2.8 Circular Plate

  Centre of Mass and Conservation of Linear Momentum

Consider a thin circular plate of mass $m$ and radius $r$. Let the centre $O$ of the circular plate be the origin of the cartesian co-ordinate.

Mass per unit area $\left( {{\lambda _A}} \right) = \frac{{{\text{Total mass}}}}{{{\text{Total area}}}} = \frac{M}{{\pi {R^2}}}\quad ...(i)$

Let us now find the position of the centre of mass about the $y$ axis.

Area of the small green rectangular strip $(dA)=2r(dy)$

As, $$r = R\cos \theta \quad ,\quad dy = Rd\theta $$ So, $$dA = 2R\cos \theta (Rd\theta ) = 2{R^2}\cos \theta d\theta $$

Therefore, $$dm = {\lambda _A}\left( {dA} \right)$$ or $$dm = {\lambda _A}2{R^2}\cos \theta d\theta \quad ...(ii)$$

Centre of mass $(COM)$ of the small rectangular strip is at point $P$ i.e. at a distance $y = R\sin \theta $ 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, $$\begin{equation} \begin{aligned} {\overrightarrow r _{CO{M_y}}} = \frac{{\int {{{\overrightarrow r }_{{P_y}}}dm} }}{{\int {dm} }} = \frac{{\int {R\sin \theta } dm}}{{\int {dm} }}\widehat j \\ {\overrightarrow r _{CO{M_y}}} = \frac{{\int {y{\lambda _A}2{R^2}\cos \theta d\theta } }}{{\int {{\lambda _A}2{R^2}\cos \theta d\theta } }}\widehat j \\\end{aligned} \end{equation} $$ or $${\overrightarrow r _{CO{M_y}}} = \frac{{\int {R\sin \theta \cos \theta d\theta } }}{{\int {\cos \theta d\theta } }}\widehat j$$

Integrating for the circular plate from point $B$ to $A$,

At point $B$, $\theta = - \frac{\pi }{2}$ & at point $A$, $\theta = + \frac{\pi }{2}$. So, the limit becomes, $$\begin{equation} \begin{aligned} {\overrightarrow r _{CO{M_y}}} = \frac{{\int\limits_{ - \frac{\pi }{2}}^{ + \frac{\pi }{2}} {R\sin \theta \cos \theta d\theta } }}{{\int\limits_{ - \frac{\pi }{2}}^{ + \frac{\pi }{2}} {\cos \theta d\theta } }} \\ {\overrightarrow r _{CO{M_y}}} = - \frac{{R\left[ {\cos \left( {2\theta } \right)} \right]_{ - \pi /2}^{\pi /2}}}{{2\left[ {\sin \theta } \right]_{ - \pi /2}^{\pi /2}}}\widehat j \\ {\overrightarrow r _{CO{M_y}}} = 0\widehat j \\\end{aligned} \end{equation} $$ Similarly, $${\overrightarrow r _{CO{M_x}}} = 0\widehat i$$

Since, the circular plate is a symmetric body so its centre of mass lies at its centre.

So, the position of the centre of mass $(COM)$ for the thin circular plate is $(0,0)$.