Each side of a rhombus has length a. Four par | vedclass

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

Question

(B) Let the masses at the vertices be $m_1 = m, m_2 = 2m, m_3 = 3m, m_4 = 4m$.Based on the geometry of the rhombus in the $x-y$ plane:$r_1 = (0, 0)$$r

Vedclass · Accepted Answer

$\left( 0.95a, \frac{\sqrt{3}}{4}a \right)$

Vedclass · Answer

(B) Let the masses at the vertices be $m_1 = m, m_2 = 2m, m_3 = 3m, m_4 = 4m$.Based on the geometry of the rhombus in the $x-y$ plane:$r_1 = (0, 0)$$r_2 = (a \cos 60^o, a \sin 60^o) = (a/2, a\sqrt{3}/2)$$r_3 = (a + a \cos 60^o, a \sin 60^o) = (3a/2, a\sqrt{3}/2)$$r_4 = (a, 0)$The center of mass $(X_{cm}, Y_{cm})$ is given by:$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4} = \frac{m(0) + 2m(a/2) + 3m(3a/2) + 4m(a)}{10m} = \frac{0 + am + 4.5am + 4am}{10m} = \frac{9.5am}{10m} = 0.95a$$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4} = \frac{m(0) + 2m(a\sqrt{3}/2) + 3m(a\sqrt{3}/2) + 4m(0)}{10m} = \frac{am\sqrt{3} + 1.5am\sqrt{3}}{10m} = \frac{2.5a\sqrt{3}}{10} = \frac{\sqrt{3}}{4}a$Thus,the center of mass is $(0.95a, \frac{\sqrt{3}}{4}a)$.

Similar Questions

$A$ rod of length $5L$ is bent at a right angle,keeping one side length as $2L$. Find the position of the center of mass of the system. (Consider $L = 10 \ cm$)

For a system consisting of two objects,when can they be treated as point particles?

Find the position of the center of mass of a system of two particles of masses $m_1$ and $m_2$ separated by a distance $L$.

Masses $m \left(\frac{1}{3}\right)^N \frac{1}{N}$ are placed at $x=N$,where $N=2, 3, 4, \ldots \infty$. If the total mass of the system is $M$,then the centre of mass is

Three point masses $m_1 = 1.6 \, kg$,$m_2 = 2.0 \, kg$,and $m_3 = 2.4 \, kg$ are placed at the corners of a thin massless rectangular sheet $(1.2 \, m \times 1.0 \, m)$ as shown. The center of mass will be located at the point ........... $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module