/*! 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 40 Multiple-Concept Example 12 revi... [FREE SOLUTION] | 91Ó°ÊÓ

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Multiple-Concept Example 12 reviews the concepts that play roles in this problem. A block (mass \(=2.0 \mathrm{~kg}\) ) is hanging from a massless cord that is wrapped around a pulley (moment of inertia \(\left.=1.1 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of \(0.040 \mathrm{~m}\) during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.

Short Answer

Expert verified
The angular acceleration of the pulley is approximately 238.10 rad/s², and the tension in the cord is approximately 6.554 N.

Step by step solution

01

Understand the System

We have a block of mass 2.0 kg hanging from a pulley. The pulley has a moment of inertia of 1.1 × 10^-3 kg·m², and the radius of the cord around the pulley is 0.040 m. The block falls, causing the pulley to rotate, and we need to find the angular acceleration of the pulley and the tension in the cord.
02

Establish the Relationship between Linear and Angular Quantities

The linear acceleration of the block (\( a \)) is related to the angular acceleration (\( \alpha \)) of the pulley by the formula \( a = \alpha r \), where \( r \) is the radius of the pulley (0.040 m).
03

Newton's Second Law for the Block

Apply Newton's second law for the block:\[ F = ma \]where \( F \) is the net force acting on the block, \( m \) is the mass (2.0 kg), and \( a \) is the linear acceleration. The forces acting on the block are gravity (\( mg \)) and tension (\( T \)) in the cord. Thus, \( mg - T = ma \).
04

Torque and Angular Acceleration

The torque (\( \tau \)) on the pulley due to the tension in the cord is given by:\[ \tau = Tr \]where \( T \) is the tension and \( r \) is the radius (0.040 m). Using the rotational analog of Newton's second law, \( \tau = I\alpha \), where \( I \) is the moment of inertia of the pulley (1.1 × 10^-3 kg·m²), we have:\[ Tr = I\alpha \]
05

Solve the Equations Simultaneously

We have two equations:1. \( mg - T = ma \)2. \( Tr = I\alpha \)Using \( a = \alpha r \) from Step 2, substitute for \( a \) in the first equation, yielding:\[ mg - T = m\alpha r \]From the torque equation: \[ T = \frac{I\alpha}{r} \]Substitute \( T \) into the force equation:\[ mg - \frac{I\alpha}{r} = m\alpha r \]
06

Solve for Angular Acceleration

Rearrange the equation to solve for \( \alpha \):\[ mg = m\alpha r + \frac{I\alpha}{r} \]Factor \( \alpha \) out:\[ mg = \alpha (mr^2 + I) \]Thus,\[ \alpha = \frac{mg}{mr^2 + I} \]Substitute \( m = 2.0 \) kg, \( g = 9.8 \) m/s², \( r = 0.040 \) m, and \( I = 1.1 \times 10^{-3} \) kg·m² to find:\[ \alpha = \frac{2.0 \times 9.8}{2.0 \times 0.040^2 + 1.1 \times 10^{-3}} \approx 238.10 \] rad/s².
07

Solve for Tension in the Cord

Using \( T = \frac{I\alpha}{r} \), substitute \( \alpha = 238.10 \) rad/s²:\[ T = \frac{1.1 \times 10^{-3} \times 238.10}{0.040} \approx 6.554 \] N.

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Key Concepts

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

Angular Acceleration
Angular acceleration (\( \alpha \)) is a measure of how quickly an object speeds up or slows down its rotational motion. It is analogous to linear acceleration in straight-line motion.
The unit of angular acceleration is radians per second squared (\( \text{rad/s}^2 \)). In the context of the pulley system, angular acceleration describes how the pulley's rotation changes as the block falls.
The relationship between linear acceleration (\( a \)) and angular acceleration is given by the formula:
  • \( a = \alpha r \)
Here, \( r \) is the radius of the pulley. This equation helps in linking the linear fall of the block with the spinning motion of the pulley. If the linear acceleration is known, this formula can easily determine the angular acceleration.
Understanding this relationship is crucial because it bridges the physical connection between different types of motion involved.
Newton's Second Law
Newton's Second Law is a fundamental principle that explains how the velocity of an object changes when it is subject to an external force. The law can be expressed as:
  • \( F = ma \)
In this formula, \( F \) is the total force applied to the object, \( m \) is the object's mass, and \( a \) is the linear acceleration.
When applied to the block in the pulley system, Newton's Second Law helps determine the forces at play. The main forces on the block are gravitational force (\( mg \)) pulling it down and the tension (\( T \)) in the cord pulling upwards.
Consequently, Newton's Second Law helps form the equation:
  • \( mg - T = ma \)
This shows the relationship between the acceleration of the block and the tension in the cord. Solving this equation is essential for finding the value of the tensile force in the cord which is required for the pulley to rotate.
Torque
Torque (\( \tau \)) is a measure of the rotational force applied to an object causing it to rotate around an axis. It can be thought of as the equivalent of force in a linear system but for rotational movement.
The mathematical expression for torque is:
  • \( \tau = Tr \)
where \( T \) is the tension force in the cord and \( r \) is the radius of the pulley.
This torque results in angular acceleration based on the moment of inertia. For the pulley system, this is expressed by the equation:
  • \( \tau = I\alpha \)
Torque is crucial for calculating how much tension is required to cause a specific angular acceleration in the pulley. Without this torque, the pulley would not be able to rotate, and therefore the block would not fall.
Moment of Inertia
Moment of inertia (\( I \)) refers to an object's resistance to change in its rotational motion. It is the rotational equivalent of mass in linear motion and depends on both the object's mass and the distribution of that mass around the axis of rotation.
For the pulley in this exercise, the given moment of inertia is \( 1.1 \times 10^{-3} \text{ kg} \cdot \text{m}^2 \). This value indicates how much effort in terms of torque is needed to change the pulley's rotational velocity.
In the formula:
  • \( I\alpha \)
moment of inertia works alongside torque to determine the angular acceleration. The greater the moment of inertia, the more torque required for the same angular acceleration.
This concept is crucial as it affects the calculations for both the angular acceleration of the pulley and the resulting tension in the cord.

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Most popular questions from this chapter

As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. (a) What applies the force to the person to create the torque causing this acceleration? What is the direction of this force? (b) According to Newton's actionreaction law, what can you say about the direction of the force applied to the carousel by the person and about the nature (clockwise or counterclockwise) of the torque that it creates? (c) Does the torque identified in part (b) increase or decrease the angular speed of the carousel?

Three objects lie in the \(x, y\) plane. Each rotates about the \(z\) axis with an angular speed of \(6.00 \mathrm{rad} / \mathrm{s} .\) The mass \(m\) of each object and its perpendicular distance \(r\) from the \(z\) axis are as follows: \((1) m_{1}=6.00 k g\) and \(r_{1}=2.00 \mathrm{~m},(2)\) \(m_{2}=4.00 \mathrm{~kg}\) and \(r_{2}=1.50 \mathrm{~m},(3) \mathrm{m}_{3}=3.00 \mathrm{~kg}\) and \(r_{3}=3.00 \mathrm{~m} .\) (a) Find the tangential speed of each object. (b) Determine the total kinetic energy of this system using the expression \(\mathrm{KE}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+\frac{1}{2} m_{3} v_{3}^{2}\) (c) Obtain the moment of inertia of the system. (d) Find the rotational kinetic energy of the system using the relation \(\frac{1}{2} I \omega^{2}\) to verify that the answer is the same as that in (b).

Consult Multiple-Concept Example 10 to review an approach to problems such as this. A \(\mathrm{CD}\) has a mass of \(17 \mathrm{~g}\) and a radius of \(6.0 \mathrm{~cm} .\) When inserted into a player, the \(\mathrm{CD}\) starts from rest and accelerates to an angular velocity of \(21 \mathrm{rad} / \mathrm{s}\) in \(0.80 \mathrm{~s}\). Assuming the \(\mathrm{CD}\) is a uniform solid disk, determine the net torque acting on it.

A uniform board is leaning against a smooth vertical wall. The board is at an angle \(\underline{\theta}\) above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is \(0.650\). Find the smallest value for the angle \(\theta\), such that the lower end of the board does not slide along the ground.

A cylindrically shaped space station is rotating about the axis of the cylinder to create artificial gravity. The radius of the cylinder is \(82.5 \mathrm{~m}\). The moment of inertia of the station without people is \(3.00 \times 10^{9} \mathrm{~kg} \cdot \mathrm{m}^{2}\). Suppose 500 people, with an average mass of \(70.0 \mathrm{~kg}\) each, live on this station. As they move radially from the outer surface of the cylinder toward the axis, the angular speed of the station changes. What is the maximum possible percentage change in the station's angular speed due to the radial movement of the people?

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