/*! 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 28 A plastic film moves over two dr... [FREE SOLUTION] | 91Ó°ÊÓ

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A plastic film moves over two drums. During a 4 -s interval the speed of the tape is increased uniformly from \(v_{0}=2 \mathrm{ft} / \mathrm{s}\) to \(v_{1}=4 \mathrm{ft} / \mathrm{s}\). Knowing that the tape does not slip on the drums, determine (a) the angular acceleration of drum \(B,(b)\) the number of revolutions executed by drum \(B\) during the 4 - interval.

Short Answer

Expert verified
Angular acceleration: \( \alpha = \frac{0.5}{r} \), Number of revolutions: \( \frac{\theta}{2\pi} \) based on radius \( r \).

Step by step solution

01

Determine Linear Acceleration

The linear acceleration of the tape can be found using the formula for constant acceleration: \[ a = \frac{v_{1} - v_{0}}{t} \]Substituting the given values, we have:\[ a = \frac{4 \ \text{ft/s} - 2 \ \text{ft/s}}{4 \ \text{s}} = 0.5 \ \text{ft/s}^2 \]
02

Relate Linear and Angular Acceleration

The angular acceleration \( \alpha \) of the drum relates to the linear acceleration \( a \) via the radius \( r \):\[ \alpha = \frac{a}{r} \]Without the radius provided, use the relation assuming \( r \) is known. Calculate \( \alpha \) when \( r \) becomes available.
03

Calculate Drum's Angular Distance

The angular distance \( \theta \) is obtained using the relation for angular motion:\[ \theta = \omega_{0} t + \frac{1}{2} \alpha t^2 \]First, find \( \omega_{0} \) (initial angular velocity):\[ \omega_{0} = \frac{v_{0}}{r} \]Use it with \( \alpha \) and \( t = 4 \ \text{s} \) to find \( \theta \).
04

Convert to Revolutions

Convert the angular distance in radians to revolutions using:\[ \text{revolutions} = \frac{\theta}{2\pi} \].Substitute \( \theta \) (from Step 3) after obtaining the computed value.

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Key Concepts

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

Understanding Linear Acceleration
Linear acceleration is a key concept in physics and engineering. It describes how quickly the velocity of an object changes over time. In this case, the tape's velocity increase from 2 ft/s to 4 ft/s over 4 seconds illustrates linear acceleration.

Linear acceleration is expressed using the formula: \[ a = \frac{\Delta v}{t} \]where \( \Delta v \) is the change in velocity and \( t \) is the time over which the change occurs. In our example, \( \Delta v = 4 \, \text{ft/s} - 2 \, \text{ft/s} \), providing us with a very steady acceleration rate:
  • Initial velocity, \( v_0 = 2 \, \text{ft/s} \)
  • Final velocity, \( v_1 = 4 \, \text{ft/s} \)
  • Time interval, \( t = 4 \, \text{s} \)
  • Resulting in acceleration, \( a = 0.5 \, \text{ft/s}^2 \)
This consistent increase helps in smoothly transitioning to understanding the relationship between linear and angular motion, crucial for systems involving rotational dynamics.
Linking Linear and Angular Motion
Angular motion occurs when an object rotates around a specific point, and it’s deeply linked to linear motion when the linear path is traced over a curve, like a drum or wheel. In the case of a drum and tape setup, the tape's motion translates into the drum’s rotational motion.

The primary point of interest is the angular acceleration \( \alpha \), which signifies how fast the angular velocity changes. To compute \( \alpha \), we use the formula:\[ \alpha = \frac{a}{r} \]Here, \( a \) is linear acceleration, and \( r \) is the drum's radius. This shows us the dependency of angular acceleration on both linear speed and the drum size, highlighting a fundamental dynamic in understanding rotating systems.
  • Linear acceleration \( (a) \) serves as a source of angular motion.
  • Drum's radius \( (r) \) affects how drastically the rotational speed changes.
  • Without \( r \), we cannot compute \( \alpha \) precisely but conceptually understand rotational impacts.
This interrelationship is pivotal when extrapolating data or designs relating to machinery or mechanical systems.
Converting Angular Distance to Revolutions
Conversion of angular motion into practical terms like revolutions is significant for understanding the extent of rotation in real-world scenarios. In this exercise, we calculated the angular distance \( \theta \), which determines how much drum \( B \) turns.

To understand how far the drum rotates, we need to convert \( \theta \), measured in radians, into revolutions. The conversion formula is:\[ \text{revolutions} = \frac{\theta}{2\pi} \]This formula defines the relationship between radians and full circles, given that \( 2\pi \) radians equals one full rotation. This process is critical in several engineering tasks and applications, such as:
  • Calibrating mechanical systems requiring precise rotational movements.
  • Assessing performance in rotating machinery (e.g., motors, turbines).
  • Evaluating motion profiles in robotics and automation.
Through this conversion, engineers and students can effectively interpret and manipulate rotational data for efficient mechanical design and analysis.

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Most popular questions from this chapter

The rotor of a gas turbine is rotating at a speed of \(6900 \mathrm{rpm}\) when the turbine is shut down. It is observed that \(4 \mathrm{min}\) is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine ( \(a\) ) the angular acceleration, \((b)\) the number of revolutions that the rotor executes before coming to rest.

The collar \(P\) slides outward at a constant relative speed \(u\) along rod \(A B,\) which rotates counterclockwise with a constant angular velocity of \(20 \mathrm{rpm} .\) Knowing that \(r=250 \mathrm{mm}\) when when \(\theta=0\) and that the collar reaches \(B\) when \(\theta=90^{\circ},\) determine the magnitude of the acceleration of the collar \(P\) just as it reaches \(B\)

Ring \(C\) has an inside radius of \(55 \mathrm{mm}\) and an outside radius of \(60 \mathrm{mm}\) and is positioned between two wheels \(A\) and \(B\), each of \(24 \mathrm{mm}\) outside radius. Knowing that wheel \(A\) rotates with a constant angular velocity of \(300 \mathrm{rpm}\) and that no slipping occurs, determine (a) the angular velocity of ring \(C\) and of wheel \(B,(b)\) the acceleration of the points on \(A\) and \(B\) that are in contact with \(C .\)

At the instant shown during deceleration, the velocity of an automobile is \(40 \mathrm{ft} / \mathrm{s}\) to the right. Knowing that the velocity of the contact point \(A\) of the wheel with the ground is \(5 \mathrm{ft} / \mathrm{s}\) to the right, determine (a) the instantaneous center of rotation of the wheel, \((b)\) the velocity of point \(B,\) (c) the velocity of point \(D\).

Rod \(A B\) of length 29 in. is connected by ball-and-socket joints to the rotating crank \(B C\) and to the collar \(A\). Crank \(B C\) is of length 8 in. and rotates in the horizontal \(x z\) plane at the constant rate \(\omega_{0}=10\) rad/s. At the instant shown, when crank \(B C\) is parallel to the \(z\) axis, determine the velocity of collar \(A .\)

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