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(A) For pure translational motion,the force must be applied at the center of mass of the body. Therefore,we need to calculate the location of the cent

Vedclass · Accepted Answer

$\frac{4}{3}l$

Vedclass · Answer

(A) For pure translational motion,the force must be applied at the center of mass of the body. Therefore,we need to calculate the location of the center of mass of the $T$-shaped object.Let the mass of rod $AB$ be $m$. Since the rod $CD$ has the same length $2l$ (implied by the geometry of the $T$-shape where $AB$ is length $l$ and $CD$ is length $2l$),let its mass be $2m$ (assuming uniform density).Let $y_1$ be the center of mass of rod $AB$ and $y_2$ be the center of mass of rod $CD$.Taking point $C$ as the origin $(0,0)$:The center of mass of rod $AB$ is at a distance $2l$ from $C$ along the vertical axis,so $\vec{r}_1 = 2l\hat{j}$.The center of mass of rod $CD$ is at a distance $l$ from $C$ along the vertical axis,so $\vec{r}_2 = l\hat{j}$.The masses are $m_1 = m$ and $m_2 = 2m$.The position of the center of mass of the system is given by:$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$$\vec{r}_{cm} = \frac{m(2l\hat{j}) + (2m)(l\hat{j})}{m + 2m} = \frac{2ml\hat{j} + 2ml\hat{j}}{3m} = \frac{4ml\hat{j}}{3m} = \frac{4}{3}l\hat{j}$.Thus,the distance of the center of mass from $C$ is $\frac{4}{3}l$.

Column-$I$	Column-$II$
$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$	$(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$	$(b)$ Position vector of the center of mass for a system of two equal masses

$A$ $T$-shaped object with dimensions shown in the figure is lying on a smooth floor. $A$ force $\vec{F}$ is applied at point $P$ parallel to $AB$,such that the object has only translational motion without rotation. Find the location of $P$ with respect to $C$.

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