/*! 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 115 A mass \(M\) is attached to a ro... [FREE SOLUTION] | 91Ó°ÊÓ

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A mass \(M\) is attached to a rope that passes over a diskshaped pulley of mass \(m\) and radius \(r .\) The mass hangs to the left side of the pulley. On the right side of the pulley, the rope is pulled downward with a force \(F .\) Find \((a)\) the acceleration of the mass, (b) the tension in the rope on the left side of the pulley, and \((c)\) the tension in the rope on the right side of the pulley. (d) Check your results in the limits \(m \rightarrow 0\) and \(m \rightarrow \infty\)

Short Answer

Expert verified
(a) \( a = \frac{F - Mg}{M + \frac{3}{2} m} \); (b) \( T_1 = M(\frac{F - Mg}{M + \frac{3}{2}m}) + Mg \); (c) \( T_2 = F - m(\frac{F - Mg}{M + \frac{3}{2}m}) \).

Step by step solution

01

Understanding the Problem

We have a mass \(M\) hanging on the left side of a pulley and a force \(F\) applied on the right side of the pulley. The pulley itself has mass \(m\) and radius \(r\). We need to find the acceleration of the mass \(M\) and the tensions in the rope on both sides of the pulley. We should consider the torque and net forces acting on the pulley.
02

Applying Newton's Second Law to the Mass

For the hanging mass \(M\), we apply Newton's second law: \( M a = T_1 - Mg \), where \(T_1\) is the tension in the rope on the left side.
03

Applying Newton's Second Law to the Pulley

The net torque on the pulley is given by \( \tau = I \alpha \), where \(I = \frac{1}{2} m r^2\) is the moment of inertia of the disk-shaped pulley, and \(\alpha = \frac{a}{r}\) is the angular acceleration. The torque caused by the tensions is \(\tau = T_1 r - T_2 r = I \alpha\). Thus, \( T_1 - T_2 = \frac{1}{2} m a \).
04

Applying Newton's Second Law to the Right Side Force

For the force \(F\) on the right side, we apply Newton's second law: \( F = T_2 - m a_2 \), where \(a_2 = a\) since the entire system moves together.
05

Solving the System of Equations

From the equations:1. \( M a = T_1 - Mg \)2. \( T_1 - T_2 = \frac{1}{2} m a \)3. \( F = T_2 + m a \)We solve for the acceleration \(a\) first, then for \(T_1\) and \(T_2\). Express each in terms of known quantities.
06

Calculating the Acceleration

Combine the equations to find the acceleration: \[ a = \frac{F - Mg}{M + \frac{3}{2} m} \]
07

Calculating the Tension on the Left

Using \( M a = T_1 - Mg \), solve for \(T_1\): \[ T_1 = M\left(\frac{F - Mg}{M + \frac{3}{2}m}\right) + Mg \]
08

Calculating the Tension on the Right

Using \( F = T_2 + m a \), solve for \(T_2\): \[ T_2 = F - m\left(\frac{F - Mg}{M + \frac{3}{2}m}\right) \]
09

Limit Analysis

In the case when \(m \rightarrow 0\), the pulley has no mass. The results simplify to:- Acceleration \( a \rightarrow \frac{F - Mg}{M} \)- Tensions become equal as \(T_1 = T_2 = Mg\).When \(m \rightarrow \infty\), the heavy pulley prevents the rope from moving:- Acceleration \( a \rightarrow 0 \)- Tensions become \(T_1 = Mg\) and \(T_2 = F\).

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Key Concepts

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

Newton's Second Law
Newton's Second Law forms the foundation of our understanding of forces acting on objects. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In formula terms, it is expressed as: \[ F = ma \] In the context of our pulley problem, this law helps us determine the acceleration and tension forces within the system. The hanging mass, represented by \( M \), is subject to gravitational force \( Mg \). This gravitational pull creates tension \( T_1 \) in the rope, and we find acceleration \( a \) by balancing these forces. Similarly, Newton's Second Law is applied to other parts of the system, evaluating the net force and subsequent tension in different sections of the rope, including when an external force \( F \) is applied on the opposite side of the pulley.
moment of inertia
The moment of inertia is an important concept in rotational motion, analogous to mass in linear motion. It quantifies an object's resistance to rotational acceleration around an axis. For our disk-shaped pulley, the moment of inertia \( I \) is calculated using the formula: \[ I = \frac{1}{2}mr^2 \] Here, \( m \) is the mass of the pulley and \( r \) is its radius. This value plays a crucial role in determining how the pulley responds to applied torques. The greater the value of the moment of inertia, the harder it is to change the pulley's rotation, directly affecting the system's behavior. In our solution, knowing the pulley's moment of inertia helps us apply and understand rotational dynamics alongside linear dynamics.
torque
Torque is essentially the rotational equivalent of force. It is an action that causes an object to rotate around an axis. The formula for torque \( \tau \) is given by: \[ \tau = rF \sin(\theta) \] Where \( r \) is the distance from the pivot point, \( F \) is the applied force, and \( \theta \) is the angle between the force and the lever arm. In our pulley scenario, torque is generated by the tensions on either side of the pulley, causing it to rotate. The net torque on the pulley is calculated by the difference in tensions multiplied by the radius, which is set equal to \( I \alpha \), where \( \alpha \) represents the angular acceleration. Understanding torque allows us to explore how forces act to change rotation, contributing to the broader equilibrium of the system.
angular acceleration
Angular acceleration represents the rate of change of angular velocity over time. It's similar to linear acceleration but applied to rotational motion. The relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) can be expressed as: \[ \alpha = \frac{a}{r} \] For the pulley system in question, this means that the linear acceleration of the rope translates into angular acceleration of the pulley. By knowing how the rope's motion affects the rotation of the pulley, we can make informed assessments about the system's dynamics.
  • Angular acceleration helps in relating the pulley's movement with the force dynamics and tensions.
  • It allows us to understand the overall rotational response of the system when forces are applied.
Incorporating angular acceleration in our analysis bridges the gap between linear and rotational motion, ensuring a comprehensive approach to solving the problem.

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