If you stand up straight the spot where you are balanced is probably right below your belly. The position is the center of mass. |

You could have seen many a times any object is balanced by a finger. This point where its balanced is called its Center of Mass.

The Center of gravity also called as Center of Mass is a point where most of the objects mass is concentrated.

The Center of gravity also called as Center of Mass is a point where most of the objects mass is concentrated.

The Center of Mass is that unique point where the whole mass of the body is concentrated. Let the two masses

**m**at point_{i}**A**and**M**_{i }at point**B**exerts force on each other and they accelerate along line joining them.These particles are at rest initially and the force acting between them is attractive.The Center of mass at initial time

**t**and distance**X**is given byX = $\frac{m_i x_1 + M_i x_2}{m_i + M_i}$

$\vec{V}$ = $\frac{dx}{dt}$ = $\frac{m_i v_1 + M_i v_2}{m_i + M_i}$The acceleration After some time t having distance x

_{1}and x_{2}changes and obvious fact is X also changes along x-axis. So the Velocity of the center of mass at any time t is**a**of the Center of mass is

_{}$\vec{a}$ = $\frac{d\vec{V_cm}}{dt}$ = $\frac{m_i \vec{a_1} + M_i \vec{a_2}}{m_i + M_i}$

If the force magnitude between these particles is F. As only force acts only on A the acceleration is

a

_{1}= $\frac{\vec{F}}{m_i}$The force acting on B is (-F) so

a

_{2}= - $\frac{\vec{F}}{M_i}$Substituting in the acceleration equation we get

$\vec{a}$ = $\frac{m_i \frac{F_1}{m_i} + M_{i} \frac{\vec{-F}}{M_i}}{m_i + M_i}$ = 0

It justifies that if the velocity of the center of mass acting does not changes as time passes then $\vec{v_1}$ = $\vec{v_2}$ = 0 then velocity V is zero. Hence the center of mass remains constant and hence will not change with time.

Consider each corner of a triangle to have a unit mass to be located at the point a from the origin. If a

_{1}a_{2}and a_{3}are the coordinate points of the corners of the triangle.Center of mass of the triangle = $\frac{1}{3}$ (a

_{1}+ a_{2}+ a_{3}).Center of Mass is the point where the whole mass is assumed to be concentrated without taking the account of weight. Center of gravity is point where net weight acts. Normally, center of mass and gravity are same, this is because Earthâ€™s gravitational field is uniform.

Now consider a situation in which gravitational field is non-uniform. Both changes from each other as in case of the rod the center of gravity and center of mass differs on each side. Hence the center of mass is called the center of gravity for a object on the point on where the force due to gravity appears to act.

Now consider a situation in which gravitational field is non-uniform. Both changes from each other as in case of the rod the center of gravity and center of mass differs on each side. Hence the center of mass is called the center of gravity for a object on the point on where the force due to gravity appears to act.

**Here are given some problems on finding center of Mass you can go through it:**

### Solved Examples

**Question 1:**A system consists of two masses of 0.5 kg wt and 1 kg wt connected by a massless rod and 2 m and 7 m respectively. At what point will be the center of mass?

**Solution:**

Given: Mass m

_{1}= 0.5 kg, Mass m

_{2}= 1 kg, Distance x

_{1}= 2m, Distance x

_{2}= 7m

The center of mass is given by

x

_{cm}= $\frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

= $\frac{0.5 \times 2 + 1 \times 7}{0.5 + 1}$

= $\frac{1+7}{1.5}$

= 5.33.

**Question 2:**Calculate the center of mass if the bodies of masses 5kg, 2kg and 4kg are separated by the distances 0.3 m, 0.7 m and 1 m.

**Solution:**

Given: Mass m

_{1 }= 5 kg, Mass m

_{2}= 2 kg, Mass m

_{3}= 4 kg,

Distance x

_{1}= 0.3 m, x

_{2 }= 0.7 m, x

_{3}= 1m

The center of mass is given by

X

_{cm}= $\frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$

= $\frac{5 \times 0.3 + 2 \times 0.7 + 4 \times 1}{5 + 2 + 4}$

= $\frac{1.5 + 1.4 + 4 }{5 + 2 + 4}$

= $\frac{6.9}{11}$

= 0.62