A T-shaped object with dimensions shown in th | vedclass

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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

$\frac{4}{3}l$

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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 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 $\frac{4}{3}l$.

$A$ $T$-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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