2.11 Questions

  Centre of Mass and Conservation of Linear Momentum

Question 5. A rod of length $L$ is placed along the $x$ axis between $x=0$ and $x=L$. The linear mass density $\lambda $ of the rod varies with distance $x$ from the origin as $\lambda = a + bx$, where $a$ & $b$ are constants. Find the position of the centre of mass of the rod.

Solution: Consider a small section of the rod of mass $dm$ and length $dx$ at a distance $x$ from one end of the rod.

So, $$\begin{equation} \begin{aligned} dm = \lambda dx \\ dm = (a + bx)dx\quad ...(i) \\\end{aligned} \end{equation} $$

Therefore, the position of the centre of mass of the rod can be given by, $${x_{COM}} = \frac{{\int {xdm} }}{{\int {dm} }} = \frac{{\int {x(a + bx)dx} }}{{\int {(a + bx)dx} }}$$

Integrating the above equation from end $O$ i.e. $x=0$ to end $A$ i.e. $x=L$

$$\begin{equation} \begin{aligned} {x_{COM}} = \frac{{\int\limits_0^L {x(a + bx)dx} }}{{\int\limits_0^L {(a + bx)dx} }} \\ {x_{COM}} = \frac{{\left[ {\frac{{a{x^2}}}{2} + \frac{{b{x^3}}}{3}} \right]_0^L}}{{\left[ {ax + \frac{{b{x^2}}}{2}} \right]_0^L}} \\ {x_{COM}} = \frac{{3aL + 2b{L^2}}}{{6a + 3bL}} \\\end{aligned} \end{equation} $$

So, the position of the centre of mass of the above rod is $\left( {\frac{{3aL + 2b{L^2}}}{{6a + 3bL}},0} \right)$

Question 6. Find the position of the centre of mass of a thin uniform disc of radius $R$ from which a small concentric disc of radius $r$ is removed as shown in the figure.

Solution: Consider a uniform semicircular disc of mass M as shown in the figure.

Area of the semicircular ring $(A) = \frac{{\pi \left( {{R^2} - {r^2}} \right)}}{2}\quad ...(i)$

Mass per unit area $\left( {{\lambda _A}} \right) = \frac{{2M}}{{\pi \left( {{R^2} - {r^2}} \right)}}\quad ...(ii)$

Consider a ring of radius $s$ and infinitesimal small thickness $ds$ having its centre at the origin and mass $dm$ at point $P$ on the disc.

Co-ordinate of the centre of mass of the ring passing through point $P$ is $\left( {0,\frac{{2s}}{\pi }} \right)$.

So, the position vector of the centre of mass $(COM)$ of ring passing through point $P$ is ${\overrightarrow r _{CO{M_P}}} = \frac{{2s}}{\pi }\widehat j$

Area of the ring is, $$dA = \left( {\pi s} \right)ds\quad ...(iii)$$ Also, $$\begin{equation} \begin{aligned} dm = {\lambda _A}dA \\ dm = {\lambda _A}\pi sds\quad ...(iv) \\\end{aligned} \end{equation} $$

Therefore, the centre of mass $(COM)$ of the semicircular disc is given by, $${\overrightarrow r _{COM}} = \frac{{\int {{{\overrightarrow r }_{CO{M_P}}}dm} }}{{\int {dm} }} = \frac{{\int {\frac{{2s}}{\pi }} dm}}{{\int {dm} }}\widehat j\quad ...(v)$$

From equation $(iv)$ & $(v)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{{\int {\frac{{2s}}{\pi }} {\lambda _A}\pi sds}}{{\int {{\lambda _A}\pi sds} }}\widehat j \\ {\overrightarrow r _{COM}} = \frac{2}{\pi }\frac{{\int {{s^2}ds} }}{{\int {sds} }}\widehat j \\\end{aligned} \end{equation} $$

Integrating for the semicircular disc from point $O$ to $A$,

At point $A$, $s=r$ & at point $B$, $s=R$, so the limit becomes, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{2}{\pi }\frac{{\int\limits_r^R {{s^2}ds} }}{{\int\limits_r^R {sds} }}\widehat j \\ {\overrightarrow r _{COM}} = \frac{2}{\pi }\frac{{\left[ {\frac{{{s^3}}}{3}} \right]_r^R}}{{\left[ {\frac{{{s^2}}}{2}} \right]_r^R}}\widehat j \\ {\overrightarrow r _{COM}} = \frac{{4\left( {{R^3} - {r^3}} \right)}}{{3\pi \left( {{R^2} - {r^2}} \right)}}\widehat j \\\end{aligned} \end{equation} $$

So, the centre of mass $(COM)$ of the disc as shown in the figure is $\left( {0,\frac{{4\left( {{R^3} - {r^3}} \right)}}{{3\pi \left( {{R^2} - {r^2}} \right)}}} \right)$