/*! 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 1 Four particles of masses \(1 \ma... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 1

Four particles of masses \(1 \mathrm{~kg}, 2 \mathrm{~kg}, 3 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are placed at the corners \(A, B, C\) and \(D\) respectively of a square \(A B C D\) of edge \(1 \mathrm{~m}\). If point \(A\) is taken as origin, edge \(A B\) is taken along \(X\) -axis and edge \(A D\) is taken along \(Y\) -axis, the co-ordinates of centre of mass in S.I. is : (a) \((1,1)\) (b) \((5,7)\) (c) \((0.5,0.7)\) (d) none of these

Short Answer

Expert verified
(0.5, 0.7)

Step by step solution

01

Understand the Problem

We need to find the center of mass of four particles located at the corners of a square. The masses are located at specific coordinates, and the coordinates must be considered with respect to the defined origin at point \(A\).
02

Assign Coordinates to the Particles

Since \( A \) is at the origin, its coordinates are \((0,0)\). \( B \) is along the x-axis and hence has coordinates \((1,0)\). \( C \) is diagonally opposite to \( A \) with coordinates \((1,1)\), and \( D \) being on the y-axis has coordinates \((0,1)\).
03

Calculate Total Mass

Calculate the total mass \( M \) of all particles: \[ M = m_A + m_B + m_C + m_D = 1 + 2 + 3 + 4 = 10 \, \text{kg} \]
04

Apply Formula for Center of Mass (X-coordinate)

The x-component of the center of mass, \( X_{cm} \), is calculated as: \[ X_{cm} = \frac{m_A \cdot x_A + m_B \cdot x_B + m_C \cdot x_C + m_D \cdot x_D}{M} \] Substituting the values: \[ X_{cm} = \frac{1 \times 0 + 2 \times 1 + 3 \times 1 + 4 \times 0}{10} = \frac{5}{10} = 0.5 \]
05

Apply Formula for Center of Mass (Y-coordinate)

The y-component of the center of mass, \( Y_{cm} \), is calculated as: \[ Y_{cm} = \frac{m_A \cdot y_A + m_B \cdot y_B + m_C \cdot y_C + m_D \cdot y_D}{M} \] Substituting the values: \[ Y_{cm} = \frac{1 \times 0 + 2 \times 0 + 3 \times 1 + 4 \times 1}{10} = \frac{7}{10} = 0.7 \]
06

Compile Center of Mass Coordinates

Combine the x and y components of the center of mass to get the final coordinates: \((X_{cm}, Y_{cm}) = (0.5, 0.7)\).

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Key Concepts

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

Calculation of Coordinates
Understanding the coordinates of objects is the first step when finding the center of mass. In this exercise, we have four particles positioned at the corners of a square. Taking point A as the origin, we define the coordinates: A as (0, 0), B as (1, 0), C as (1, 1), and D as (0, 1). These coordinates align perfectly with the edges of the square along the X and Y axes.

This representation helps visualize their positions in the coordinate system. Each point has either an X or Y value of 1 or 0 because they reside on axis with the 1-meter edge square. Identifying coordinates in this systematic manner is essential to accurately calculate further properties such as the center of mass.
Mass Distribution
The distribution of mass across objects in a system significantly impacts where the center of mass is located. In this particular problem, each corner of the square has a particle with its own mass: A has 1 kg, B has 2 kg, C has 3 kg, and D has 4 kg.

The masses are not uniformly distributed since each corner carries a different weight. The heavier mass at D will pull the center of mass towards its coordinates more than the lighter masses at other points. Therefore, understanding how mass affects the overall balance is crucial in considering how to manage or predict the center of mass in real-world scenarios.

In formulas, mass distribution allows us to weight the coordinates effectively, making it possible to calculate central properties using a simple sum.
Coordinate System Analysis
Analyzing a coordinate system is essential when determining the center of mass. Once coordinates and mass distribution are established, the center of mass calculation begins. This involves treating the weighted sum of the coordinates, multiplied by their respective masses.

Using the formula: \[ X_{cm} = \frac{m_A \cdot x_A + m_B \cdot x_B + m_C \cdot x_C + m_D \cdot x_D}{M} \]and\[ Y_{cm} = \frac{m_A \cdot y_A + m_B \cdot y_B + m_C \cdot y_C + m_D \cdot y_D}{M} \],we discover the coordinates (0.5, 0.7) as the center of mass. Here, M is the total mass which sums up the weights of all particles.

This approach shows how manipulating a coordinate system and using basic mathematical operations of addition and division allows us to determine critical information such as the center of mass in a conveniently depicted space. In real-world applications, this can be used to balance objects like bridges or vehicles, enhancing stability and performance.

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Most popular questions from this chapter

In classical system: (a) the varying mass system is not considered (b) the varying mass system must be considered (c) the varying mass system may be considered (d) only varying of mass due to velocity is considered

In which of the following cases the centre of mass of a rod is certainly not at its geometrical centre? (a) The density continuously decreases from left to right (b) The density continuously increases from left to right (c) The density decreases from left to right upto the centre and then increases (d) Both (a) and (b) are correct

A body is dropped and observed to bounce a height greater than the dropping height. Then (a) the collision is elastic (b) there is additional source of energy during collision (c) it is not possible (d) this type of phenomenon does not occur in nature

Two bodies of mass \(M\) and \(m\) are moving with same kinetic energy. If they are stopped by same retarding force, then: (a) both bodies cover same distance before coming to rest (b) if \(M>m\), the time taken to come to rest for body of mass \(M\) is more than that of body of mass \(m\) (c) if \(m>M\), then body of mass \(m\) has more momentum than that of mass \(M\)

A circular hole of radius \(1 \mathrm{~cm}\) is cut off from a disc of radius \(6 \mathrm{~cm}\). The centre of hole is \(3 \mathrm{~m}\) from the centre of the disc. The position of centre of mass of the remaining disc from the centre of disc is : (a) \(-\frac{3}{35} \mathrm{~cm}\) (b) \(\frac{1}{35} \mathrm{~cm}\) (c) \(\frac{3}{10} \mathrm{~cm}\) (d) None of these
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