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

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 $\mathrm{&lsquo;T&rsquo;}$

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

$C=\frac{4}{3} \ell$

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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