Motion of Centre of Mass

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.

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.

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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

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.