on the centre of mass of the body. So we have to
calculate the location of centre of mass of $\mathrm{‘T’}$
shaped object.
Let mass of rod $\mathrm{AB}$ in $\mathrm{m}$ so the mass of $\mathrm{rod} \mathrm{CD}$
will be $2 \mathrm{m}$
Let $y_{1}$ is the centre of mass of rod $A B$ and $y_{2}$ is
the centre of mass of rod $CD.$ We can consider
that whole mass of the rod is placed at their respect
at their respective centre of mass i.e.. mass $\mathrm{m}$ is
placed at $y_{1}$ and mass $2 m$ is placed at $y_{2}$
Taking point $C$ at the origin position vector
of point $y_{1}$ and $y_{2}$ can be written as
$\overrightarrow {{f_1}} = \ell \hat j,\overrightarrow {{f_2}} = \ell \hat j.\,\,and\,\,{m_1} = m\,\,and\,\,{m_2} = 2m$
Position vector of centre of mass of the system
${\overrightarrow f _{cm}} = \frac{{{m_1}\overrightarrow {{r_1}} + {m_2}\overrightarrow {{r_2}} }}{{{m_1} + {m_2}}} = \frac{{{m_2}\ell \hat j + 2m\ell \hat j}}{{m + 2m}} = \frac{{4m\ell \hat j}}{{3m}} = \frac{4}{3}\ell \hat j$
Hence the distance of centre of mass from
$C=\frac{4}{3} \ell$
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