Question

A uniform thin rod of mass 3kg has a length of 1m. If a point mass of 1kg is attached to it at a distance of 40cm from its center,the center of mass shifts by a distance of: 

• A. 2.5cm
• B. 5cm
• C. 8cm
• D. 10cm
• E. 20cm

Answer 1

The Correct Option is D

Given:

• Mass of the rod, \( M = 3 \, \text{kg} \) (uniform, length 1 m)
• Point mass, \( m = 1 \, \text{kg} \), attached at 40 cm from the rod's center

Step 1: Define Coordinate System

Place the rod along the x-axis with its original center of mass (COM) at \( x = 0 \).

The point mass is attached at \( x = 40 \, \text{cm} = 0.4 \, \text{m} \).

Step 2: Calculate New COM Position

The new COM (\( x_{\text{COM}} \)) is given by:

\[ x_{\text{COM}} = \frac{M \cdot 0 + m \cdot 0.4}{M + m} = \frac{3 \times 0 + 1 \times 0.4}{3 + 1} = \frac{0.4}{4} = 0.1 \, \text{m} = 10 \, \text{cm} \]

Step 3: Determine Shift in COM

The shift in COM is equal to the new COM position since the original COM was at 0:

\[ \text{Shift} = 10 \, \text{cm} \]

Conclusion:

The center of mass shifts by 10 cm.

Answer: \(\boxed{D}\)

Answer 2

Given:
- Mass of the rod (\( m_1 \)) = 3 kg
- Length of the rod (\( L \)) = 1 m
- Mass of the point mass (\( m_2 \)) = 1 kg
- Distance of the point mass from the center of the rod (\( d \)) = 0.4 m

Center of mass of the rod before attaching the point mass:
The center of mass of the uniform rod lies at its midpoint, so initially, the center of mass of the rod is at \( x_1 = 0 \) (taking the center of the rod as the origin).

Position of the point mass:
The point mass is attached at a distance of 40 cm (or 0.4 m) from the center of the rod. Let’s assume it is attached to the right of the center.

Center of mass of the system:
To find the new center of mass (\( x_{\text{cm}} \)) of the system consisting of the rod and the point mass, we use the formula:
\[ x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]

Here,
- \( x_1 \) is the center of mass of the rod, which is 0 (since we took the center of the rod as the origin).
- \( x_2 \) is the position of the point mass, which is 0.4 m.
- \( m_1 \) is the mass of the rod.
- \( m_2 \) is the mass of the point mass.

Calculation:
\[ x_{\text{cm}} = \frac{(3 \, \text{kg} \cdot 0 \, \text{m}) + (1 \, \text{kg} \cdot 0.4 \, \text{m})}{3 \, \text{kg} + 1 \, \text{kg}} \]
\[ x_{\text{cm}} = \frac{0 + 0.4}{4} \]
\[ x_{\text{cm}} = \frac{0.4}{4} \]
\[ x_{\text{cm}} = 0.1 \, \text{m} \]

Conclusion:
The center of mass of the system shifts by \( 0.1 \, \text{m} \) or \( 10 \, \text{cm} \) from the original center of mass of the rod.

Hence, the shift in the center of mass is \( 10 \, \text{cm} \).