The identical spheres each of mass 2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is \(\frac{4 \sqrt2} x\) , where the value of x is ___________
Correct Answer: 3
Approach Solution - 1
Given three identical spheres of mass \(2M\) placed at the corners of a right-angled triangle. The sides of the triangle are 4 m each. Let the point of intersection of the two sides be the origin \((0, 0)\). The position vectors of the masses are:
- \(m_1 = 2M\), \(r_1 = (0, 0)\)
- \(m_2 = 2M\), \(r_2 = (4, 0)\)
- \(m_3 = 2M\), \(r_3 = (0, 4)\)
The position vector of the center of mass is given by:
\[ r_{\text{com}} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3}{m_1 + m_2 + m_3} \]
Substituting the values:
\[ r_{\text{com}} = \frac{2M \times (0, 0) + 2M \times (4, 0) + 2M \times (0, 4)}{6M} = \left(\frac{4}{3}, \frac{4}{3}\right) \]
Magnitude of \(r_{\text{com}}\):
\[ |r_{\text{com}}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{\frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{32}{9}} = \frac{4\sqrt{2}}{3} \]
Thus, \(x = 3\).
Approach Solution -2
This problem requires finding the magnitude of the position vector of the center of mass (COM) for a system of three identical spheres placed at the vertices of a right-angled isosceles triangle.
Concept Used:
The position vector of the center of mass, \( \vec{R}_{COM} \), for a system of \( n \) particles is given by the weighted average of their position vectors:
\[ \vec{R}_{COM} = \frac{\sum_{i=1}^{n} m_i \vec{r}_i}{\sum_{i=1}^{n} m_i} \]In terms of coordinates, the x and y coordinates of the center of mass are:
\[ X_{COM} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \] \[ Y_{COM} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} \]The magnitude of the position vector \( \vec{R}_{COM} = X_{COM} \hat{i} + Y_{COM} \hat{j} \) is calculated as:
\[ |\vec{R}_{COM}| = \sqrt{X_{COM}^2 + Y_{COM}^2} \]Step-by-Step Solution:
Step 1: Define the coordinate system and locate the positions of the spheres.
Let the origin (0, 0) be the vertex of the right angle. The two mutually perpendicular sides of length 4 m each lie along the positive x-axis and positive y-axis.
The three identical spheres, each of mass \( m = 2M \), are placed at the corners of this triangle. Let's label them as follows:
- Sphere 1 is at the origin: \( (x_1, y_1) = (0, 0) \).
- Sphere 2 is at the end of the side along the x-axis: \( (x_2, y_2) = (4, 0) \).
- Sphere 3 is at the end of the side along the y-axis: \( (x_3, y_3) = (0, 4) \).
The mass of each sphere is \( m_1 = m_2 = m_3 = 2M \).
Step 2: Calculate the total mass of the system.
The total mass \( M_{total} \) is the sum of the individual masses:
\[ M_{total} = m_1 + m_2 + m_3 = 2M + 2M + 2M = 6M \]Step 3: Calculate the x and y coordinates of the center of mass.
Using the formula for \( X_{COM} \):
\[ X_{COM} = \frac{(2M)(0) + (2M)(4) + (2M)(0)}{6M} = \frac{0 + 8M + 0}{6M} = \frac{8M}{6M} = \frac{4}{3} \text{ m} \]Using the formula for \( Y_{COM} \):
\[ Y_{COM} = \frac{(2M)(0) + (2M)(0) + (2M)(4)}{6M} = \frac{0 + 0 + 8M}{6M} = \frac{8M}{6M} = \frac{4}{3} \text{ m} \]So, the position vector of the center of mass is \( \vec{R}_{COM} = \frac{4}{3} \hat{i} + \frac{4}{3} \hat{j} \).
Step 4: Calculate the magnitude of the position vector of the center of mass.
\[ |\vec{R}_{COM}| = \sqrt{X_{COM}^2 + Y_{COM}^2} = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} \] \[ |\vec{R}_{COM}| = \sqrt{2 \times \left(\frac{4}{3}\right)^2} = \sqrt{2 \times \frac{16}{9}} = \frac{4\sqrt{2}}{3} \text{ m} \]Final Computation & Result:
The magnitude of the position vector of the center of mass is \( \frac{4\sqrt{2}}{3} \) m.
The problem states that this magnitude is equal to \( \frac{4\sqrt{2}}{x} \).
\[ \frac{4\sqrt{2}}{3} = \frac{4\sqrt{2}}{x} \]By comparing the two expressions, we can see that:
\[ x = 3 \]The value of x is 3.
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