Seven identical homogeneous bricks,each of le | vedclass

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

Question

(D) Let the mass of each brick be $m$. The $x$-coordinate of the centre of mass of each individual brick is at its geometric centre.Taking the origin

Vedclass · Accepted Answer

$\frac{22L}{35}$

Vedclass · Answer

(D) Let the mass of each brick be $m$. The $x$-coordinate of the centre of mass of each individual brick is at its geometric centre.Taking the origin $O$ at the left edge of the bottom brick (brick $1$):$x_1 = \frac{L}{2}$$x_2 = \frac{L}{2} + \frac{L}{10}$$x_3 = \frac{L}{2} + \frac{2L}{10}$$x_4 = \frac{L}{2} + \frac{3L}{10}$$x_5 = \frac{L}{2} + \frac{2L}{10}$$x_6 = \frac{L}{2} + \frac{L}{10}$$x_7 = \frac{L}{2}$Since there are $7$ identical bricks,the total mass $M = 7m$.The $x$-coordinate of the centre of mass of the system is given by:$x_{cm} = \frac{\sum m_i x_i}{\sum m_i} = \frac{m(x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7)}{7m}$$x_{cm} = \frac{1}{7} [(\frac{L}{2}) + (\frac{L}{2} + \frac{L}{10}) + (\frac{L}{2} + \frac{2L}{10}) + (\frac{L}{2} + \frac{3L}{10}) + (\frac{L}{2} + \frac{2L}{10}) + (\frac{L}{2} + \frac{L}{10}) + (\frac{L}{2})]$$x_{cm} = \frac{1}{7} [7(\frac{L}{2}) + 2(\frac{L}{10} + \frac{2L}{10} + \frac{3L}{10})]$$x_{cm} = \frac{1}{7} [\frac{7L}{2} + 2(\frac{6L}{10})] = \frac{1}{7} [\frac{7L}{2} + \frac{6L}{5}]$$x_{cm} = \frac{1}{7} [\frac{35L + 12L}{10}] = \frac{47L}{70}$Wait,re-evaluating the displacement pattern from the image: The bricks are stacked symmetrically. The middle brick (brick $4$) is shifted by $3L/10$. The calculation above is correct based on the symmetry shown.

Seven identical homogeneous bricks,each of length $L$,are arranged as shown in the figure. Each brick is displaced with respect to the one in contact by $\frac{L}{10}$. Calculate the $x$-coordinate of the centre of mass of this system relative to the origin $O$ as shown.

Similar Questions

Two particles whose masses are $10\,kg$ and $30\,kg$ and their position vectors are $\hat{i} + \hat{j} + \hat{k}$ and $-\hat{i} - \hat{j} - \hat{k}$ respectively would have the centre of mass at:

Particles of masses $m, 2m, 3m, \ldots, nm$ grams are placed on the same line at distances $l, 2l, 3l, \ldots, nl$ cm from a fixed point. The distance of the centre of mass of the particles from the fixed point in centimetres is:

The position of the center of mass of three masses $2 \text{ kg}$, $3 \text{ kg}$, and $15 \text{ kg}$ placed with respect to the midpoint $(p)$ of the base, as shown in the figure, is . . . . . . .

The coordinates of the positions of particles of mass $7 \, g$,$4 \, g$,and $10 \, g$ are $(1, 5, -3) \, cm$,$(2, 5, 7) \, cm$,and $(3, 3, -1) \, cm$ respectively. The position of the centre of mass of the system would be:

Five masses are placed in a plane as shown in the figure. The coordinates of the centre of mass are nearest to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module