Question

Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass:

1. Show p_i = p’_i + m_iV Where p_i is the momentum of the i^th particle (of mass m_i)  and p′ _i = m_i v′ _i. Note v′ _i is the velocity of the i^th particle relative to the centre of mass. Also, prove using the definition of the centre of mass `sum"p""'"_"t" = 0`

2. Show K = K′ + 1/2MV²

   where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the
   system when the particle velocities are taken with respect to the centre of mass and MV²/2 is the
   kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the
   system). The result has been used in Sec. 7.14.

3. Show where `"L""'" = sum"r""'"_"t" xx "p""'"_"t"` is the angular momentum of the system about the centre of mass with
   velocities taken relative to the centre of mass. Remember `"r"_"t" = "r"_"t" - "R"`; rest of the notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
4. Show `"dL"^"'"/"dt" = ∑"r"_"i"^"'" xx "dP"^"'"/"dt"`
   Further show that `"dL"^'/"dt" = τ_"ext"^"'"`
   Where `"τ"_"ext"^"'"` is the sum of all external torques acting on the system about the centre of mass.
   (Hint: Use the definition of centre of mass and third law of motion. Assume the internal forces between any two particles act along the line joining the particles.)

Answer 1

(a) Take a system of i moving particles.

Mass of the i^th particle = m_i

Velocity of the i^th particle = v_i

Hence, momentum of the i^th particle, p_i = m_i v_i

Velocity of the centre of mass = V

The velocity of the i^th particle with respect to the centre of mass of the system is given as:

v’_i = v_i – V … (1)

Multiplying m_i throughout equation (1), we get:

m_i v’_i = m_i v_i – m_i V

p’_i = p_i – _­m_i V

Where,

p_i’ = m_iv_i’ = Momentum of the i^th particle with respect to the centre of mass of the system

∴p_i = p’_{i ­}+ m_i V

We have the relation: p’_i = m_iv_i’

Taking the summation of the momentum of all the particles with respect to the centre of mass of the system, we get:

`sum_("i") "p"_"t" = sum_("i")"m"
_"iv"_"t" = sum_("i")"m"_"i" ("dr"_"t")/"dt"`

Where

`"r""'"_"t" `= Position vector of the ith particle with respect to the centre of mass

`"v"_"t" =  ("dr"_"t")/"dt"`

As per the definition of the centre of mass, we have

`sum_("i") "m"_"ir"_"t" = 0`

`:. sum_("i") "m"_"i" ("dr"_"t")/"dt" = 0`

`sum_("i") "p"_"i" = 0`

(b)  We have the relation for the velocity of the i^th particle as:

v_i = v’_i + V

`sum_("i")"m"_"i" "v"_"i" =  sum_("i") "m"_"i" "v"_"t" + sum_("i")"m"_"i" "V"` ....(2) 

Taking the dot product of equation (2) with itself, we get: 

`sum_("i")"m"_"i" "v"_"i" .sum_("i") "m"_"iv"_"i" = sum_("i") "m"_"i"("v"_"i" + "V").sum_("i") "m"_"i"("v"_"i" + "V")`

`"M"^2sum_("i") "v"_"i"^2 = "M"^2 sum_("i") "v"_"i"^2 + "M"^2 sum_("i") "v"_"i". "v"_"i"+ "M"^2 sum_("i") "v"_"i"."v"_"i" + "M"^2"V"^2`

Here, for the centre of mass of the system of particle

`sum_("i")"v"_"i"."v"_"i" =-sum_("i")"v"_"i"."v"_"i"`

`"M"^2sum_("i") "v"_"i"^2 =  "M"^2 sum_("i")"v"^2_"i" + "M"^2"V"^2`

`1/2 "M" sum_("i") "v"_"i"^('2) = 1/2 "M" sum_("i") "v"_"i"^('2) + 1.2 "MV"^2 `

`"K" = "K"  + 1/2 "MV"^2`

Where

`"K" = 1/2 "M"sum_("i")"v"_"i"
^2` = Total kinetic energy of the system of particles

`"K""'" = 1/2 "M" sum_("i")"v"_"i"^(2)` = Total kinetic energy of the system of particles with respect to the centre of mass

`1/2 "MV"^2` = Kinetic energy of the translation of the system as a whole.

(c) Position vector of the i^th particle with respect to origin = r_i

Position vector of the i^th particle with respect to the centre of mass = r’_i

Position vector of the centre of mass with respect to the origin = R

It is given that:

r’_i = r_i – R

r_i = r’_i + R

We have from part (a),

p_i = p’_{i ­}+ m_i V

Taking the cross product of this relation by r_i, we get:

`sum_("i")"r"_"i" xx "p"_"i" =  sum_("i")"r"_"i"xx"p""'"_"i" + sum_("i") "r"_"i" xx "m"_"i" "V"`

`"L" = sum_("i") ("r""'"_"i" + "R" )xx "p""'"_"i" + sum_("i")("r""'"_"i" +  "R") xx "m"_"iV"`

`= sum_("i")"r""'"_"i" xx "p""'"_"i" + sum_("i") ("r""'"_"i"+ "R") xx "m"_"iV"`

`= "L""'" + sum_("i") + sum_("i")"R" xx "p""'"_"i" + sum_("i") "r""'"_"i" xx "m"_"iV"+ sum_("i")"R" xx "m"_"iV"`

Where

`"R" xx sum_("i") "p""'"_"i" = 0` and  `(sum_("i")"r""'"_"i") xx "MV" = 0`

`sum_("i") "m"_"i" = "M"`

`:."L" = "L""'" + "R" xx "MV"`

(d) Show `("dL""'")/"dt" = sum "r'"_"t" xx  ("dp""'")/"dt"`

Further, show that

`("dL'")/("dt") = tau"'"_"ext"`

where `tau"'"_"ext"` is the sum of all external torques acting on the system about the
centre of mass.
(Hint : Use the definition of centre of mass and Newton’s Third Law. Assume the internal forces between any two particles act along the line joining the particles.)