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(A) For an object to undergo pure translational motion,the net force must be applied at its centre of mass $(CM)$.Let the horizontal bar $AB$ have mas

Vedclass · Accepted Answer

$\frac {4l}{3}$

Vedclass · Answer

(A) For an object to undergo pure translational motion,the net force must be applied at its centre of mass $(CM)$.Let the horizontal bar $AB$ have mass $m$ and the vertical bar have mass $2m$ (since its length is $2l$ compared to $l$ for $AB$).Let the junction of the two bars be the origin $(0,0)$.The centre of mass of the horizontal bar is at $(0,0)$.The centre of mass of the vertical bar is at $(0, -l)$.The total mass of the object is $M = m + 2m = 3m$.The $y$-coordinate of the centre of mass is given by:$Y_{CM} = \frac{m(0) + 2m(-l)}{3m} = \frac{-2ml}{3m} = -\frac{2l}{3}$.This means the centre of mass is at a distance of $\frac{2l}{3}$ from the junction towards $C$.Since the total length of the vertical bar is $2l$,the distance of the centre of mass from $C$ is $2l - \frac{2l}{3} = \frac{4l}{3}$.

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