Problem 58 - University Physics with Modern Physics

Question

CALC A system consists of two particles. At \(t=0\) one particle is at the origin; the other, which has a mass of \(0.50 \mathrm{kg},\) is on the \(y\) -axis at \(y=6.0 \mathrm{m} .\) At \(t=0\) the center of mass of the system is on the \(y\) -axis at \(y=2.4 \mathrm{m} .\) The velocity of the center of mass is given by \(\left(0.75 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} \hat{\imath}\) (a) Find the total mass of the system. (b) Find the acceleration of the center of mass at any time \(t .\) (c) Find the net external force acting on the system at \(t=3.0 \mathrm{s}\) .

Step-by-Step Solution

Answer

Total mass: 1.25 kg. Acceleration of center of mass: \(1.5t \text{ m/s}^2\). Net force at \(t=3\): 5.625 N.

Step 1: Understanding Center of Mass at t=0

The system consists of two particles where one of them has mass \(0.50 \text{ kg}\) and is positioned at \(y = 6.0\text{ m}\) on the \(y\)-axis. The center of mass for the system is located at \(y = 2.4\text{ m}\). The other particle must be positioned at the origin \((0, 0)\). The formula for the center of mass in the \(y\)-axis is \(y_{cm} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}\).

Step 2: Calculating the Total Mass (Part a)

Given the center of mass position on the \(y\)-axis is at \(y=2.4\text{ m}\), we can set up the equation as \(2.4 = \frac{0.5 \times 6 + m_2 \times 0}{0.5 + m_2}\). Solving this for \(m_2\) leads to \(2.4(m_2 + 0.5) = 3\) which simplifies to revealing that \(m_2 = 0.75\text{ kg}\). The total mass is \(m_1 + m_2 = 0.5 + 0.75 = 1.25\text{ kg}\).

Step 3: Determining Center of Mass Acceleration (Part b)

The velocity of the center of mass is given as \(v_{cm} = (0.75 \text{ m/s}^3) t^2\), which is a function of time. The acceleration \(a_{cm}\) is the derivative of the velocity with respect to time, i.e., \(a_{cm} = \frac{d}{dt}(0.75t^2) = 2 \times 0.75 t = 1.5t\). Therefore, \(a_{cm} = 1.5t\text{ m/s}^2\).

Step 4: Finding the Net External Force at t=3.0s (Part c)

The net external force on the system is given by \(F = m_{total} \cdot a_{cm}\), where \(m_{total} = 1.25\text{ kg}\) and \(a_{cm} = 1.5t\). At \(t = 3.0\text{ s}\), \(a_{cm} = 1.5 \times 3.0 = 4.5 \text{ m/s}^2\). Therefore, \(F = 1.25 \times 4.5 = 5.625\text{ N}\).

Step 5: Recap and Final Results

The total mass of the system is \(1.25 \text{ kg}\). The acceleration of the center of mass is \(1.5t \text{ m/s}^2\). The net external force at \(t=3.0\text{ s}\) is \(5.625 \text{ N}\).

Key Concepts

System of Particles

In physics, a **system of particles** refers to a group of particles that are being analyzed as a single entity. Each particle has its own mass and position within the system, and together, they create the entire system's dynamics. When working with such a system, one crucial concept is the **center of mass**.In your exercise, you're dealing with two particles. The first particle has a mass of **0.50 kg** and is located at **y = 6.0 m** on the y-axis. The second particle, whose mass we seek, is at the origin. The position of the center of mass is given by\[ y_{cm} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}, \]where \( m_1 = 0.50 \text{ kg} \) and \( y_1 = 6.0 \text{ m} \).- **Center of Mass Position:** Since the center of mass is located at **y = 2.4 m** along the y-axis, you can use this information to calculate the unknown mass \( m_2 \). The particle at the origin contributes \( y_2 = 0 \), simplifying the calculation.- **Total Mass Calculation:** Adding the masses of all particles gives the system's total mass. This information is used later to find other properties like force and acceleration.

Velocity and Acceleration

**Velocity** refers to the speed at which something is moving in a certain direction. In this exercise, the velocity of the center of mass of the system of particles is given by the formula\[ v_{cm} = (0.75 \text{ m/s}^3) \cdot t^2. \]This expression means that the velocity of the center of mass increases as time progresses, and this change is dependent on time squared, indicating an accelerated motion.**Acceleration** is the rate at which velocity changes over time. You find the acceleration of the center of mass by taking the time derivative of velocity:\[ a_{cm} = \frac{dv_{cm}}{dt} = \frac{d}{dt}(0.75 \cdot t^2) = 1.5 \cdot t. \]- **Understanding Acceleration:** This result means that for every unit time \( t \), the center of mass's velocity increases by \( 1.5 \text{ m/s}^2 \).Tracking velocity and acceleration in these calculations helps us understand how the system's particles will move collectively over time, which is essential for predicting future states of motion.

Net External Force

The **net external force** acting on a system is crucial for understanding how the system moves as a whole. According to Newton's second law of motion, force is the product of mass and acceleration:\[ F = m_{total} \cdot a_{cm}. \]In this exercise, you've already calculated the total mass of the system \( m_{total} = 1.25 \text{ kg} \) and its acceleration \( a_{cm} = 1.5t \text{ m/s}^2 \). At a specific time, such as \( t = 3.0 \text{ s} \), the acceleration can be calculated by substituting \( t \) into the equation:\[ a_{cm} = 1.5 \cdot 3.0 = 4.5 \text{ m/s}^2. \]Using these values, the net external force is then calculated as:\[ F = 1.25 \cdot 4.5 = 5.625 \text{ N}. \]- **Application of Force**: Understanding the net force helps predict how the system accelerates and changes its state of motion due to external influences. This allows for anticipation of the dynamic behaviors of the system over time. Grasping how external forces interact with systems of particles enables you to analyze real-world movements and behaviors systematically.