The position vector of the centre of mass vec | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

The figure shows a composite system of two uniform rods of lengths as indicated. The coordinates of the centre of mass of the system of rods are ...........

$A$ uniform square plate of side $a$ has a circular hole of diameter $a$ cut out from it as shown in the figure. The centre of the hole is at the centre of the square. The centre of mass of the remaining part is:

$A$ uniform square plate has a side of length $2R$. $A$ circular piece of maximum possible area is cut and removed from one of the quadrants of the plate as shown in the figure. Calculate the shift in the centre of mass of the plate.

Vedclass Test Series

Exam Paper Generator

Online Exam Module