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Chapter 18: Problem 4

Calculate the position of the centre of mass of \(2 \mathrm{~kg}\) placed at \(x=1,3 \mathrm{~kg}\) placed at \(x=4,1 \mathrm{~kg}\) placed at \(x=6\) and \(6 \mathrm{~kg}\) placed at \(x=-5\).

Short Answer

Expert verified
The center of mass is at \(-\frac{5}{6}\).

Step by step solution

01

Understand the Centre of Mass Formula

The center of mass (COM) for a system of particles can be found using the formula: \( x_{\text{COM}} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + \cdots}{m_1 + m_2 + m_3 + \cdots} \). Here, \( m_i \) are the masses and \( x_i \) are their positions on a one-dimensional line.
02

Assign Given Values

List the masses and their corresponding positions: \( m_1 = 2 \) kg at \( x_1 = 1 \), \( m_2 = 3 \) kg at \( x_2 = 4 \), \( m_3 = 1 \) kg at \( x_3 = 6 \), \( m_4 = 6 \) kg at \( x_4 = -5 \).
03

Calculate the Numerator

Using the formula, calculate the sum of the products of masses and positions: \( m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4 = (2 \times 1) + (3 \times 4) + (1 \times 6) + (6 \times -5) \).
04

Compute the Numerator

Calculate the above expression: \( 2 + 12 + 6 - 30 = -10 \).
05

Calculate the Denominator

Find the total mass: \( m_1 + m_2 + m_3 + m_4 = 2 + 3 + 1 + 6 \).
06

Compute the Denominator

Calculate the total mass: \( 2 + 3 + 1 + 6 = 12 \).
07

Calculate Centre of Mass

Use the values obtained to find the center of mass: \( x_{\text{COM}} = \frac{-10}{12} \).
08

Simplify the Result

Simplify the fraction: \( x_{\text{COM}} = -\frac{10}{12} = -\frac{5}{6} \).

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Key Concepts

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

Understanding the Centre of Mass Formula
The centre of mass formula is a fundamental tool in physics and mathematics. It helps us find the point where the mass of a system is concentrated. This point is known as the centre of mass (COM). The formula used to calculate the centre of mass for a system of particles is:
  • \[ x_{\text{COM}} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + \cdots}{m_1 + m_2 + m_3 + \cdots} \]
In this equation:
  • \(m_i\) represents the mass of each particle in the system.
  • \(x_i\) represents the position of each mass on the line.
By calculating the sum of the products of masses and their positions, we can determine where the centre of mass lies. This simplifies the study of complex systems by concentrating the mass at a single point.
Exploring the System of Particles
A system of particles refers to multiple masses that interact or are related to each other in some context. In the given problem, we have a system consisting of four particles with different masses and positions on a one-dimensional line:
  • 2 kg at position 1.
  • 3 kg at position 4.
  • 1 kg at position 6.
  • 6 kg at position -5.
To find the centre of mass of this system, we need to consider both the individual masses and their respective positions. Each mass contributes to the overall centre of mass proportionally to its quantity and location. By understanding these interactions, we can predict system behavior more accurately.
Navigating the One-Dimensional Line
When calculating the centre of mass, we often consider a one-dimensional line, making it simpler to map positions. A one-dimensional line is like a straight number line where each point is defined by a single coordinate or position value. In our problem:
  • The particles are positioned at defined points: 1, 4, 6, and -5.
  • These positions help us determine where each mass is located relative to the others.
The simplicity of a one-dimensional line allows us to focus on the linear arrangement of particles. It minimizes the complication that could arise from three-dimensional space. We simply analyze the line to find out how the masses are distributed along it, which aids in the centre of mass calculation.
Engaging in Mathematics Problem Solving
Mathematics problem solving is both an art and science that requires a clear understanding of concepts and a methodical approach. Solving for the centre of mass involves several crucial steps:
  • Listing all given data such as masses and positions.
  • Applying the centre of mass formula to compute the sum of mass-position products and total mass.
  • Solving for \( x_{\text{COM}} \) from these values.
  • Simplifying the results to find the precise location of the centre of mass.
Practicing these steps enhances problem-solving skills and helps students understand the application of theoretical concepts. Breaking down each step also makes the solution process more digestible and less overwhelming. This structured approach is essential in mathematics and aids in building critical thinking.

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Most popular questions from this chapter

Find the volume of the solid generated when the curve \(y=3 x^{2}\) for \(0 \leq x \leq 1\) is rotated around the \(y\) axis.

Calculate the average value of the given functions across the specified interval: (a) \(f(t)=1+t\) across \([0,2]\) (b) \(f(x)=2 x-1\) across \([-1,1]\) (c) \(f(t)=t^{2}\) across \([0,1]\) (d) \(f(t)=t^{2}\) across \([0,2]\) (e) \(f(z)=z^{2}+z\) across \([1,3]\)

Calculate the average value of the given functions over the specified interval: (a) \(f(x)=x^{3}\) across \([1,3]\) (b) \(f(x)=\frac{1}{x}\) across \([1,2]\) (c) \(f(t)=\sqrt{t}\) across \([0,2]\) (d) \(f(z)=z^{3}-1\) across \([-1,1]\) (e) \(f(t)=\frac{1}{t^{2}}\) across \([-3,-2]\)

The moment of inertia about a diameter of a sphere of radius \(1 \mathrm{~m}\) and mass \(1 \mathrm{~kg}\) is found by evaluating the integral $$ \frac{3}{8} \int_{-1}^{1}\left(1-x^{2}\right)^{2} \mathrm{~d} x $$ Show that the moment of inertia of the sphere is \(\frac{2}{5} \mathrm{~kg} \mathrm{~m}^{2}\).

Calculate the moment of inertia of a uniform thin rod of mass \(M\) and length \(l\) about a perpendicular axis of rotation at its end.
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