First slide

Centre of mass(com)

Question

A wire of uniform cross-section is bent in the shape shown in the figure. The coordinates of the center of mass of each side are shown in the figure. The coordinates of the center of mass of the system are

Easy

A

$(\frac{15 l}{14}, \frac{6 l}{7})$
B

$(\frac{15 l}{14}, l)$
C

$(l, \frac{l}{2})$
D

$(l, l)$

Solution

$\begin{array}{l} X_{cm} = \frac{m \times 0 + m l + m \times 2 l + \frac{m}{2} \frac{3 l}{2}}{3.5 m} \\ \Rightarrow X_{cm} = \frac{15 ml}{4 \times \frac{7}{2} m} = \frac{15 l}{14} \\ Y_{cm} = \frac{m \times 0 + m l + m l + \frac{m}{2} 2 l}{\frac{7}{2} m} \Rightarrow Y_{cm} = \frac{3 ml}{\frac{7 m}{2}} = \frac{6 l}{7} \end{array}$

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