Problem 57 - College Physics

Question

One mass, \(m_{1}=0.215 \mathrm{~kg},\) of an ideal Atwood machine (see Fig. 4.42) rests on the floor \(1.10 \mathrm{~m}\) below the other mass, \(m_{2}=0.255 \mathrm{~kg},\) (a) If the masses are released from rest, how long does it take \(m_{2}\) to reach the floor? (b) How high will mass \(m_{1}\) ascend from the floor? (Hint: When \(m_{2}\) hits the floor, \(m_{1}\) continues to move upward.)

Step-by-Step Solution

Answer

(a) 1.62 s, (b) 1.19 m.

Step 1: Understanding the Problem

The Atwood machine consists of two masses, \(m_1 = 0.215\, \mathrm{kg}\) and \(m_2 = 0.255\, \mathrm{kg}\), connected by a string over a pulley. \(m_2\) starts 1.10 m above, and we need to find out how long it takes to reach the ground and how high \(m_1\) ascends after \(m_2\) hits the ground.

Step 2: Calculate the Net Acceleration

Using the Atwood machine formula for acceleration, \( a = \frac{(m_2 - m_1)g}{m_1 + m_2} \), where \(g = 9.81\, \mathrm{m/s^2}\), substitute the given masses:\[ a = \frac{(0.255 - 0.215) \times 9.81}{0.215 + 0.255} = \frac{0.04 \times 9.81}{0.47} \approx 0.834\, \mathrm{m/s^2}. \]

Step 3: Calculate Time for m_2 to Reach the Floor

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(s = 1.10\, \mathrm{m}\), \(u = 0\, \mathrm{m/s}\), and \(a = 0.834\, \mathrm{m/s^2}\), solve for \(t\):\(1.10 = 0 + \frac{1}{2} \times 0.834 \times t^2\).\[ t^2 = \frac{2 \times 1.10}{0.834} \approx 2.64, \]\[ t \approx \sqrt{2.64} \approx 1.62\, \mathrm{s}. \]

Step 4: Calculate the Distance m_1 Ascends

When \(m_2\) hits the floor, the initial speed of \(m_1\) is the final speed of \(m_2\). First, calculate this speed using \(v = at\):\(v = 0.834 \times 1.62 \approx 1.35\, \mathrm{m/s}.\)Now use it to find the additional ascent using the equation \(v^2 = u^2 + 2as\), where final \(v = 0\):\(0 = 1.35^2 + 2 \times (-9.81) \times s\).\[ s = \frac{1.35^2}{2 \times 9.81} \approx 0.093\, \mathrm{m}. \]

Step 5: Total Ascend of m_1

Since \(m_1\) initially travels the same distance \(m_2\) falls (1.10 m) and then continues after \(m_2\) hits the ground, total ascent = \(1.10 + 0.093 = 1.193\, \mathrm{m}.\)

Key Concepts

Net Acceleration

The concept of net acceleration in an Atwood machine is central to understanding how this pulley system operates. When two masses are connected over a pulley, as in an ideal Atwood machine, gravity acts on both masses, creating a difference in force. This difference causes the entire system to accelerate.For the Atwood machine, net acceleration (\(a\)) is calculated using the formula:\[a = \frac{(m_2 - m_1)\cdot g}{m_1 + m_2}\]

• \(m_1\) and \(m_2\) are the two masses,\(g\) is the acceleration due to gravity, approximately \(9.81\, \mathrm{m/s^2}\).
• This formula effectively balances the gravitational forces acting on both masses to determine a single acceleration value for the system.

In our problem, substituting the given masses results in an acceleration of approximately \(0.834 \mathrm{m/s^2}\). This is the net acceleration with which the entire system moves. Understanding this acceleration is key to analyzing the motion of both masses in the Atwood apparatus.

Equations of Motion

The equations of motion are employed to describe how objects move under certain forces. For this Atwood machine problem, they are essential in predicting how long it takes one mass to reach the ground and how far the other ascends afterward.1. **Time of Descent**: For mass \(m_2\) starting from rest, the time \(t\) it takes to fall is calculated from the equation: \[s = ut + \frac{1}{2}at^2\] where \(s\) is the distance, \(u\) is the initial velocity (0 in this case), and \(a\) is acceleration. Solving gives\(t \approx 1.62 \mathrm{s}\) to reach the floor.2. **Further Ascent of \(m_1\)**: After \(m_2\) lands, \(m_1\) continues moving upward. We can find the additional height using: \[v^2 = u^2 + 2as\] where \(v\) is the final velocity (0 when it stops ascending), and \(s\) is the distance moved upward. These equations offer a mathematical framework to detail the dynamic behavior of the masses in the Atwood machine.

Kinematics

Kinematics deals with the motion of objects without considering the forces that cause them. In the context of the Atwood machine, kinematic equations describe how the two masses move through space over time.

• The kinematic analysis provided insights on velocity, displacement, and time. For example, understanding initial and final speeds of the masses as they move.
• At \(t = 0\), both masses start from rest, but as they proceed, the velocity and position change according to time and acceleration.

Kinematics thus simplifies the complex motion of the entire system into a few easy-to-understand variables — acceleration (\(a\)), initial velocity (\(u\)), time (\(t\)), and distance (\(s\)). This helps in precisely determining how each mass behaves at any given moment during the experiments.

Pulley System

A pulley system is central to the Atwood machine, enabling it to illustrate fundamental physics principles such as force balance and motion.

• The pulley facilitates the change in direction of the tension force in the string connecting the masses, allowing for vertical motion.
• The efficiency of an ideal pulley (frictionless and weightless) assumes all forces are transmitted without loss, simplifying calculations.

In this exercise, understanding how the pulley distributes weight between the two masses allows us to calculate net acceleration and predict how the system evolves over time. The Atwood machine's configuration as a pulley system effectively captures the interplay between gravity, mass, and acceleration. This makes it a staple in teaching introductory physics concepts.