The position vector of the centre of mass vec | vedclass

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(A) The bar can be divided into three segments,each treated as a point mass located at its respective geometric center.$1$. The vertical segment of le

Vedclass · Accepted Answer

$\vec{r}_{cm} = \frac{13}{8}L\hat{i} + \frac{5}{8}L\hat{j}$

Vedclass · Answer

(A) The bar can be divided into three segments,each treated as a point mass located at its respective geometric center.$1$. The vertical segment of length $2L$ has mass $2m$ and its center of mass is at $(L, L)$.$2$. The horizontal segment from $x=2L$ to $x=3L$ has mass $m$ and its center of mass is at $(2.5L, 0)$.$3$. Wait,looking at the figure,the vertical part is from $(0,0)$ to $(0,L)$ and $(0,L)$ to $(2L,L)$. Let's re-evaluate: The horizontal bar is from $(0,L)$ to $(2L,L)$ (mass $2m$,$CM$ at $(L,L)$). The vertical bar is from $(2L,0)$ to $(2L,L)$ (mass $m$,$CM$ at $(2L, L/2)$). The horizontal extension is from $(2L,0)$ to $(3L,0)$ (mass $m$,$CM$ at $(2.5L, 0)$).Total mass $M = 2m + m + m = 4m$.$\vec{r}_{cm} = \frac{2m(L\hat{i} + L\hat{j}) + m(2L\hat{i} + 0.5L\hat{j}) + m(2.5L\hat{i} + 0\hat{j})}{4m}$$\vec{r}_{cm} = \frac{(2L + 2L + 2.5L)\hat{i} + (2L + 0.5L + 0)\hat{j}}{4}$$\vec{r}_{cm} = \frac{6.5L\hat{i} + 2.5L\hat{j}}{4} = \frac{13/2 L\hat{i} + 5/2 L\hat{j}}{4} = \frac{13}{8}L\hat{i} + \frac{5}{8}L\hat{j}$.

The position vector of the centre of mass $\vec{r}_{cm}$ of an asymmetric uniform bar of negligible area of cross-section as shown in the figure is:

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