/*! 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 11 A big olive \((m=0.50 \mathrm{~k... [FREE SOLUTION] | 91Ó°ÊÓ

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A big olive \((m=0.50 \mathrm{~kg})\) lies at the origin of an \(x y\) coordinate system, and a big Brazil nut \((M=1.5 \mathrm{~kg})\) lics at the point \((1.0,2.0) \mathrm{m} .\) At \(t=0,\) a force \(\vec{F}_{o}=(2.0 \mathrm{i}+3.0 \mathrm{j}) \mathrm{N}\) begins to act on the olive, and a force \(\vec{F}_{n}=(-3.0 \mathrm{i}-2.0 \mathrm{j}) \mathrm{N}\) begins to act on the nut. In unit-vector notation, what is the displacement of the center of mass of the olive-nut system at \(t=4.0 \mathrm{~s}\), with respect to its position at \(t=0 ?\)

Short Answer

Expert verified
The displacement of the center of mass is \(-4\, \mathrm{i} + 4\, \mathrm{j}\) m.

Step by step solution

01

Identify Initial Positions

The olive is at position \((0, 0)\) and the Brazil nut is at position \((1.0, 2.0)\).
02

Determine Initial Center of Mass Position

The formula for the initial center of mass \((x_{cm}, y_{cm})\) is \(x_{cm} = \frac{m \cdot x_o + M \cdot x_n}{m + M}\) and \(y_{cm} = \frac{m \cdot y_o + M \cdot y_n}{m + M}\). Using \(m = 0.50\, \mathrm{kg}\), \(M = 1.5\, \mathrm{kg}\), \(x_o = 0\), \(y_o = 0\), \(x_n = 1.0\), and \(y_n = 2.0\), we find: \(x_{cm} = \frac{0.5 \cdot 0 + 1.5 \cdot 1.0}{0.5+1.5} = 0.75 \) m and \(y_{cm} = \frac{0.5 \cdot 0 + 1.5 \cdot 2.0}{0.5+1.5} = 1.5\) m.
03

Calculate Acceleration of Each Object

For the olive, \(\vec{a}_o = \frac{\vec{F}_o}{m} = \frac{(2.0\, \mathrm{i} + 3.0\, \mathrm{j})}{0.50}\) resulting in \(\vec{a}_o = (4.0\, \mathrm{i} + 6.0\, \mathrm{j}) \mathrm{m/s}^2\). For the Brazil nut, \(\vec{a}_n = \frac{\vec{F}_n}{M} = \frac{(-3.0\, \mathrm{i} - 2.0\, \mathrm{j})}{1.5}\) resulting in \(\vec{a}_n = (-2.0\, \mathrm{i} - 1.33\, \mathrm{j}) \mathrm{m/s}^2\).
04

Determine Displacement of Each Object

The displacement \(\vec{d}\) due to constant acceleration is given by \(\vec{d} = \frac{1}{2} \vec{a} t^2\). For the olive, \(\vec{d}_o = \frac{1}{2} (4.0\, \mathrm{i} + 6.0\, \mathrm{j})(4^2)\), resulting in \(\vec{d}_o = 32\, \mathrm{i} + 48\, \mathrm{j} \mathrm{m}\). For the Brazil nut, \(\vec{d}_n = \frac{1}{2} (-2.0\, \mathrm{i} - 1.33\, \mathrm{j})(4^2)\), resulting in \(\vec{d}_n = -16\, \mathrm{i} - 10.67\, \mathrm{j} \mathrm{m}\).
05

Calculate Center of Mass Displacement

The change in the center of mass position \(\Delta x_{cm}\) and \(\Delta y_{cm}\) is found using \(\Delta x_{cm} = \frac{m \cdot \Delta x_o + M \cdot \Delta x_n}{m + M}\) and \(\Delta y_{cm} = \frac{m \cdot \Delta y_o + M \cdot \Delta y_n}{m + M}\). So, \(\Delta x_{cm} = \frac{0.5 \cdot 32 + 1.5 \cdot (-16)}{2} = -4\) m and \(\Delta y_{cm} = \frac{0.5 \cdot 48 + 1.5 \cdot (-10.67)}{2} = 4\) m.

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Key Concepts

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

Displacement
Displacement is a fundamental concept in physics that describes how far an object has moved from its initial position. Unlike distance, displacement is a vector quantity. This means it has both magnitude and direction. So, when we talk about the displacement of the center of mass of a system, we're looking at how the system's midpoint changes position over time.

In our exercise, we're interested in the displacement of the center of mass of an olive-nut system after 4 seconds, due to forces acting on the objects. To find this, we calculate the individual displacements of both the olive and the nut, then determine the overall movement of their center of mass. This approach synthesizes the contribution of each object's displacement, based on their mass and how far they each move due to the forces applied.
Unit-Vector Notation
Unit-vector notation is a compact way to express vectors, which are quantities that have both a magnitude and direction. In two dimensions, we commonly use unit vectors \(oldsymbol{i}\) and \(oldsymbol{j}\) to represent the x and y components of a vector, respectively.

For example, in our problem, the forces are given as \(oldsymbol{F}_{o} = (2.0 \boldsymbol{i} + 3.0 \boldsymbol{j})\) and \(oldsymbol{F}_{n} = (-3.0 \boldsymbol{i} - 2.0 \boldsymbol{j})\). This notation clearly shows the direction and strength of the force in each axis.

Using unit vectors helps simplify vector calculations since all vectors can be split into their \(oldsymbol{i}\) and \(oldsymbol{j}\) components, making it easier to add them or perform other mathematical operations as needed in physics calculations.
Forces and Acceleration
Forces cause objects to accelerate. This relationship is described by Newton's Second Law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this is expressed as \(\vec{a} = \frac{\vec{F}}{m}\).

In our context, we apply this formula to determine the accelerations due to the forces acting on the olive and the nut. These are:
  • For the olive: \(\vec{a}_o = (4.0 \ \boldsymbol{i} + 6.0 \ \boldsymbol{j})\ \mathrm{m/s^2}\).
  • For the nut: \(\vec{a}_n = (-2.0 \ \boldsymbol{i} - 1.33 \ \boldsymbol{j})\ \mathrm{m/s^2}\).
These accelerations tell us how the velocities of the olive and nut change over time, which we use to calculate their individual displacements over the 4 seconds of interest.
Coordinate Systems
A coordinate system is a framework used to define positions in space. In this problem, we use a two-dimensional (2D) Cartesian coordinate system where each point in the plane is determined by an ordered pair of numbers: its x-coordinate and y-coordinate.

Initially, the olive is located at the origin \(0,0\), while the nut is at \(1.0, 2.0\). As forces are applied, these coordinates will change based on the calculated displacements. The beauty of the coordinate system lies in its ability to help break down complicated movements into manageable parts.

By tracking changes in x and y coordinates, we can identify shifts in position and thus compute the center of mass displacement accurately. This structured approach ensures precision when working with multiple moving parts, especially in dynamics where several forces and movements interact.

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