Particles of masses 8 , kg, 2 , kg, 4 , kg,an | vedclass

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

Question

(B) Let vertex $A$ be at the origin $(0, 0)$. The side length of the square $s$ is related to the diagonal $d = 80 \, cm$ by $s = d / \sqrt{2} = 80 /

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) Let vertex $A$ be at the origin $(0, 0)$. The side length of the square $s$ is related to the diagonal $d = 80 \, cm$ by $s = d / \sqrt{2} = 80 / \sqrt{2} = 40\sqrt{2} \, cm$.The coordinates of the vertices are: $A(0, 0)$,$B(40\sqrt{2}, 0)$,$C(40\sqrt{2}, 40\sqrt{2})$,and $D(0, 40\sqrt{2})$.The masses are $m_A = 8 \, kg, m_B = 2 \, kg, m_C = 4 \, kg, m_D = 2 \, kg$. Total mass $M = 8 + 2 + 4 + 2 = 16 \, kg$.The $x$-coordinate of the center of mass is $X_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{M} = \frac{8(0) + 2(40\sqrt{2}) + 4(40\sqrt{2}) + 2(0)}{16} = \frac{240\sqrt{2}}{16} = 15\sqrt{2} \, cm$.The $y$-coordinate of the center of mass is $Y_{cm} = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{M} = \frac{8(0) + 2(0) + 4(40\sqrt{2}) + 2(40\sqrt{2})}{16} = \frac{240\sqrt{2}}{16} = 15\sqrt{2} \, cm$.The distance from $A(0, 0)$ is $R = \sqrt{X_{cm}^2 + Y_{cm}^2} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{450 + 450} = \sqrt{900} = 30 \, cm$.

Particles of masses $8 \, kg, 2 \, kg, 4 \, kg$,and $2 \, kg$ are placed at the vertices $A, B, C$,and $D$ of a square respectively. The distance of the center of mass of the system from vertex $A$ is $....... \, cm$. The length of the diagonal of the square is $80 \, cm$.

Similar Questions

Three identical spheres each of diameter $2\sqrt{3} \text{ m}$ are kept on a horizontal surface such that each sphere touches the other two spheres. If one of the spheres is removed,then the shift in the position of the centre of mass of the system is

The coordinates of the center of mass of the triangular lamina shown in the figure are ....... .

Two objects $A$ and $B$ of masses $M$ and $m$ respectively,where $M > m$,are separated by a distance. They move towards each other under the influence of equal forces. At what point will they collide?

Three particles of masses $1\,kg$,$\frac{3}{2}\,kg$,and $2\,kg$ are located at the vertices of an equilateral triangle of side $a$. The $x, y$ coordinates of the centre of mass are

The centre of mass is a point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module