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Chapter 8: Problem 28

A big olive \((m=0.50 \mathrm{~kg})\) lies at the origin and a big Brazil nut \((M=1.5 \mathrm{~kg})\) lies at the point \((1.0,2.0) \mathrm{m}\) in an \(x y\) plane. At \(t_{1}=0\), a force \(\vec{F}_{o}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}+(3.0 \mathrm{~N}) \hat{\mathrm{j}}\) begins to act on the olive, and a force \(\vec{F}_{n}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-2.0 \mathrm{~N}) \hat{\mathrm{j}}\) 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 \(t_{2}=4.0 \mathrm{~s}\), with respect to its position at \(t_{1}=0\) ?

Short Answer

Expert verified
The displacement of the center of mass of the olive-nut system at \( t_2 = 4.0 \) seconds is \( -4 \boldsymbol{i} + 4 \boldsymbol{j} \) meters.

Step by step solution

01

Determine Initial Position of Center of Mass

The initial positions of the olive and the nut are given as \(0, 0\) and \(1.0, 2.0\) meters respectively. The formula for the center of mass \( (x_{cm}, y_{cm}) \) for two masses is: \[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]\[ y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} \]Plugging in the values, we get:\[ x_{cm} = \frac{0.5 \times 0 + 1.5 \times 1.0}{0.5 + 1.5} = \frac{1.5}{2.0} = 0.75 \text{ meters} \]\[ y_{cm} = \frac{0.5 \times 0 + 1.5 \times 2.0}{0.5 + 1.5} = \frac{3.0}{2.0} = 1.5 \text{ meters} \]Thus, the initial position of the center of mass is \( (0.75, 1.5) \text{ meters} \).
02

Calculate Accelerations of Olive and Nut

Using Newton's second law, acceleration \( \boldsymbol{a} \) is given by \( \boldsymbol{F} = m \boldsymbol{a} \. \) For the olive:\[ \boldsymbol{a}_o = \frac{\boldsymbol{F}_o}{m} = \frac{ (2.0 \boldsymbol{i} + 3.0 \boldsymbol{j})}{0.50} = 4.0 \boldsymbol{i} + 6.0 \boldsymbol{j} \text{ m/s}^2 \]For the nut:\[ \boldsymbol{a}_n = \frac{\boldsymbol{F}_n}{M} = \frac{ (-3.0 \boldsymbol{i} - 2.0 \boldsymbol{j})}{1.50} = -2.0 \boldsymbol{i} - \frac{4}{3} \boldsymbol{j} \text{ m/s}^2 \]
03

Determine Displacement of Olive and Nut

Since both the olive and the nut start from rest, their displacement \( \boldsymbol{s} \) can be calculated using \boldsymbol{s} = \frac{1}{2} \boldsymbol{a} t^2 \. The displacement for the olive is:\[ \boldsymbol{s}_o = \frac{1}{2} \boldsymbol{a}_o t^2 = \frac{1}{2} (4.0 \boldsymbol{i} + 6.0 \boldsymbol{j}) (4.0)^2 \]\[ \boldsymbol{s}_o = \frac{1}{2} (4.0 \boldsymbol{i} + 6.0 \boldsymbol{j}) 16 = 32 \boldsymbol{i} + 48 \boldsymbol{j} \text{ meters} \]The displacement for the nut is:\[ \boldsymbol{s}_n = \frac{1}{2} \boldsymbol{a}_n t^2 = \frac{1}{2} (-2.0 \boldsymbol{i} - \frac{4}{3} \boldsymbol{j}) (4.0)^2 \]\[ \boldsymbol{s}_n = \frac{1}{2} (-2.0 \boldsymbol{i} - \frac{4}{3} \boldsymbol{j}) 16 = -16 \boldsymbol{i} - \frac{32}{3} \boldsymbol{j} = -16 \boldsymbol{i} - 10.\bar{6} \boldsymbol{j} \text{ meters} \]
04

Calculate the Center of Mass Displacement

The displacement of the center of mass \( \boldsymbol{s}_{cm} \) is given by:\[ \boldsymbol{s}_{cm} = \frac{m_1 \boldsymbol{s}_1 + m_2 \boldsymbol{s}_2}{m_1 + m_2} \]Substitute \( m_1 = 0.50 \text{ kg}, \ \boldsymbol{s}_1 = 32 \boldsymbol{i} + 48 \boldsymbol{j}, \ m_2 = 1.5 \text{ kg}, \ \boldsymbol{s}_2 = -16 \boldsymbol{i} - 10.\bar{6} \boldsymbol{j} \):\[ \boldsymbol{s}_{cm} = \frac{0.50(32 \boldsymbol{i} + 48 \boldsymbol{j}) + 1.5(-16 \boldsymbol{i} - 10.\bar{6} \boldsymbol{j})}{0.50 + 1.5} \]\[ \boldsymbol{s}_{cm} = \frac{ (16 \boldsymbol{i} + 24 \boldsymbol{j}) + (-24 \boldsymbol{i} - 16 \boldsymbol{j})}{2} \]\[ \boldsymbol{s}_{cm} = \frac{ -8 \boldsymbol{i} + 8 \boldsymbol{j} }{2} = -4 \boldsymbol{i} + 4 \boldsymbol{j} \text{ meters} \]

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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 of motion relates the force acting on an object to its mass and acceleration. This law can be expressed with the formula: \( \vec{F} = m \vec{a} \). This equation is pivotal in calculating the acceleration of an object when a specific force is applied. By rearranging the formula, we get: \( \vec{a} = \frac{\vec{F}}{m} \), where \( \vec{a}\) is the acceleration, \( \vec{F}\) is the force, and \( m \) is the mass of the object.
Understanding how force and mass relate to acceleration helps solve problems involving the motion of objects under various forces.
Vector Notation
Vector notation is crucial when dealing with physical quantities that have both magnitude and direction, such as force, displacement, and acceleration. In our problem, forces acting on the olive and the nut are represented using vectors: \( \vec{F}_{o} = (2.0 \mathrm{N}) \hat{\mathrm{i}} + (3.0 \mathrm{N}) \hat{\mathrm{j}} \) and \( \vec{F}_{n} = (-3.0 \mathrm{N}) \hat{\mathrm{i}} + (-2.0 \mathrm{N}) \hat{\mathrm{j}} \).
Each vector component corresponds to a specific direction: \( \hat{\mathrm{i}} \) for the x-axis and \( \hat{\mathrm{j}} \) for the y-axis.
Vectors allow us to perform operations like addition and subtraction easily. For example, to find the resultant displacement, we add or subtract the vector components of the displacements of the olive and nut.
Displacement Calculation
Displacement refers to the change in position of an object and is a vector quantity. For an object starting from rest, displacement can be calculated using the formula: \( \vec{s} = \frac{1}{2} \vec{a} t^{2} \).
In this problem, we calculate the displacement of both the olive and the nut over a given time period.
For the olive: \( \vec{s}_{o} = \frac{1}{2} (4.0 \hat{\mathrm{i}} + 6.0 \hat{\mathrm{j}}) (4.0)^2 = 32 \hat{\mathrm{i}} + 48 \hat{\mathrm{j}} \) meters
For the nut: \( \vec{s}_{n} = \frac{1}{2} (-2.0 \hat{\mathrm{i}} - \frac{4}{3} \hat{\mathrm{j}}) (4.0)^2 = -16 \hat{\mathrm{i}} - 10.\bar{6} \hat{\mathrm{j}} \) meters
The center of mass displacement can then be determined using the weighted average of these displacements.
Acceleration
Acceleration is the rate of change of velocity and is influenced by the net force acting on an object and its mass. According to Newton's second law, \( \vec{a} = \frac{\vec{F}}{m} \).
In the given exercise, we first calculate the accelerations of both the olive and the nut due to the applied forces.
For the olive: \( \vec{a}_{o} = \frac{\vec{F}_{o}}{m} = \frac{(2.0 \hat{\mathrm{i}} + 3.0 \hat{\mathrm{j}})}{0.50} = 4.0 \hat{\mathrm{i}} + 6.0 \hat{\mathrm{j}} \) meters per second squared
For the nut: \( \vec{a}_{n} = \frac{\vec{F}_{n}}{M} = \frac{(-3.0 \hat{\mathrm{i}} - 2.0 \hat{\mathrm{j}})}{1.5} = -2.0 \hat{\mathrm{i}} - \frac{4}{3} \hat{\mathrm{j}} \) meters per second squared
Understanding these accelerations allows us to calculate the subsequent displacements and the overall center of mass displacement.

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Most popular questions from this chapter

Ricardo, of mass \(80 \mathrm{~kg}\), and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a \(30 \mathrm{~kg}\) canoe. When the canoe is at rest in the placid water, they exchange seats, which are \(3.0 \mathrm{~m}\) apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moves \(40 \mathrm{~cm}\) relative to a submerged log during the exchange and calculates Carmelita's mass, which she has not told him. What is it?

Two skaters, one with mass \(65 \mathrm{~kg}\) and the other with mass \(40 \mathrm{~kg}\), stand on an ice rink holding a pole of length \(10 \mathrm{~m}\) and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the \(40 \mathrm{~kg}\) skater move?

A stone is dropped at \(t_{1}=0 .\) A second stone, with twice the mass of the first, is dropped from the same point at \(t_{2}=100 \mathrm{~ms}\). (a) How far below the release point is the center of mass of the two stones at \(t_{3}=300 \mathrm{~ms}\) ? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the twostone system moving at that time?

At a certain instant, four particles have the \(x y\) coordinates and velocities given in the following table. At that instant, what are (a) the coordinates of their center of mass and (b) the velocity of their center of mass? $$\begin{array}{cccc} \text { Particle } & \text { Mass (kg) } & \text { Position (m) } & \text { Velocity (m/s) } \\ \hline 1 & 2.0 & 0,3.0 & -9.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{j}} \\ 2 & 4.0 & 3.0,0 & 6.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{i}} \\ 3 & 3.0 & 0,-2.0 & 6.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{j}} \\ 4 & 12 & -1.0,0 & -2.0 \mathrm{~m} / \mathrm{s} \hat{\mathrm{i}} \\ \hline \end{array}$$

A \(4.0 \mathrm{~kg}\) particle-like object is located at \(x=0, y=2.0 \mathrm{~m} ;\) a \(3.0 \mathrm{~kg}\) particle-like object is located at \(x=\) \(3.0 \mathrm{~m}, y=1.0 \mathrm{~m}\). At what (a) \(x\) and (b) \(y\) coordinates must a \(2.0 \mathrm{~kg}\) particle-like object be placed for the center of mass of the threeparticle system to be located at the origin?
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