Question

A uniform metre rod is bent into L shape with the bent arms at 90^@ to each other. The distance of the centre of mass from the bent point is

Answer 1

Given:

• Rod of metre rod is bent at midpoint into L shape.
• Arms of the L are at 90 degrees to each other.

To find:

• Distance of centre of mass from bent point.

Solution:

• For ease of solving, lets consider two arms of the L as two different rods.
• Length of entire rod was 1 m so the length of each arm will be 0.5 m since rod is bent from midpoint.
• Suppose Mass of rod was M then mass of each arm is m/2.
• Now, the centre of mass of each arm individually will be at midpoint of each arm.
• Hence, if 0.5 m is length of arm, centre of mass of arm of L individually will be at 0.25 metre from bent point.
• Now, lets say bent point is the origin and one of the arms of L is along positive X-axis and other arm is along positive Y-axis.
• Centre of masses are at (0.25m , 0) and (0 , 0.25 m)
• Formula for centre of mass of the entire L shape in X direction is
• $CM_x = \frac{m_1*x_1 + m_2*x_2}{m_1 + m_2}$  
• Substituting x1 = 0.25 m and x2 = 0 m and m1=m2=m/2
• $CM_x = \frac{(m/2)*0.25 + (m/2)*0}{m/2 + m/2}$  
• CM =  0.125 m     in the X direction.
• Similarly, centre of mass of the entire L shaped system in Y direction is
• $CM_y = \frac{m_1*y_1 + m_2*y_2}{m_1 + m_2}$  
• $CM_y = \frac{(m/2)*0 + (m/2)*0.25}{m/2 + m/2}$  
• CM = 0.125 m in the Y-direction.
• Now, distance from bent point will be
• $\sqrt{CM_x^2 + CM_y^2}$
• $\sqrt{0.125^2 + 0.125^2}$
• $\sqrt{2(0.125^2)}$  
• Distance from bent point = √2  * 0.125
• Distance from bent point = 0.1768 m.

Answer:

• The centre of mass is at a distance of $0.125 \sqrt{2} m$ or 0.1678 metre from the bent point.