Question

A rod of length $ 5L $ is bent at a right angle, keeping one side length as $ 2L $. The position of the centre of mass of the system (Consider $ L = 10 $ cm):

• A. \( 2\hat{i} + 3\hat{j} \)
• B. \( 3\hat{i} + 7\hat{j} \)
• C. \( 5\hat{i} + 8\hat{j} \)
• D. \( 4\hat{i} + 9\hat{j} \)

Hint:

To find the center of mass of a system of objects, calculate the weighted average of the positions of each object. The weights are the masses of the objects in the system.

Answer 1

The Correct Option is D

A rod is bent into two segments of lengths \( 2L \) and \( 3L \) at a right angle, with the joint at the origin. The \( 2L \) segment is along the x-axis, and the \( 3L \) segment is along the y-axis.

• The center of mass for the \( 2L \) segment is at \( (L, 0) \).
• The center of mass for the \( 3L \) segment is at \( (0, 1.5L) \).

The center of mass is calculated using the formula: \[x_{\text{cm}} = \frac{(2L)(L) + (3L)(0)}{2L + 3L} = \frac{2L^2}{5L} = \frac{2}{5}L, \quady_{\text{cm}} = \frac{(2L)(0) + (3L)(1.5L)}{2L + 3L} = \frac{4.5L^2}{5L} = \frac{9}{10}L\] Substituting \( L = 10 \) cm: \[x_{\text{cm}} = \frac{2}{5} \times 10 = 4 \text{ cm}, \quady_{\text{cm}} = \frac{9}{10} \times 10 = 9 \text{ cm}\] Therefore, the center of mass vector is: \[\vec{r}_{\text{cm}} = 4\hat{i} + 9\hat{j}\]