Question

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 ___________

Answer 1

Correct Answer: 3

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, \(x = 3\).

Answer 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.