/*! 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 A cord is wrapped around the rim... [FREE SOLUTION] | 91Ó°ÊÓ

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A cord is wrapped around the rim of a wheel \(0.250 \mathrm{~m}\) in radius, and a steady pull of \(40.0 \mathrm{~N}\) is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of incrtia of the wheel about this shaft is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Compute the angular acceleration of the wheel.

Short Answer

Expert verified
The angular acceleration of the wheel is \(2.00 \mathrm{~rad/s}^{2}\).

Step by step solution

01

Identify the Known Variables

First, let's identify what information we have: the radius of the wheel \( r = 0.250 \mathrm{~m} \), the force exerted on the cord \( F = 40.0 \mathrm{~N} \), and the moment of inertia of the wheel \( I = 5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} \).
02

Relate Force to Torque

The torque \( \tau \) exerted by the force can be calculated using the formula \( \tau = F \cdot r \). Substitute the known values: \( \tau = 40.0 \mathrm{~N} \times 0.250 \mathrm{~m} = 10.0 \mathrm{~Nm} \).
03

Apply Newton’s Second Law for Rotation

Newton’s second law for rotation states \( \tau = I \cdot \alpha \), where \( \alpha \) is the angular acceleration. We can solve for \( \alpha \): \( \alpha = \frac{\tau}{I} = \frac{10.0 \mathrm{~Nm}}{5.00 \mathrm{~kg} \cdot \mathrm{m}^{2}} \).
04

Calculate Angular Acceleration

Substitute the torque and moment of inertia into the equation to find the angular acceleration: \( \alpha = \frac{10.0}{5.00} = 2.00 \mathrm{~rad/s}^{2} \).

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Key Concepts

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

Torque
Torque is a fundamental concept in rotational dynamics that describes the tendency of a force to cause an object to rotate around an axis. It's crucial when determining how forces affect rotational motion. Imagine pushing a door open; the farther you apply the force from the hinge, the easier it is to open the door. This is because torque (\( \tau \)) is the product of the force applied (\( F \)) and the distance (\( r \)) from the axis of rotation. The formula is given by:
  • \( \tau = F \cdot r \)
In our exercise, a force of \( 40.0 \mathrm{~N} \) is applied at a distance of \( 0.250 \mathrm{~m} \) from the center of a wheel, which generates a torque of \( 10.0 \mathrm{~Nm} \). Torque is what initiates the wheel's rotation.
This concept is similar to how a seesaw works, where applying force farther from the center results in greater rotational effect. Torque is key to understanding how rotational motion starts and changes.
Moment of Inertia
The moment of inertia, often denoted as \( I \), is a measure of an object's resistance to rotate about a specific axis. It plays a similar role in rotation as mass does in linear motion. The larger the moment of inertia, the harder it is to change the rotational state of the object.
Factors influencing the moment of inertia include:
  • The mass of the object
  • The distribution of that mass relative to the axis\( I = \sum m_i r_i^2 \)
Essentially, the further the mass is distributed from the axis, the higher the moment of inertia. In the exercise, the wheel's moment of inertia is given as \( 5.00 \mathrm{~kg \cdot m^2} \). This indicates how much torque is needed to achieve a certain angular acceleration. Understanding moment of inertia helps in designing systems where efficient rotational motion is desired, such as wheels, gears, or turbines.
Newton's Second Law for Rotation
Newton's second law for rotation provides a critical relationship between torque, moment of inertia, and angular acceleration. It mirrors Newton's second law for linear motion, which relates force, mass, and acceleration. The rotational form of this law is expressed as:
  • \( \tau = I \cdot \alpha \)
Here, \( \tau \) is the torque, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. By rearranging the formula, we can solve for angular acceleration: \( \alpha = \frac{\tau}{I} \).
This allows us to see how a specific torque will affect the angular speed of an object, considering its moment of inertia. In the wheel exercise, with a torque of \( 10.0 \mathrm{~Nm} \) and a moment of inertia of \( 5.00 \mathrm{~kg \cdot m^2} \), the angular acceleration comes out to be \( 2.00 \mathrm{~rad/s^2} \).
This demonstrates how increasing torque or reducing moment of inertia will increase the rate at which an object spins, illustrating a clear path from applied forces to rotational movement.

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Most popular questions from this chapter

A small block on a frictionless horizontal surface has a mass of \(0.0250 \mathrm{~kg}\). It is attached to a mascless cord passing through a hole in the surface. (Sce Figure \(10.58 .\) ) The block is originally revolving at a distance of \(0.300 \mathrm{~m}\) from the hole with an angular speed of \(1.75 \mathrm{rad} / \mathrm{s}\). The cord is then pulled from below, shortening the radius of the circle in which the block revolves to \(0.150 \mathrm{~m}\). You may treat the block as a particle. (a) Is angular momentum conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinctic encrgy of the block. (d) How much work was done in pulling the cord?

A large turntable rotates about a fixed vertical axis, making one revolution in 6.00 s. The moment of inertia of the turntable about this axis is \(1200 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A child of mass \(40.0 \mathrm{~kg}\). initially standing at the center of the turntable, runs out along a radius. What is the angular speed of the turntable when the child is \(2.00 \mathrm{~m}\) from the center, assuming that you can treat the child as a particle?

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of \(5,00 \mathrm{~N} \cdot \mathrm{m}\) is applied to the tire for \(2.00 \mathrm{~s}\), the angular speed of the tire increases from 0 to 100 rev \(/\) min. The external torque is then removed, and the wheel is brought to rest in \(125 \mathrm{~s}\) by friction in its bearings. Compute (a) the moment of inertia of the wheel about the axis of rotation, (b) the friction torque, and (c) the total number of revolutions made by the wheel in the \(125 \mathrm{~s}\) time interval.

The moment of inertia of the empty tumtable is \(1.5 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m}\). the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s}\). What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. A. \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) B. \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) C. \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) D. \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\)

If the competitor leans slightly farther back (increasing the angle between his body and the vertical) but remains stationary in this new position, which statement is true? Assume that the rope remains horizontal. A. The difference between \(T_{1}\) and \(T_{2}\) increases, balancing the increased torque about his feet that his weight produces when he leans farther back. B. The difference between \(T_{1}\) and \(T_{2}\) decreases, balancing the increased torque about his feet that his weight produces when he leans farther back. C. Neither \(T_{1}\) nor \(T_{2}\) changes because no other forces are changing. D. Both \(T_{1}\) and \(T_{2}\) change, but the difference between them remains the same.

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