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

$A$ uniform rectangular thin sheet $ABCD$ of mass $M$ has length $a$ and breadth $b$,as shown in the figure. If the shaded portion $HBGO$ is cut off,the coordinates of the centre of mass of the remaining portion will be

$A$ uniform square plate is shown in the figure. Four identical squares are removed from its corners. Where does the center of mass lie?

The figure shows a uniform square plate from which four identical small squares have been removed from the corners. If square $1$ is removed,where will the center of mass $(C.M.)$ be located?

$A$ small disc of radius $2\, cm$ is cut from a disc of radius $6\, cm$. If the distance between their centres is $3.2\, cm$,what is the shift in the centre of mass of the disc in $cm$?

From a circular disc of radius $R$,a square is cut out with a radius as its diagonal. The center of mass of the remaining part is at a distance (from the centre) of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module