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10 Questions around this concept.
A rod '' has non-uniform linear mass density given by where a and b are constants and The value of x for the center of mass of the rod is at :
Two plates of the same thickness and same material are arranged as shown in the diagram. The dimensions are as shown in the diagram. Find the position of the centre of mass with reference to the origin.
A square shaped hole of side is carved out at a distance from the centre 'O' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is value of X (to the nearest integer)is .......
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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.
Rectangular plate
Square plate
Circular plate
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, $m=\rho . A . t$
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.
Position of centre of mass when some mass is removed
$\vec{r}_{c m}=\frac{m_1 \overrightarrow{r_1}-m_2 \overrightarrow{r_2}}{m_1-m_2}$
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