2.6 Rectangular Plate

  Centre of Mass and Conservation of Linear Momentum

Consider a thin rectangular plate $OABC$ of mass $m$ and sides of the rectangle are $a$ & $b$. So, the corresponding area $(A)=ab$.

Let the corner $O$ of the rectangle be the origin of the cartesian co-ordinate.

Mass per unit area $\left( {{\lambda _A}} \right) = \frac{{{\text{Total mass}}}}{{{\text{Total area}}}} = \frac{M}{{ab}}\quad ...(i)$

Now, we will find the centre of mass $(COM)$ about $x$ & $y$ axis separately.

$COM$ about $y$ axis

Area of the small green rectangular strip $(dA)=a(dy)$

Therefore, $$dm = {\lambda _A}(dA)$$ or $$dm = {\lambda _A}a(dy)\quad ...(ii)$$

Centre of mass $(COM)$ of the small rectangular strip is at point $P$ i.e. at a distance $y$ from the $x$ axis.

Note: The small section at point $P$ is infinitesimally small, so, the centre of mass of that section is assumed to be at point $P$ only.

So, for the centre of mass $(COM)$ of the rectangular plate along the $y$ axis is given by, $$\begin{equation} \begin{aligned} {\overrightarrow r _{CO{M_y}}} = \frac{{\int {{{\overrightarrow r }_{{P_y}}}dm} }}{{\int {dm} }} = \frac{{\int {ydm} }}{{\int {dm} }}\widehat j \\ {\overrightarrow r _{CO{M_y}}} = \frac{{\int {y{\lambda _A}ady} }}{{\int {{\lambda _A}ady} }}\widehat j \\\end{aligned} \end{equation} $$ or $${\overrightarrow r _{CO{M_y}}} = \frac{{\int {ydy} }}{{\int {dy} }}\widehat j$$ or $${\overrightarrow r _{CO{M_y}}} = \frac{{\int {ydy} }}{{\int {dy} }}\widehat j$$

Integrating for the rectangular plate along the $y$ axis from point $A$ to $B$.

At point $A$, $y=0$ & at point $B$, $y=b$, so the limit becomes,

$$\begin{equation} \begin{aligned} {\overrightarrow r _{CO{M_y}}} = \frac{{\int\limits_0^b {ydy} }}{{\int\limits_0^b {dy} }}\widehat j \\ {\overrightarrow r _{CO{M_y}}} = \frac{{\left[ {\frac{{{y^2}}}{2}} \right]_0^b}}{{\left[ y \right]_0^b}}\widehat j \\ {\overrightarrow r _{CO{M_y}}} = \frac{b}{2}\widehat j \\\end{aligned} \end{equation} $$

$COM$ about $x$ axis

Area of the small green rectangular strip $(dA)=b(dx)$

Therefore, $$dm = {\lambda _A}(dA)$$ or $$dm = {\lambda _A}b(dx)\quad ...(ii)$$

Centre of mass $(COM)$ of the small rectangular strip is at point $Q$ i.e. at a distance $x$ from the $x$ axis.

Note: The small section at point $Q$ is infinitesimally small, so, the centre of mass of that section is assumed to be at point $Q$ only.

So, for the centre of mass $(COM)$ of the rectangular plate along the $x$ axis is given by, $$\begin{equation} \begin{aligned} {\overrightarrow r _{CO{M_x}}} = \frac{{\int {{{\overrightarrow r }_{{Q_x}}}dm} }}{{\int {dm} }} = \frac{{\int {xdm} }}{{\int {dm} }}\widehat i \\ {\overrightarrow r _{CO{M_x}}} = \frac{{\int {x{\lambda _A}bdx} }}{{\int {{\lambda _A}ady} }}\widehat i \\\end{aligned} \end{equation} $$or $${\overrightarrow r _{CO{M_x}}} = \frac{{\int {xdx} }}{{\int {dx} }}\widehat i$$

Integrating for the rectangular plate along the $x$ axis from point $O$ to $A$.

At point $O$, $x=0$ & at point $A$, $x=a$, so the limit becomes,

$$\begin{equation} \begin{aligned} {\overrightarrow r _{CO{M_x}}} = \frac{{\int\limits_0^b {xdx} }}{{\int\limits_0^b {dx} }}\widehat i \\ {\overrightarrow r _{CO{M_x}}} = \frac{{\left[ {\frac{{{x^2}}}{2}} \right]_0^b}}{{\left[ x \right]_0^b}}\widehat i \\ {\overrightarrow r _{CO{M_x}}} = \frac{a}{2}\widehat i \\\end{aligned} \end{equation} $$

So, the position of centre of mass $(COM)$ for the thin rectangular plate is $\left( {\frac{a}{2},\frac{b}{2}} \right)$.

Note: Thin mean negligible thickness