Question

Four particles of masses m, 2m, 3m and 4m are kept at 4 corners of a

square of side 1m. Find the centre of mass of the system about the

mass m placed at origin.​

Answer 1

GIVEN:

The masses of four particles are

• First particle (m₁) = m
• Second particle (m₂) = 2m
• Third particle (m₃) = 3m
• Fourth particle (m₄) = 4m

TO FIND:

• The centre of mass of the system about the mass m placed at the origin.

The co-ordinated in which the particles are lying are,

→ (0,0) ; (1,0) ; (1,1) ; (0,1)

$\boxed{\boxed{\bf x_1 = 0,\ x_2 = 1,\ x_3 = 1,\ x_4 = 0 }}$

$\boxed{\boxed{\bf y_1 = 0,\ y_2 = 0,\ y_3 = 1,\ y_4 = 1 }}$

Now,

• Applying for the formula for the X-component of the centre of mass,

$\boxed{X_{cm} = \dfrac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}+m_{4}x_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}}$

Putting the respective values of, we get,

$X_{cm} = \dfrac{m(0)+2m(1)+3m(1)+4m(0)}{m+2m+3m+4m}$

$X_{cm} = \dfrac{2m+3m}{10m}$

$X_{cm} = \dfrac{5\not{m}}{10 \not{m}}$

Both numerator and denominator can be divided by 2.

$\boxed{\boxed{X_{cm} = \dfrac{1}{2}}}$

• Applying for the formula for the Y-component of the centre of mass,

$\boxed{Y_{cm} = \dfrac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}+m_{4}y_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}}$

Putting the respective values of, we get,

$Y_{cm} = \dfrac{m(0)+2m(0)+3m(1)+4m(1)}{m+2m+3m+4m}$

$Y_{cm} = \dfrac{3m+4m}{10m}$

$X_{cm} = \dfrac{7\not{m}}{10\not{m}}$

$X_{cm} = \dfrac{7}{10}$

Hence,

The coordinates for the centre of mass should be written as,

$\boxed{(X_{cm}, Y_{cm})=(\frac{1}{2},\frac{7}{10})}$