Problem 67 - University Physics with Modern Physics

Question

Atwood's Machine. Figure Pl0.67 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C,\) and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be 4.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) , respectively, the moment of inertia of the wheel about its axis be \(0.300 \mathrm{kg} \cdot \mathrm{m}^{2},\) and the radius of the wheel be 0.120 \(\mathrm{m} .\)

Step-by-Step Solution

Answer

The accelerations are 2.00 m/s² for the blocks and 16.67 rad/s² for the wheel. Tensions are 47.2 N and 15.6 N.

Step 1: Identify the Forces

In an Atwood's machine, two blocks are connected by a cord running over a pulley. Each block experiences a gravitational force: \( F_{A} = m_A imes g \) and \( F_{B} = m_B imes g \), where \( g = 9.8 \, \text{m/s}^2 \). Calculate these forces: \( F_{A} = 4.00 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 39.2 \, \text{N} \) and \( F_{B} = 2.00 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \).

Step 2: Set Up the Equations for Acceleration

Using Newton's second law for both blocks, consider the tensions: \( T_1 \) and \( T_2 \). For block \( A \): \( m_Aa = T_1 - m_Ag \). For block \( B \): \( m_Ba = m_Bg - T_2 \). Assuming both blocks have the same linear acceleration \( a \), and using force constraints due to no slipping, related linear and angular accelerations by \( a = r\alpha \).

Step 3: Equation for the Pulley

The torque \( \tau \) on the pulley relates to the tensions and angular acceleration: \( r(T_1 - T_2) = I \alpha \), where \( I = 0.300 \, \text{kg} \cdot \text{m}^2 \). Using \( \alpha = \frac{a}{r} \), substitute to get: \( r(T_1 - T_2) = I \frac{a}{r} \), leads to \( a = \frac{r^2(T_1 - T_2)}{I} \).

Step 4: Solve for Linear Acceleration

Substitute expressions for \( T_1 \) and \( T_2 \) from the previous equations: - For block \( A \): \( T_1 = m_Aa + m_Ag \),- For block \( B \): \( T_2 = m_Bg - m_Ba \).Substitute these into the pulley torque equation:\[ a = \frac{r^2((m_Aa + m_Ag) - (m_Bg - m_Ba))}{I} \] Simplifying gives \[ a = \frac{r^2(m_A + m_B)g}{m_A + m_B + \frac{I}{r^2}} \].Calculate \( a \): \( a = \frac{(0.120)^2(4.00 + 2.00) \times 9.8}{4.00 + 2.00 + \frac{0.300}{(0.120)^2}} = 2.00 \, \text{m/s}^2 \).

Step 5: Calculate Angular Acceleration

Given \( a = r \alpha \), solve for \( \alpha \) using \( \alpha = \frac{a}{r} = \frac{2.00}{0.120} = 16.67 \text{rad/s}^2 \).

Step 6: Find the Tensions

Substitute the acceleration \( a = 2.00 \, \text{m/s}^2 \) back into the tension equations:\( T_1 = 4.00 \, \text{kg} \times (2.00 + 9.8) = 47.2 \, \text{N} \),\( T_2 = 2.00 \, \text{kg} \times (9.8 - 2.00) = 15.6 \, \text{N} \).

Key Concepts

Newton's Second Law

In any physical scenario, Newton's second law is a fundamental principle that describes how the motion of objects is influenced by forces. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The equation for Newton's second law is expressed as \( F = ma \), where \( F \) is the net force, \( m \) is the mass of the object, and \( a \) is the acceleration.
In the context of Atwood’s machine, we apply this law to both blocks to find their accelerations. For each block, we set up an equation using the gravitational force (weight) and the tension in the cord. By balancing these forces using Newton's second law, we can determine how each block will accelerate.
This law helps us understand the relationship between the forces exerted by the tensions in the strings and the actual motion of the blocks.

Angular Acceleration

Angular acceleration is a key concept when dealing with rotating objects like the pulley in Atwood's machine. It describes how quickly the angular velocity of the wheel changes with time. The equation for angular acceleration \( \alpha \) is given by \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega \) is the change in angular velocity and \( \Delta t \) is the time period.
For Atwood’s machine, the angular acceleration of the pulley is related to the linear acceleration of the blocks. This connection is made by the equation \( a = r\alpha \), where \( a \) is the linear acceleration, \( r \) is the radius of the pulley, and \( \alpha \) is the angular acceleration.
Therefore, the angular acceleration of the wheel can be calculated once the linear acceleration of the blocks is known, offering a mirror to understanding the rotational dynamics interconnected with linear motion.

Moment of Inertia

Moment of inertia is the resistance of an object to changes in its rotational motion. It acts like mass in linear motion. The greater the moment of inertia, the harder it is to change the state of rotation.
For the Atwood's machine, the moment of inertia \( I \) is important for calculating the rotational effects on the pulley. It depends on both the mass and the shape of the wheel. Mathematically, it is often defined as \( I = \sum m_ir_i^2 \), where \( m_i \) are the masses and \( r_i \) are the distances to the axis of rotation.
In the exercise, this value is used to compute the relationship between the angular acceleration and the torques exerted by the different tensions on the pulley, enabling us to solve for the linear accelerations.

Torque

Torque is essentially the rotational equivalent of linear force. It describes how much a force acting on an object causes that object to rotate. The formula for torque \( \tau \) is \( \tau = r \times F \), where \( r \) is the distance from the axis of rotation to where the force is applied, and \( F \) is the force.
In Atwood’s machine, the torque experienced by the pulley due to the tensions can be calculated via the difference in tensions multiplied by the radius of the pulley: \( \tau = r(T_1 - T_2) \).
This torque is related to the angular acceleration of the wheel through the equation \( \tau = I \alpha \), thereby linking the linear and angular analyses. By using this torque relationship, one can determine how the tensions influence the rotation and, subsequently, the linear acceleration of the system.

Linear Acceleration

Linear acceleration refers to how the velocity of an object changes with time. It is a vector quantity, meaning it has both magnitude and direction. The standard formula for linear acceleration \( a \) is \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity over time \( \Delta t \).
For the Atwood's Machine, both masses experience linear acceleration due to gravity and the constraints of the pulley system. The machine highlights how linear acceleration of the blocks is influenced by both gravitational force and tension.
The calculation of linear acceleration for the system allows us to find out more about how both masses move in sync, maintaining the system's equilibrium. Through various forces acting in the system, we compute a net acceleration that describes how fast the blocks ascend or descend.