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Motion of Centre of Mass

Motion of Centre of Mass

Vishal kumarUpdated on 02 Jul 2025, 05:43 PM IST

Assume for the moment that the place at which the mass of a body or system of bodies is meant to be concentrated during its entire motion is known as the centre of mass. Another way to put it is that the mean location of mass distribution in space where the force is typically applied is the average position of all the system components, leading to a linear acceleration devoid of any rotational acceleration. The body's or system's condition of motion or rest will not change if an external force is applied to the centre of mass of the body or system of bodies.

This Story also Contains

  1. Velocity of the Centre of Mass
  2. Solved Examples Based On Motion of the Centre of Mass
  3. Summary
Motion of Centre of Mass
Motion of Centre of Mass

In this article, we will cover the concept of velocity of the centre of mass and acceleration of the centre of mass under the topic of motion of the centre of mass. This topic falls under the broader category of Rotational Motion, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), more than five questions have been asked on this concept. It's also an important topic from NEET's point of view.

Let's read this entire article to gain an in-depth understanding of the motion of the centre of mass.

Velocity of the Centre of Mass

$
\vec{v}_{C M}=\frac{m_1 \overrightarrow{v_1}+m_2 \vec{v}_2 \ldots \ldots \ldots}{m_1+m_2 \ldots \ldots}
$
where: $\mathrm{m}_1, \mathrm{~m}_2 \ldots$ - - are the mass of all the particles and $\overrightarrow{v_1}, \overrightarrow{v_2} \ldots \ldots$ are velocities of all the particles.

Similarly momentum of the system $=P_{\text {sys }}=M v_{c m}$

Acceleration of Centre of Mass

$
\vec{a}_{C M}=\frac{m_1 \overrightarrow{a_1}+m_2 \vec{a}_2 \ldots \ldots .}{m_1+m_2 \ldots \ldots}
$
$\mathrm{m}_1, \mathrm{~m}_2$ are the mass of all the particles $\overrightarrow{a_1}, \overrightarrow{a_2} \cdots$ are their respective acceleration.

Similarly Net force on the system $=F_{n e t}=M a_{c m}$
And $F_{n e t}=\overrightarrow{F_{e x t}}+\overrightarrow{F_{i n t}}$

And we know that both the action and reaction of an internal force must be within the system. In this way, vector summation will cancel all internal forces and hence net internal force on the system is zero.

So $\overrightarrow{F_{n e t}}=M \overrightarrow{a_{c m}}$
If the External Force $=0$
$\vec{F}_{e x t}=0 \Rightarrow M \vec{a}_{c m}=0 \Rightarrow \vec{a}_{c m}=0$
if $\vec{a}_{c m}=0 \Rightarrow v_{c m}=$ constant
If $v_{c m}=$ constant $\Rightarrow P_{\text {sys }}=$ constant

So it implies that the total momentum of the system must remain constant.

i.e. if no external force is acting on the system, the net momentum of the system remains constant. This is nothing but the principle of conservation of momentum in the absence of external forces. Which says ìf the resultant external force is zero on the system, then the net momentum of the system must remain constant.

Special Case

If External Force = 0 and Velocity of Centre of Mass = 0

The centre of mass remains at rest. Individual components of a system may move and have non-zero momentum due to mutual forces but the net momentum of the system remains zero.

Recommended Topic Video

Solved Examples Based On Motion of the Centre of Mass

Example 1: identical particles move towards each other with velocities 2v and v respectively. The velocity of the centre of mass is :

1) $\nu$
2) $\nu / 3$
3) $\nu / 2$
4) zero

Solution:

$\begin{aligned} & v_c=\frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\ & o r \quad v_c=\frac{m(2 v)+m(-v)}{m+m}=\frac{v}{2}\end{aligned}$

Hence, the answer is option (3).

Example 2: The arrangement of two masses $m_1$ and $m_2$ are as shown in the fig Assume strings and pulley are massless and frictionless. Find the acceleration of the centre of mass of the system let $m_1=3 \mathrm{Kg}$ and $m_2=1 \mathrm{Kg}$.

1) g/2

2) g/3

3) g/4

4) g

Solution:

Acceleration of centre of mass -
$
\vec{a}_{C M}=\frac{m_1 \vec{a}_1+m_2 \vec{a}_2 \ldots \ldots \cdots}{m_1+m_2 \ldots \ldots .}
$
- wherein $\mathrm{m}_1, \mathrm{~m}_2$ are the mass of all the particles $\overrightarrow{a_1}, \overrightarrow{a_2} \cdots$ are their respective acceleration.
acceleration of system
$
a=\frac{\left(m_1-m_2\right)}{\left(m_1+m_2\right)} g
$
$\vec{a}_{c m}=\frac{\left(m_1 \vec{a}_1+m_2 \vec{a}_2\right)}{\left(m_1+m_2\right)}$
$
\vec{a}_{c m}=\frac{\left(m_1(-a)+m_2 a\right)}{\left(m_1+m_2\right)} \Rightarrow-\left(\frac{m_1-m_2}{m_1+m_2}\right) a
$

The magnitude of acceleration of the centre of mass is
$
a_{c m}=\left(\frac{m_1-m_2}{m_1+m_2}\right) a
$

$\begin{aligned} & a=\frac{(3-1)^2}{(3+1)^2} g \\ a= & g / 4\end{aligned}$

Example 3: A body A of mass $M$ while falling vertically downwards under gravity breaks into two parts; a body $B$ of mass $\frac{1}{3} M$ and another body $C$ of mass $\frac{2}{3} M$. The centre of mass of bodies $B$ and $C$ taken together shifts compared to that of the body $A$ towards

1) body $C$

2) body B

3) depends on the height of the breaking

4) does not shift

Solution:

If External Force = 0 and Velocity of Centre of Mass = 0

The centre of mass remains at rest. Individual components of a system may move and have non-zero momentum due to mutual forces but the net momentum of the system remains zero.

The centre of mass of bodies B and C taken together does not shift as no external force is applied horizontally.

Hence, the answer is option (4).

Example 4: A man of mass 80 Kg stands on a plank of mass 40 Kg. The plank is lying on a smooth horizontal floor. Initially, both are at rest. The man starts walking on the plank towards the north and stops after moving a distance of 6m on the plank. The plank will move

1) 6m, south

2) 4m, north

3) 4m, south

4) 6m, north

Solution:

As we have learned

If External Force = 0 -

$\vec{P}_{\text {total }}=\text { constant }$

wherein

i.e. if no external force is acting on the system, the net momentum of the system remains constant.

As

\begin{aligned}
& F_{\text {ext }}=0 \\
& \text { so } \vec{a}_{c m}=0 \quad \vec{v}_{c m}=\text { cost }=0 \\
& \vec{v}_{c m}=\frac{m_1 \vec{v}_1+m_2 \vec{v}_2}{m_1+m_2}=0 \\
& \text { or } m_1 v_1+m_2 v_2=0 \\
& m_1 \frac{d x}{d t}+m_2 \frac{d x_2}{d t}=0 \\
& \text { or } m_1 \vec{x}_1+m_2 \vec{x}_2=0
\end{aligned}

\begin{aligned}
& \quad x_1=(6-x) \\
& x_2=-x \\
& m_1(6-x)-m_2 x=0 \\
& 80(6-x)=40 x \\
& x=4 m, \text { south }
\end{aligned}

Example 5: A projectile of mass M is fired so that the horizontal range is 4 Km. At the highest point, the projectile explodes in two parts of masses M/4 and 3M/4 respectively and the heavier part starts falling down vertically with zero initial speed. The horizontal range (distance from the point of firing )of the lighter part is :

1) 16 Km

2) 1 Km

3) 10 Km

4) 2 Km

Solution:

\begin{aligned}
& X_{C O M}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\
& R=\frac{\frac{M}{4} x+\frac{3 M}{4} \times \frac{R}{2}}{M} \\
& \Rightarrow x=10 \mathrm{Km}
\end{aligned}

Hence, the answer is the option 3.

Summary

The forces that the system's particles apply to one another are known as internal forces; but, according to Newton's third rule, these internal forces only exist in pairs that are equal in magnitude and directed in opposing directions. Thus, they have a zero net sum.

Newton’s third law states that for every action, there is an equal and opposite reaction; therefore, the gravitational forces between them are identical in magnitude but opposite in direction.

Frequently Asked Questions (FAQs)

Q: How does the motion of the center of mass relate to the concept of normal modes in vibrating systems?
A:
In systems with multiple vibrating components, like molecules or coupled oscillators, the motion can be decomposed into normal modes. The center of mass motion often corresponds to one of these modes, typically the lowest frequency mode representing the overall translation of the system.
Q: What is the role of the center of mass in understanding the dynamics of star clusters?
A:
The center of mass of a star cluster moves through the galaxy under the influence of galactic tides and other external forces. Internally, stars orbit around this common center of mass, with their collective behavior determining the cluster's stability and evolution.
Q: What is the significance of the center of mass in analyzing the motion of tethered satellites?
A:
In the analysis of tethered satellite systems, the motion of the center of mass of the entire system (including the tether) follows a simple orbit, while the relative motion of the satellites about the center of mass can be complex and is influenced by the tether forces.
Q: How does the concept of center of mass apply to particle physics experiments?
A:
In particle physics, collisions are often analyzed in the center of mass frame. This frame simplifies the analysis of particle creation and decay processes, and is crucial for interpreting data from particle accelerators.
Q: How does the motion of the center of mass relate to the virial theorem in physics?
A:
The virial theorem, which relates the average kinetic energy of a stable system to its average potential energy, is often applied in the center of mass frame. This application is particularly useful in astrophysics for understanding the stability of star clusters and galaxies.
Q: How does the concept of center of mass apply to the field of robotics?
A:
In robotics, tracking and controlling the center of mass is crucial for maintaining balance, especially in bipedal robots. The center of mass trajectory is a key factor in planning stable walking gaits and other complex movements.
Q: What is the role of the center of mass in understanding the formation of planetary systems?
A:
The center of mass concept is essential in modeling the formation and evolution of planetary systems. It helps explain how planets and other bodies coalesce from rotating disks of gas and dust, and how they settle into stable orbits.
Q: How does the motion of the center of mass relate to the concept of pseudoforces in non-inertial frames?
A:
In non-inertial frames, like a rotating reference frame, pseudoforces appear to act on objects. These forces can be understood by considering the motion of the center of mass in an inertial frame and then transforming to the non-inertial frame.
Q: What is the significance of the center of mass in analyzing the motion of asteroids and comets?
A:
The center of mass motion of asteroids and comets determines their orbits around the Sun. However, their rotation about their center of mass, which can be complex due to irregular shapes, affects phenomena like outgassing in comets and the Yarkovsky effect in asteroids.
Q: What is the role of the center of mass in understanding the dynamics of black hole mergers?
A:
In the merging of black holes, the center of mass of the system emits gravitational waves. The motion of this center of mass, along with the complex orbital dynamics of the merging black holes, is crucial for predicting and interpreting gravitational wave signals.
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