Question

3 particles of masses 2 kg each are placed such
that one lies on -ve x-axis, 2nd one lies on -ve
x-axis and the third one lies on +ve Z-axis at
distances of 2m, 3m and 1 m respectively from
the origin. Then the square of the distance of
centre of mass of the system from the origin is:
(a) 1.55 m (b) 1.55 m (c) 1.55 m (d) 1.25 m2​

Answer 1

$d^2=2\times \frac{1}{9}$ is the square of the distance between the center of mass and origin.

Explanation:

• Let the masses be, $m_1, m_2\ \&\ m_3$ each of 2 kg.

• coordinates of mass $m_1$, $(x_1,z_1)=(-2,0)$

• coordinates of mass $m_2$, $(x_2,z_2)=(3,0)$

• coordinates of mass $m_3$, $(x_3,z_3)=(0,1)$

Now, for the x-coordinate of the center of mass of this system:

$\bar{x}=\frac{m_1.x_1+m_2.x_2+m_3.x_3}{m_1+m_2+m_3}$

$\bar{x}=\frac{2(-2+3+0)}{6}$

$\bar{x}=\frac{1}{3}$

Now, for the z-coordinate of the center of mass of this system:

$\bar{z}=\frac{m_1.z_1+m_2.z_2+m_3.z_3}{m_1+m_2+m_3}$

$\bar{z}=\frac{2(0+0+1)}{6}$

$\bar{z}=\frac{1}{3}$

Now, the square of the distance of  center of mass of the system from the origin:

$d^2=(\frac{1}{3}-0 )^2+(\frac{1}{3}-0 )^2$

$d^2=2\times \frac{1}{9}$

TOPIC: center of mass

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