Three particles of masses 50, g,100, g,and 15 | vedclass

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

Question

(D) The coordinates of the three particles are:$m_1 = 50\, g$ at $(0, 0)$$m_2 = 100\, g$ at $(1, 0)$$m_3 = 150\, g$ at $(0.5, \frac{\sqrt{3}}{2})$The

Vedclass · Accepted Answer

$\left( \frac{7}{12}\,m, \frac{\sqrt{3}}{4}\,m \right)$

Vedclass · Answer

(D) The coordinates of the three particles are:$m_1 = 50\, g$ at $(0, 0)$$m_2 = 100\, g$ at $(1, 0)$$m_3 = 150\, g$ at $(0.5, \frac{\sqrt{3}}{2})$The $x$-coordinate of the center of mass is:$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} = \frac{50(0) + 100(1) + 150(0.5)}{50 + 100 + 150} = \frac{100 + 75}{300} = \frac{175}{300} = \frac{7}{12}\, m$The $y$-coordinate of the center of mass is:$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} = \frac{50(0) + 100(0) + 150(\frac{\sqrt{3}}{2})}{300} = \frac{75\sqrt{3}}{300} = \frac{\sqrt{3}}{4}\, m$Thus,the coordinates of the center of mass are $\left( \frac{7}{12}\,m, \frac{\sqrt{3}}{4}\,m \right)$.

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

Similar Questions

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:

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:

Three identical spheres of mass $1 \ kg$ each are arranged as shown in the figure. The centers of the spheres,which are touching each other,lie on a straight line. If their centers are denoted by $P, Q,$ and $R$,what is the distance of the center of mass of the system from $P$?

$A$ square plate and a circular disc of uniform density are arranged as shown in the figure. The center of mass of this combined system will be .........

Vedclass Test Series

Exam Paper Generator

Online Exam Module