We know that the **center of mass** of a body is a point inside it where the entire mass of the body is situated. For a uniform and symmetric body, its center of mass is exactly its middle point. Now, when the body moves, its center of mass also moves. So, the center of mass can have velocity and acceleration depending upon the type of motion of the body. In this article, we are going to **explain the velocity of center of mass of an object**.

**Contents in this article:**

**Velocity of Center of Mass****Velocity of center of mass equation****How to find velocity of center of mass?****Calculate velocity of center of mass with an example**

**Velocity of center of mass**

When an object moves, its center of mass also moves with the same velocity and acceleration that of the object. Then we can have the concept of the velocity of center of mass.

Velocity of center of mass has same unit and dimension as that of the normal velocity we know. But the presentation or the equation is different from that. For a pure **translational motion** in a straight line, the velocity of the center of mass and the velocity of other points in that body are same. The velocity of center of mass of a body changes after the collision.

**Velocity of center of mass equation**

The formula of velocity center of mass depends on the system of mass – whether it is continuous or discrete. Its equations for both types of system of mass are as followings –

**Equation of velocity center of mass for a discrete system of mass**

We consider a discrete system of masses **m _{1}, m_{2}, m_{3}, ……, m_{n}** moving with uniform speeds

**v**respectively. Then the formula for velocity center of mass of the discrete mass system is,

_{1}, v_{2}, v_{3}, …., v_{n}\color{Blue}v_{c}=\frac{m_{1}v_{1} + m_{2}v_{2} + m_{3}v_{3} + ....}{m_{1} + m_{2} + m_{3} + ...}………..(1)

It will be more accurate if we write the **velocities in vector form**. For a two-body system, we need to write up to two masses and so on.

**Formula of velocity center of mass for a continuous body**

For a continuous body, we need to use **integral equation**. Let velocity of an elementary mass (**dm**) of a body is **v**. This velocity can be a constant or a function of position on the body.

Then the equation for the velocity center of mass of the continuous body is,

\color{Blue}v_{c}=\frac{1}{M}\int v.dm……..(2)

Here, **M** is the mass of entire body.

**The equation of velocity center of mass is consistent with the law of conservation of linear momentum**

We know that the linear momentum (p) = mass × velocity.

That means, **p _{1} = m_{1}v_{1}**,

**p**, and so on.

_{2}= m_{2}v_{2}The from equation-(1) we get, \color{Blue}v_{c}=\frac{p_{1}+p_{2}+p_{3} + ....}{M}.

Here, **M** is the mass of entire system which is the sum of all masses.

Now, if we multiply both side by **M** then we get, \color{Blue}Mv_{c}=p_{1}+p_{2}+p_{3} + ...

or, \color{Blue}p_{c}=p_{1}+p_{2}+p_{3} + .....

This shows that the linear momentum of the center of mass of a system is equal to the sum of the linear momentum of all masses in that system. This is the concept of law of conservation of linear momentum. Above equation is sometime called the center of mass momentum equation.

**How to find velocity center of mass?**

One can calculate the velocity of the central point of a body by using above two equations. Equation-(1) is for a discrete mass system and equation-(2) is for a continuous body.

There is an example at below that shows the calculation of center of mass velocity using above formulae.

**Calculation of velocity center of mass with example**

**Question: **

**Two blocks of masses 6 kg and 3 kg are connected by a spring of negligible mass and placed on a frictionless table. An impulse gives a velocity of 10 m/s to the 6 kg mass in the direction of 3 kg mass. Find the velocity of the center of the mass.**

**Answer:**

This is an example of two-body discrete system. The masses are **m _{1} = 6 kg** and

**m**and their velocities are

_{2}= 3 kg**v**and

_{1}= 10 m/s**v**respectively. Then using equation-(1) we get,

_{2}= 0 m/sthe velocity of the center of mass of the spring-block system is, \color{Blue}v_{c}=\frac{6×10+3×0}{6+3}

or, \color{Blue}v_{c}=6.67 m/s. (**Answer**)

This is all from this article. If you have any doubt on this topic you can ask me in the comment section.

Thanks for reading this article and keep visiting this website.

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