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Four particles of mass \(m_{1}=2 m, m_{2}=4 m, m_{3}=m\) and \(m_{4}\) arc placed at four corners of a square. What should be the value of \(m_{4}\) so that the centre of mass of all the four particles are exactly at the centre of the square? (a) \(2 \mathrm{~m}\) (b) \(8 \mathrm{~m}\) (c) \(6 \mathrm{~m}\) (d) none

Short Answer

Expert verified
The value of \(m_4\) should be \(8m\).

Step by step solution

01

Express Center of Mass Coordinates

The center of mass coordinates in a system is given by\[x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}\quad \text{and}\quady_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}\]where \((x_i, y_i)\) are the coordinates of each particle.
02

Assign Particles to Square Corners

Let the square have side length \(a\). Position the particles with coordinates: - \(m_1\) at \((0,0)\),- \(m_2\) at \((a,0)\),- \(m_3\) at \((0,a)\),- \(m_4\) at \((a,a)\).
03

Calculate Center of Mass Coordinates for Symmetry

For the center of mass to be at the center of the square, these coordinates should be \((\frac{a}{2}, \frac{a}{2})\). Substitute this alongside the respective coordinates into the center of mass formula from Step 1.
04

Obtain Equations for \(x_{cm}\) and \(y_{cm}\)

Using the known coordinates, the equation for \(x_{cm}\) becomes:\[\frac{2m \cdot 0 + 4m \cdot a + m \cdot 0 + m_4 \cdot a}{2m + 4m + m + m_4} = \frac{a}{2}\]Similarly, \(y_{cm}\) becomes:\[\frac{2m \cdot 0 + 4m \cdot 0 + m \cdot a + m_4 \cdot a}{2m + 4m + m + m_4} = \frac{a}{2}\]
05

Simplify and Solve for \(m_4\)

Both equations from Step 4 are the same, hence solving one of them gives us\[\frac{4ma + m_4a}{7m + m_4} = \frac{a}{2}\]Cancelling \(a\) from numerator and denominator and multiplying both sides by \((7m + m_4)\), this equation simplifies to\[4m + m_4 = \frac{7m + m_4}{2}\]Rearranging and simplifying:\[8m + 2m_4 = 7m + m_4\]Yielding:\[m_4 = 8m\]
06

Verify Solution

Plug \(m_4 = 8m\) back into one of the center of mass equations to confirm symmetry:\[\frac{4ma + 8ma}{7m + 8m} = \frac{a}{2} \leftrightarrow \frac{12m}{15m} = \frac{1}{2}\]. This confirms our solution is correct.

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Key Concepts

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

Mass Distribution
In physics, mass distribution refers to how mass is spread across an object or a system. In problems involving point masses, such as the one described here, each particle or mass resides at a specific location. For the system to be balanced, the center of mass must be positioned accurately to account for how these masses are spread across the structure.

In the given problem, we have four masses located at the corners of a square. The distribution is not uniform; each mass has a different value, affecting the overall center of mass. Understanding mass distribution is key to predicting where the center of mass will naturally lie.

In this scenario, once the mass values are substituted into the center of mass formula, they reflect how each mass's position and magnitude contribute to the overall balance. This concept helps in determining conditions that need to be satisfied, like what specific mass value (for the mass not given) will result in a balanced or specific arrangement of a system.
Coordinate Geometry
Coordinate geometry is a fundamental tool in determining the center of mass. It allows us to specify each particle's location using coordinates on a plane. In this problem, the coordinates are assigned based on the location of each mass at the corners of a square.

The problem situates the square in a coordinate system where the bottom-left corner corresponds to the origin, \(0, 0\). From there, other corners are determined based on the square's side length, making it easy to plug in these coordinates into the formula for the center of mass.

By using coordinate geometry, it becomes straightforward to visualize the problem. You can easily see how each mass's position impacts the calculated center of mass by considering where it resides in the coordinate plane. This is particularly useful in symmetric shapes, such as the square in this scenario, where achieving perfect symmetry in mass distribution helps solve the problem effectively.
Physics Problem Solving
Solving physics problems involves a deep understanding of the relationships between physical quantities and how they affect each other within the context of the problem. In the case of the center of mass, it's a matter of balancing equations derived from both mass and their coordinates to find an unknown variable.

This type of physics problem begins by defining what you know and clearly stating the goal, such as making the center of mass align with a particular point. The problem then breaks down into manageable parts. For example, setting up balance equations like those for \(x_{cm}\) and \(y_{cm}\), representing specific outcomes based on the inputs or conditions of the scenario.

Through this step-by-step approach, you're employing logical reasoning and mathematical manipulation. Simplifying equations and systematically solving for unknowns like \(m_4\) demonstrate advanced problem-solving skills, ensuring you understand both the theoretical and computational elements of physics.

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Most popular questions from this chapter

Mass is non uniformly distributed over the rod of length \(l\). Its linear mass density varies linearly with length as \(\lambda=k x\). The position of centre of mass is given by: (a) \(\frac{2 l}{5}\) (b) \(\frac{1}{3}\) (c) \(\frac{3 l}{4}\) (d) \(\frac{2 l}{3}\)

Two blocks 1 and 2 of masses \(m\) and \(2 m\) respectively are connected by a spring of force constant \(k\). The masses are moving to the right with uniform velocity \(v\) cach, the heavicr mass, leading the lighter one. 'lhe spring is of natural length in the motion. Block 2 collides head on with a third block 3 of mass \(m\), at rest, the collision being completely inclastic. Determine the velocity of blocks at the instant of maximum compression of the spring.

Match the following: Column-I (a) In the absence of cxternal forces, the internal forces (b) Centre of mass of a body can be (c) The kinetic energy of the system of two blocks (d) Centre of mass of solid cylinder is Column-II (p) insidc the material of the body (q) may affect the motion of individuals (r) do not affect the motion of centre of mass (s) outside the material of the body

An isolated particle of mass \(m\) is moving in a horizontal plane \((x, y)\) along the \(x\) -axis at a certain height above the ground. It suddenly explodes into two fragments of masses \(\frac{m}{4}\) and \(\frac{3 m}{4}\). An instant later, the smaller fragment is at \(y=+15 \mathrm{~cm}\). The larger fragment at this instant is at: (a) \(y=-5 \mathrm{~cm}\) (b) \(y=+20 \mathrm{~cm}\) (c) \(y=+5 \mathrm{~cm}\) (d) \(y=-20 \mathrm{~cm}\)

Statement-1: Impulsive force on a particle may change its kinctic cnergy and its momentum Statement-2: Momentum of a particle changes only when kinetic energy of the particle changes.

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