Question

The position vector of the centre of mass $\vec r$ cm of a symmetric uniform bar of negligible area of cross section as shown in the figure is-


A. $\vec rcm = \dfrac{{13}}{8}L\hat x + \dfrac{5}{8}L\hat y$
B. $\vec rcm = \dfrac{{11}}{8}L\hat x + \dfrac{3}{8}L\hat y$
C. $\vec rcm = \dfrac{{13}}{8}L\hat x + \dfrac{{11}}{8}L\hat y$
D. $\vec rcm = \dfrac{5}{8}L\hat x + \dfrac{{13}}{8}L\hat y$

Answer 1

Hint: First we will find out the position of the centre of the mass on X-axis and then the position of centre of the mass on Y-axis respectively. Then with the help of the formula mentioned in the solution below, we will solve this further. Refer to the figure as well.
Formula used: $\dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{M}$, $\dfrac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}}}{M}$

Complete Step-by-Step solution:
Let the position of the centre of mass be ${X_{cm}}$ on X axis.
Let the mass ${m_1}$ be 2m. Its centre of mass will be on the X-axis which will be equal to L (as can be seen in the figure). Let this centre of mass be ${x_1}$.
Let the mass ${m_2}$ be m. Its centre of mass will be on the X-axis which will be equal to 2L (as can be seen in the figure). Let this centre of mass be ${x_2}$.
Let the mass ${m_3}$ be m. Its centre of mass will be on the X-axis which will be equal to $\dfrac{5}{2}L$ (as can be seen in the figure). Let this centre of mass be ${x_3}$.
Let the total mass be M which is equal to $2m + m + m = 4m$.
Now, applying the formula $\dfrac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{M}$. Where M is the total mass, we will solve it further.
Putting all the values in the above formula, we get-
$
   \Rightarrow {X_{cm}} = \dfrac{{2m \times L + m \times 2L + m \times \dfrac{5}{2}L}}{{4m}} \\
    \\
   \Rightarrow {X_{cm}} = \dfrac{{4mL + \dfrac{{5Lm}}{2}}}{{4m}} \\
$
Taking m common-
$
   \Rightarrow {X_{cm}} = \dfrac{{4mL + \dfrac{{5Lm}}{2}}}{{4m}} \\
    \\
   \Rightarrow {X_{cm}} = \dfrac{{m\left( {4L + \dfrac{{5L}}{2}} \right)}}{{4m}} \\
    \\
   \Rightarrow {X_{cm}} = \dfrac{{\left( {4L + \dfrac{{5L}}{2}} \right)}}{4} \\
    \\
   \Rightarrow {X_{cm}} = \dfrac{{13L}}{8} \\
$
Now, let the position of the centre of mass be ${Y_{cm}}$ on the Y axis.
Let the mass ${m_1}$ be 2m. Its centre of mass will be on the Y-axis which will be equal to L (as can be seen in the figure). Let this centre of mass be ${y_1}$.
Let the mass ${m_2}$ be m. Its centre of mass will be on the Y-axis which will be equal to $\dfrac{L}{2}$ (as can be seen in the figure). Let this centre of mass be ${y_2}$.
Let the mass ${m_3}$ be m. Its centre of mass will be on the Y-axis which will be equal to $0$ (as can be seen in the figure). Let this centre of mass be ${y_3}$.
Let the total mass be M which is equal to $2m + m + m = 4m$.
Now, applying the formula $\dfrac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3}}}{M}$. Where M is the total mass, we will solve it further.
Putting all the values in the above formula, we get-
$
   \Rightarrow {Y_{cm}} = \dfrac{{2m \times L + m \times \dfrac{L}{2} + m \times 0}}{{4m}} \\
    \\
   \Rightarrow {Y_{cm}} = \dfrac{{5mL}}{{8m}} \\
$
$ \Rightarrow {Y_{cm}} = \dfrac{5}{8}L$
Therefore, r centre of mass $\vec r$ when written in vector form will give us-
$ \Rightarrow \vec rcm = \dfrac{{13}}{8}L\hat x + \dfrac{5}{8}L\hat y$

Hence, option A is the correct option.

Note: Position vector, straight line with one end fixed to one body and the other end connected to a point of movement and used to define the point location relative to the body. The position vector may change in length or direction, or both in length and direction as the point travels.