Centre Of Mass And Centre Of Gravity In Different Bodies (NEET LEVEL)

Motion of the centre of mass

The equations of motion of the centre of mass are as follows:

\overrightarrow{{{{V}_{{CM}}}\,}}=\frac{{\overrightarrow{{dR\,}}}}{{dt}}

M\frac{{\overrightarrow{{d{{V}_{{CM}}}\,}}}}{{dt}}=M\frac{{{{d}^{2}}\overrightarrow{{{{R}_{{CM}}}\,}}}}{{d{{t}^{2}}}}  

or M\frac{{\overrightarrow{{d{{V}_{{CM}}}\,}}}}{{dt}}=\overrightarrow{{F\,}} 

Working in terms of the centre of mass of the system gives us the advantage of ignoring the internal forces, internal motion and internal structure. We need to take into account only the external forces acting on the system.

Let us now establish a relation between the momentum of the system and the momentum of centre of mass of the system.

Let us consider a system consisting of n particles of masses m₁, m₂ …m_n. The total mass M of the system is given by

M={{m}_{1}}+{{m}_{2}}+\ldots +{{m}_{n}}

The position vector of the centre of mass is given by

\overrightarrow{{{{R}_{{CM}}}\,}}=\frac{{{{m}_{1}}\overrightarrow{{{{r}_{1}}}}+{{m}_{2}}\overrightarrow{{{{r}_{2}}}}+\ldots +{{m}_{n}}\overrightarrow{{{{r}_{n}}}}}}{{{{m}_{1}}+{{m}_{2}}+\ldots +{{m}_{n}}}}

or         \overrightarrow{{{{R}_{{CM}}}\,}}=\frac{{{{m}_{1}}\overrightarrow{{{{r}_{1}}}}+{{m}_{2}}\overrightarrow{{{{r}_{2}}}}+\ldots +{{m}_{n}}\overrightarrow{{{{r}_{n}}}}}}{M}

Differentiating both sides with respect to time t, we get

\frac{d}{{dt}}({{\overrightarrow{R}}_{{CM}}})=\frac{1}{M}\left[ {\frac{d}{{dt}}({{m}_{1}}\overrightarrow{{{{r}_{1}}}})+\frac{d}{{dt}}({{m}_{2}}\overrightarrow{{{{r}_{2}}}})+\ldots +\frac{d}{{dt}}({{m}_{n}}\overrightarrow{{{{r}_{n}}}})} \right]

or   M\frac{{\overrightarrow{{d{{R}_{{CM}}}\,}}}}{{dt}}={{m}_{1}}\frac{{\overrightarrow{{d{{r}_{1}}}}}}{{dt}}+{{m}_{2}}\frac{{\overrightarrow{{d{{r}_{2}}}}}}{{dt}}+\ldots +{{m}_{n}}\frac{{\overrightarrow{{d{{r}_{n}}}}}}{{dt}}

But      \frac{{\overrightarrow{{d{{R}_{{CM}}}\,}}}}{{dt}}=\overrightarrow{{{{V}_{{CM}}}\,}}      (velocity of centre of mass)

\frac{{\overrightarrow{{d{{r}_{1}}}}}}{{dt}}=\overrightarrow{{{{v}_{1}}}},\frac{{\overrightarrow{{d{{r}_{2}}}}}}{{dt}}=\overrightarrow{{{{v}_{2}}}},\ldots ,\frac{{\overrightarrow{{d{{r}_{n}}}}}}{{dt}}=\overrightarrow{{{{v}_{n}}}}

Where \,\overrightarrow{{{{v}_{1}}}},\,\overrightarrow{{{{v}_{2}}}},\ldots \,\overrightarrow{{{{v}_{n}}}}  are the velocities of the individual particles of the system.

M\,\overrightarrow{{{{V}_{{CM}}}\,}}={{m}_{1}}\overrightarrow{{{{v}_{1}}}}+{{m}_{2}}\overrightarrow{{{{v}_{2}}}}+\ldots +{{m}_{n}}\overrightarrow{{{{v}_{n}}}}             …(i)

or   M\,\overrightarrow{{{{V}_{{CM}}}\,}}=\sum\limits_{{i=1}}^{n}{{{{m}_{i}}\overrightarrow{{{{v}_{i}}}}}}

We conclude from this equation that the total momentum of the system is the same as the moment tum of the centre of mass   

(ii)    

Now let usestablish a relation between the force acting on the centre of mass and the forces acting on the individual particles of the system.

Differentiating both sides of equation (1) w.r.t. time, we get

\frac{d}{{dt}}(M\overrightarrow{{{{V}_{{CM}}}\,}})=\frac{d}{{dt}}({{m}_{1}}\overrightarrow{{{{v}_{1}}}})+\frac{d}{{dt}}({{m}_{2}}\overrightarrow{{{{v}_{2}}}})+\ldots +\frac{d}{{dt}}({{m}_{n}}\overrightarrow{{{{v}_{n}}}})

or   M\frac{d}{{dt}}(\overrightarrow{{{{V}_{{CM}}}}})={{m}_{1}}\frac{d}{{dt}}(\overrightarrow{{{{v}_{1}}}})+{{m}_{2}}\frac{d}{{dt}}(\overrightarrow{{{{v}_{2}}}})+\ldots +{{m}_{n}}\frac{d}{{dt}}(\overrightarrow{{{{v}_{n}}}})

But      \frac{d}{{dt}}(\overrightarrow{{{{V}_{{CM}}}}})=\overrightarrow{{{{A}_{{CM}}}\,}}    (acceleration of centre of mass)

\frac{d}{{dt}}(\overrightarrow{{{{v}_{1}}}})=\overrightarrow{{{{a}_{1}}}},\frac{d}{{dt}}(\overrightarrow{{{{v}_{2}}}})=\overrightarrow{{{{a}_{2}}}},\ldots \frac{d}{{dt}}(\overrightarrow{{{{v}_{n}}}})=\overrightarrow{{{{a}_{n}}}}

Where \overrightarrow{{{{a}_{1}}}},\overrightarrow{{{{a}_{2}}}}\ldots \overrightarrow{{{{a}_{n}}}}   are the acceleration of the individual particles of the system.

M\overrightarrow{{{{A}_{{CM}}}\,}}={{m}_{1}}\overrightarrow{{{{a}_{1}}}}+{{m}_{2}}\overrightarrow{{{{a}_{2}}}}+\ldots +{{m}_{n}}\overrightarrow{{{{a}_{n}}}}

 Applying Newton’s 2^nd law of motion,

\overrightarrow{{F\,}}=\overrightarrow{{{{F}_{1}}}}+\overrightarrow{{{{F}_{2}}}}+\ldots +\overrightarrow{{{{F}_{n}}}}

Here \overrightarrow{{F\,}} is the force acting on the centre of mass. This is clearly equal to the vector sum of the forces acting on the individual particles of the system.

Momentum conservation and motion of the centre of mass

The equation of motion of centre of mass is

\overrightarrow{{F\,}}=\frac{d}{{dt}}(M{{\vec{V}}_{{CM}}})

Where \overrightarrow{{F\,}} is the net external force acting on the system, M is the mass of the system and \displaystyle {{\vec{V}}_{{cm}}} is the velocity of the centre of mass of the system.

If no external force acts on the system, then \overrightarrow{{F\,}}=0.

\frac{d}{{dt}}(M{{\vec{V}}_{{CM}}})=0  

\displaystyle M{{\vec{V}}_{{CM}}}=\text{constant}                                       …(i)

{{m}_{1}}\overrightarrow{{{{v}_{1}}}}+{{m}_{2}}\overrightarrow{{{{v}_{2}}}}+\ldots +{{m}_{n}}\overrightarrow{{{{v}_{n}}}}=\text{constant}                        …(ii)

It is clear from equations (i) and (ii) that if no external force acts on the system, then the total momentum of the system is constant. This is law of conservation of momentum.

It is further clear from equation (i) that

{{\vec{V}}_{{CM}}}=\text{constant}    (QM is constant)

Thus, if no external force acts on a system, then its centre of mass moves uniformly in a straight line. This is Newton’s first law of motion.

Illustration

In the arrangement shown in figure, {{m}_{A}}=2\mathbf{kg} and {{m}_{B}}=1\mathbf{kg} String is light and inextensible. Find the acceleration of centre of mass of both the blocks. Neglect friction everywhere.