every \(34.0 \mathrm{~s}\). Calculate the coefficient of static friction between the bag and the carousel.

A \(40.0\) -kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of \(18.0 \mathrm{~m}\). (a) What is the centripetal acceleration of the child? (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? (c) What force does the seat exert on the child at the highest point of the ride? (d) What force does the seat exert on the child when the child is halfway between the top and bottom?

Suppose a 1800 -kg car passes over a bump in a roadway that follows the arc of a circle of radius \(20.4 \mathrm{~m}\), as in Figure \(\mathrm{P} 7.65 .\) (a) What force does the road exert on the car as the car passes the highest point of the bump if the car travels at \(8.94 \mathrm{~m} / \mathrm{s} ?\) (b) What is the maximum speed the car can have without losing contact with the road as it passes this highest point?

A satellite has a mass of \(100 \mathrm{~kg}\) and is located at \(2.00 \mathrm{X}\) \(10^{6} \mathrm{~m}\) above the surface of Earth. (a) What is the potential energy associated with the satellite at this location? (b) What is the magnitude of the gravitational force on the satellite?

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