Find the center of mass (x, y, z) of the foll | vedclass

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

Question

(C) First,we identify the center of mass of each individual cube. Since each cube has a side length of $1$ unit and they are placed along the axes,the

Vedclass · Accepted Answer

$(3/4, 3/4, 3/4)$

Vedclass · Answer

(C) First,we identify the center of mass of each individual cube. Since each cube has a side length of $1$ unit and they are placed along the axes,their centers are located at:Cube $1$ (at origin): $(0.5, 0.5, 0.5)$Cube $2$ (along $x$-axis): $(1.5, 0.5, 0.5)$Cube $3$ (along $y$-axis): $(0.5, 1.5, 0.5)$Cube $4$ (along $z$-axis): $(0.5, 0.5, 1.5)$Since all four cubes are identical,we can treat the center of mass of each cube as a point particle of equal mass $m$ located at its respective center.The coordinates of the center of mass of the system are given by:$x_{	ext{COM}} = \frac{m(0.5) + m(1.5) + m(0.5) + m(0.5)}{4m} = \frac{3.0}{4} = 0.75$$y_{	ext{COM}} = \frac{m(0.5) + m(0.5) + m(1.5) + m(0.5)}{4m} = \frac{3.0}{4} = 0.75$$z_{	ext{COM}} = \frac{m(0.5) + m(0.5) + m(0.5) + m(1.5)}{4m} = \frac{3.0}{4} = 0.75$Thus,the center of mass is $(0.75, 0.75, 0.75)$ or $(3/4, 3/4, 3/4)$.

Find the center of mass $(x, y, z)$ of the following structure of four identical cubes if the length of each side of a cube is $1$ unit.

Similar Questions

Two particles of masses $5 \,g$ and $3 \,g$ are separated by a distance of $40 \,cm$. The centre of mass of the system of these two particles

Three particles of masses $50\, g$,$100\, g$,and $150\, g$ are placed at the vertices of an equilateral triangle of side $1\, m$ (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be

Two particles of masses $1\,kg$ and $3\,kg$ have position vectors $2\hat{i} + 3\hat{j} + 4\hat{k}$ and $-2\hat{i} + 3\hat{j} - 4\hat{k}$ respectively. The position vector of the centre of mass is:

The position vectors of three particles of masses $1\, kg, 2\, kg$,and $3\, kg$ are $\vec{r_1} = (\hat{i} + 4\hat{j} + \hat{k})\,m$,$\vec{r_2} = (\hat{i} + \hat{j} + \hat{k})\,m$,and $\vec{r_3} = (2\hat{i} - \hat{j} - 2\hat{k})\,m$ respectively. The position vector of their centre of mass is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module