Four particles of masses m_1 = 2m,m_2 = 4m,m_ | vedclass

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

Question

(D) Let the corners of the square be $A(0,0)$,$B(a,0)$,$C(a,a)$,and $D(0,a)$.Given masses are $m_1=2m$ at $(0,0)$,$m_2=4m$ at $(a,0)$,$m_3=m$ at $(a,a

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let the corners of the square be $A(0,0)$,$B(a,0)$,$C(a,a)$,and $D(0,a)$.Given masses are $m_1=2m$ at $(0,0)$,$m_2=4m$ at $(a,0)$,$m_3=m$ at $(a,a)$,and $m_4$ at $(0,a)$.The centre of mass $(X_{cm}, Y_{cm})$ is given by:$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4} = \frac{2m(0) + 4m(a) + m(a) + m_4(0)}{2m + 4m + m + m_4} = \frac{5ma}{7m + m_4}$For the centre of mass to be at the centre of the square,$X_{cm} = \frac{a}{2}$.$\frac{5ma}{7m + m_4} = \frac{a}{2} \implies 10m = 7m + m_4 \implies m_4 = 3m$.Now check for $Y_{cm}$:$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4} = \frac{2m(0) + 4m(0) + m(a) + m_4(a)}{2m + 4m + m + m_4} = \frac{ma + m_4a}{7m + m_4}$For $Y_{cm} = \frac{a}{2}$:$\frac{a(m + m_4)}{7m + m_4} = \frac{a}{2} \implies 2m + 2m_4 = 7m + m_4 \implies m_4 = 5m$.Since the values of $m_4$ required for $X_{cm}$ and $Y_{cm}$ are different,it is impossible for the centre of mass to be at the centre of the square for any value of $m_4$. Thus,the correct option is $D$.

Similar Questions

Two bodies of mass $1 \ kg$ and $3 \ kg$ have position vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$,respectively. The centre of mass of this system has a position vector:

The centre of mass of three particles of masses $1 \ kg, 2 \ kg$ and $3 \ kg$ is at $(2, 2, 2)$. The position of the fourth mass of $4 \ kg$ to be placed in the system so that the new centre of mass is at $(0, 0, 0)$ is:

To find the centre of mass of a rigid body,why is it not possible to know $\sum m_i \vec{r}_i$ for all the particles?

Two particles of masses $200 \ g$ and $500 \ g$ are moving with velocities $10 \ \hat{i} \ m/s$ and $3 \ \hat{i} + 5 \ \hat{j} \ m/s$ respectively. The velocity of the center of mass of the system is:

The centre of mass of a triangle shown in the figure has coordinates

Vedclass Test Series

Exam Paper Generator

Online Exam Module