Center Of Mass Of The Uniform Rod MCQ - Practice Questions with Answers

Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at x = L

Mass per unit length of the rod =

Take a small dx length of rod at a distance x from x=0

So mass of that dx element is =  

Therefore, x-coordinate of COM of the rod will be 

So x coordinate of centre of mass of Uniform rod of length L 

 At a distance    from one of the ends  of the rod.   

Similarly  , 

And  y-coordinate is zero for all particles of rod

So,  

Similarly,  

And  z-coordinate is zero for all particles of rod

So,  

So the coordinates of COM of the rod are  

Means it lies at the centre of the rod.

1. Rectangular plate

2. Square plate

3. Circular plate

4. For a 2-dimensional body with uniform negligible thickness formulae for finding the position of centre of mass can be rewritten  as

$$
r_{c m}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2 \ldots}{m_1+m_2 \ldots}=\frac{\rho A_1 t \vec{r}_1+\rho A_2 t \vec{r}_2 \ldots}{\rho A_1 t+\rho A_2 t \ldots}=\frac{A_1 \vec{r}_1+A_2 \vec{r}_2 \ldots}{A_1+A_2 \ldots}
$$

Where, $m=\rho$.A. $t$

5. Centre of mass when some mass is added to the body

$$
\vec{r}_{c m}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2}{m_1+m_2}
$$

Where $m_{1 \&}, \overrightarrow{r_1}$ are mass and position of the centre of mass for the whole body. $m_2 \& \overrightarrow{r_2}$ are mass and position of the centre of mass of added mass.

6. Position of centre of mass when some mass is removed

$$
\vec{r}_{c m}=\frac{m_1 \overrightarrow{r_1}-m_2 \vec{r}_2}{m_1-m_2}
$$

Where $m_{1 i s}$ the value of the whole mass and $\overrightarrow{r_1}$ is the position of the centre of mass for whole mass. Similarly $m_2 \& \overrightarrow{r_2}$ are values for mass which has been removed.