Question

Four bodies of masses 2m, 3m, m and 2m are arranged at the four corner of a rectangle a and b. Find the center of mass of the system if one corner is at the origin of the coordinate system​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Masses are arranged in 2m, 3m ,m ,2m.

Let the length be "a" and breadth be "b".

The coordinate will come as:-

For 2m = (0,0)

For 3m = (a,0)

For m = (a,b)

For 2m = (0,b)

$\huge{\bold{\underline{Explanation:-}}}$

#refer the attachment for figure.

Centre of mass will be having two coordinates "x" and "y".

Let's Find X - coordinate Center of mass.

Applying Centre of mass formula.

$\large{\bold{X_{COM} = \frac{M_1X_1 + M_2X_2 + M_3X_3 + M_4X_4}{M_1 + M_2 + M_3 + M_4}}}$

Substituting the values.

$\large{X _{COM} = \frac{2m \times 0 + 3m \times a + m \times a + 2m \times 0}{2m + 3m + m + 2m}}$

$\large{X_{COM} = \frac{0 + 3ma + ma + 0}{8m}}$

$\large{X_{COM} = \frac{4ma}{8m}}$

$\large{X_{COM} = \frac{ \cancel{4m} a}{ \cancel{8m}}}$

$\large{\boxed{ X_{COM} = \frac{a}{2}}}$

${ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

Let's Find Y - coordinate Center of mass.

Applying Centre of mass formula.

$\large{\bold{Y_{COM} = \frac{M_1Y_1 + M_2Y_2 + M_3Y_3 + M_4Y_4}{M_1 + M_2 + M_3 + M_4}}}$

Substituting the values.

$\large{Y _{COM} = \frac{2m \times 0 + 3m \times 0 + m \times b + 2m \times b}{2m + 3m + m + 2m}}$

$\large{Y_{COM} = \frac{0 + 0 + mb + 2mb}{8m}}$

$\large{Y_{COM} = \frac{3mb}{8m}}$

$\large{Y_{COM} = \frac{ 3\cancel{m} a}{ 8\cancel{m}}}$

$\large{\boxed{ Y_{COM} = \frac{3b}{8}}}$

So,the coordinates of Centre of mass is

$\huge{\boxed{\boxed{( \frac{a}{2}, \frac{3b}{8})}}}$

Answer 2

Answer:

Explanation:Masses are arranged in 2m, 3m ,m ,2m.

Let the length be "a" and breadth be "b".

The coordinate will come as:-

For 2m = (0,0)

For 3m = (a,0)

For m = (a,b)

For 2m = (0,b)