A T shaped object with dimensions shown in th | vedclass

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Question

For translatory motion the force should be applied on the centre of mass of the body, So we have to

calculate the location of centre of mass of $\m

Vedclass · Accepted Answer

$\frac {4}{3}l$

Vedclass · Answer

For translatory motion the force should be applied on the centre of mass of the body, So we have to

calculate the location of centre of mass of $\mathrm{T}^{\prime}$ shaped object.

Let mass of rod $A B$ in $m$ so the mass of rod $CD$

will be $2 \mathrm{m}$

Let $y_{1}$ is the centre of mass of tod $A B$ and $y_{2}$ is

the centre of mass of rod $\mathrm{CD}$. We can consider

that whole mass of the rod is placed at their respect

at their respective centre of mass i $e$. mass $m$ is

placed at $Y_{1}$ and mass $2 \mathrm{nt}$ is placed at $\mathrm{Y}_{2}$

Taking point $^{\prime} C^{\prime}$ at the origin position vector of point $y_{1}$ and $y_{2}$ can be written as

$\overrightarrow {{f_1}}  = 2\ell \hat j,\overrightarrow {{f_2}}  = \ell \hat j,\,\,and\,{m_1} = m\,\,and\,\,{m_2} = 2\,m$

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

$C=\frac{4}{3} \ell$.

A $T$ shaped object with dimensions shown in the figure, is lying a smooth floor. A force $\vec {'F'}$ is applied at the point $P$ parallel to $AB$, such that the object has only the translational motion without rotation. Find the location of $P$ with respect to $C$

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