Center Of Mass Of The Uniform Rod - Practice Questions & MCQ

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 = $\mu=\frac{M}{L}$

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

So mass of that dx element is $=d m=\mu \cdot d x$
Therefore, x -coordinate of COM of the rod will be

$
\begin{aligned}
& x_{c m}=\frac{\int_0^L x \cdot d m}{\int_0^M d m} \\
& x_{c m}=\frac{\int_0^L x \cdot \mu \cdot d x}{\int_0^M d m}=\frac{\int_0^L x \cdot \frac{M}{L} \cdot d x}{\int_0^M d m} \\
& x_{c m}=\frac{\frac{M}{L} \int_0^L x d x}{M}=\frac{1}{L} \int_0^L x \cdot d x=\frac{L}{2}
\end{aligned}
$

So $x$ coordinate of centre of mass of Uniform rod of length $L$
At a distance $\frac{L}{2}$ from one of the ends of the rod.
Similarly, $y_{\mathrm{cm}}=\frac{\int y d m}{\int d m}$
And $y$-coordinate is zero for all particles of rod
So, $y_{c m}=0$

Similarly,

$
z_{c m}=\frac{\int z d m}{\int d m}
$

And $z$-coordinate is zero for all particles of rod
So, $z_{c m}=0$
So the coordinates of COM of the rod are $\left(\frac{L}{2}, 0,0\right)$

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 in the body

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

                  $\vec{r}_{c m}=\frac{m_1 \overrightarrow{r_1}-m_2 \overrightarrow{r_2}}{m_1-m_2}$