Question

Derive formula for :

(i) Centre of mass of a two particle system.

(ii) Centre of mass of a system of n particles.

(iii) Co-ordinates of the centre of mass.

(iv) Momentum conservation and motion of the centre of mass.

(v) Equation of rotational motion.

(vi) Expression for torque in cartesian co-ordinates from rotation of a particle in a plane : physical meaning of torque.

Answer 1

(i) Centre of mass of two particles system : Let O be the origin of a rectangular co-ordinates system XYZ.

Consider a system of two particles of masses m₁ and m₂ located at A and B respectively.

\(\vec{OA}=\vec{r_1}\)

and \(\vec{OB}=\vec{r_2}\)

Let C be the position of centre of mass of the system of two particles. It would lie on the line joining A and B. Let \(\vec{OC}=\vec{r}\) be the position vector of mass.

To evaluate \(\vec{r}\), suppose \(\vec{v_1}\)&\(\vec{v_2}\) be the velocities of particles m₁ and m₂ respectively at any instant t

then, \(v_1=\frac{dr_1}{dt}\)

and \(v_2=\frac{dr_2}{dt} ......(1)\)

Let

f₁ = external force on m₁

f₂ = external force on m₂

F₁₂ = internal force of m₁ due to m₂

F₂₁ = internal force on m₂ due to m₁

Linear momentum of particle m₁

\(\vec{p_1}=m_1\vec{v_1} .......(2)\)

According to Newton’s second law total force acting on this particle which is (\(\vec{f_1}+\vec{F_{12}}\))

Using (2), \(\frac{d}{dt}(m_1\vec{v_1})= \vec{f_1}+\vec{F_{12}} ........(3)\)

Similarly, for second particle

\(\frac{d}{dt}(m_2\vec{v_2})= \vec{f_2}+\vec{F_{21}} ........(4)\)

Adding (3) and (4),

\(\vec{f_1}+\vec{f_2}=\vec{f} .......(5)\)

where \(\vec{f}\) = total external force on the system of two particles.

Using (1),

Multiplying numerator and denominator of left side by (m₁ + m₂),

Let us put

\((m_1+m_2)\frac{d^2}{dt^2}\vec{r}=\vec{f} .....(8)\)

This is the equation of motion of total mass (m₁ + m₂) supposed to be concentrated at a point whose position of vectors is \(\vec{r}\) under the effect of total force \(\vec{f}.\)

Now from (7),

\((m_1+m_2)\vec{r}=m_1\vec{r_1}+m_2\vec{r_2}\)

(ii) Centre of mass of a system of n particles :

Let O be the origin of rectangular co-ordinate system particles of masses m₁, m₂, m₃, . … . , m_n whose position in the coordinate system are given respectively by the position vectors

\(\vec{r_1},\vec{r_2},\vec{r_3}........\vec{r_n}.\)

Let \(\vec{r}\) be the position vector of centre of mass of this system of n particles. Let \(\vec{v_1},\vec{v_2},\vec{v_3}........\vec{r_n}\) be the instantaneous velocities of these particles. Suppose F₁, F₂, F₃, … . , F_n are the external force acting on the particles of the system.

Total force acting on i^nt particle of the system may be written as

According to Newton’s second law, the equation of motion of i^th particle of the system may be writes as

\(\frac{d}{dt}(m_i\vec{v_i})=\vec{F_i}+\vec{f_i} .......(9)\)

Where, \(\vec{f_i}\) is the external force acting on i^th particle.

We can write such equation of motion for all the n particles of the system. On adding these n particles, we get

The internal forces on all the particles of the system cancel out in pairs, i.e.,

Multiplying and dividing left hand side of (12) by M = \(∑^{i=n}_{i=1}m_i\) = total mass of all the n particles of the system,

Centre of mass of n particles :

From equation (13),

(iii) Co-ordinates of the centre of mass : We have learnt about the position vector r of a system of n particles

As each position vector \(\vec{r}\)(of i^th particle) can be expressed in terms of its components x_i, y_i , z_i in the form :

\(\vec{r_i}=\hat{i}x_i+\hat{j}y_i+\hat{k}z_i\)

where i = 1, 2, 3, 00000, n …(16)

If x, y, z are the co-ordinates of the centre of mass of the system,

\(\vec{r}=\hat{i}x+\hat{j}y+\hat{k}z\,......(17) \)

Using equations (16) and (17) in (15),

Equating the component separately,

\(x=\frac{1}{M}(m_1x_1+m_1y_1+m_1z_1).........(18)\)

\(y=\frac{1}{M}(m_2x_2+m_2y_2+m_2z_2).........(19)\)

\(z=\frac{1}{M}(m_nx_n+m_ny_n+m_nz_n).........(20)\)

Equations (18), (19), (20) can be rewritten as

Answer 2

(iv) Momentum conservation and motion of the centre of the mass : When a system of n particles is under the action of a total force f, then

according to Newton’s second law,

If no external force acts on the system, then total force \(\vec{f}=\vec{0}\).

Then from eqn. 24 we obtain

.............(25)

If M is the total mass of the system concentrated at the centre of mass, whose position vector is r, then

from eqn. (12)

Or 

But \(\frac{d\vec{r}}{dt}=\vec{v}_{cm}\)

= velocity of centre of mass of the system

∴ .............(26)

If no external force is acting then \(\vec{f}=\vec{0}\).

Then from eqn. (26)

.........(27)

(v) Equations of rotational motion :

(i) ω = ω₀ + αt

(ii) θ = ω₀t + 1/2 αt²

(i) \(ω^2_2 − _0^2 = 2.\)

(i) ω = ω0 + αt

Suppose a rigid body is rotating about a given axis with uniform angular acceleration α.

We know that,

\(α=\frac{dω}{dt}\)

dω = α dt ..........(1)

At t = 0, let ω = ω₀, t = 0, let ω = ω

Integrating (1) within proper limits,

...........(2)

(ii) If ω is angular velocity of the rigid body at any time t, then we know that

\(ω=\frac{dθ}{dt}\)

dθ = ωdt ..........(3)

t = 0, let θ = 0, t = t, let θ = θ

Integrating (3)

Using (2)

(iii) \(ω^2-ω^2_0=2αθ\)

We know that

or ωdω = αdθ ........(5)

When θ = 0, ω = ω₀, initial angular velocity and when θ = θ, ω = ω, final angular velocity.

∴ Integrating (5)

or \(ω^2-ω^2_0=2αθ\)

(vi) Expression for torque in cartesian co-ordinates from rotation of a particle in a plane :

Physical meaning of torque :

Consider a particle of mass m rotating in plane XY about the origin O. Let P be the position of the particle at any instant, where \(\vec{OP}=\vec{r}\) and ∠XOP = θ. Let the rotation occur under the action of a force \(\vec{F}\) applied at P, along \(\vec{PA}\).

In a small time dt, let the particle at P reach Q, where

\(\vec{OP}=\vec{r}+d\vec{r}\)

and ∠POQ = dθ

In vector triangle OPQ,

\(\vec{OP}+\vec{PQ}=\vec{OQ}\)

\(\vec{PQ}=\vec{OQ}+\vec{OP}\)

\((\vec{r}+\vec{dr})-\vec{r}\)

\(\vec{PQ}=\vec{dr}\)

Small amount of work done in rotating the particle from P to Q is :

dW = \(\vec{F}.d\vec{r} ...........(i)\)

If F_x, F_y are rectangular components of force F and d_x, d_y are rectangular component of displacement of \(\vec{dr},\), then

From equation (i), we get

..........(ii)

Let the co-ordinates of the point P be (x, y) as is clear from fig.

x = r cos θ ...........(iii)

y = r sin θ ...........(iv)

Differentiating equation (iii) w.r.t. θ

\(\frac{dx}{dθ} = \frac{d}{dθ}(rcos\,θ)\)

\(= r\frac{d}{dθ}(cos\,θ)\)

= r(-sin θ) = -r sin θ

= -y .............(v)

dx = -ydθ .....(vi)

Again, differentiating equation (iv) w.r.t. θ,

\(\frac{dy}{dθ} = \frac{d}{dθ}(rsin\,θ)\)

= r cos θ

= x ........(vii)

∴ dy = xdθ .........(viii)

Substitute equation (ii)

τ is called torque.