Physics MOTION OF CENTRE OF MASS

Velocity of Centre of Mass of a System

If particles of masses `m_1, m_2, m_3`......... are moving with velocities `vecv_1, vecv_1, vecv_3`.........respectively, then velocity of centre of mass is given by

`vecV_(cm) = (m_1 vecv_1 + m_2 vecv_2 + ............)/(m_1 + m_2 + .................)`

= `(sum m_i vecv_i)/(sum m_i) = text(Total linear momentum of the system)/text(Total mass)`

Now components of velocity of centre of mass along X, Y, and Z axes can be written as.

`V_(cm_x) = (m_1v_(x_1) + m_2v_(x_2) + ............)/(m_1 + m_2 + .............)`

= `(sum m_i vecV_(x_i))/(sum m_i) = (sum P_(x_i))/(sum m_i)`

`V_(cm_y) = (m_1v_(y_1) + m_2v_(y_2) + .........)/(m_1 + m_2 + ..........) = (sum m_iV_(y_i))/(sum m_i) = (sum P_(y_i))/(sum m_i)`

`V_(cm_z) = (m_1v_(z_1) + m_2v_(z_2) +........)/(m_1 + m_2 +..........) = (sum m_iV_(z_i))/(sum m_i) = (sum P_(z_i))/(sum m_i)`

If `vecv_1 = vecv_2=........... = vecv` then `vecV_(cm) = vecv`

Note :
(a) Motion of centre of mass is not affected by the internal forces. Therefore, if a shell moving under gravity explodes into pieces moving in different direction, even then the centre of mass moves along the same (previous )path.

(b) Due to mutual interaction forces, velocity of CM does not change.

(c) If no external force acts on the body, state of motion of its centre of mass remains the same i.e. if it is moving with some velocity then it keep on moving in the same direction with same speed and if CM is at rest, it will remain at rest.

Acceleration of centre of mass of a system

If particles of system of masses `m_1, m_2` are moving with accelerations `veca_1, veca_2`....... then acceleration of CM is given by

`veca_(cm) = (m_1 veca_1 + m_2 veca_2 +..........)/(m_1 + m_2 +.........) = (sum m_ia_i)/(sum m_i) = (vecF_(ext))/M`

Components of acceleration of CM along X,Y and Z axes can be written as

`a_(cm_x) = (m_1a_(x_1) + m_2a_(x_2) + ............)/(m_1 + m_2 + .............)` = `(sum m_i veca_(x_i))/(sum m_i) = (sum a_(x_i))/(sum m_i)`

`a_(cm_y) = (m_1a_(y_1) + m_2a_(y_2) + .........)/(m_1 + m_2 + ..........) = (sum m_ia_(y_i))/(sum m_i)`

`a_(cm_z) = (m_1a_(z_1) + m_2a_(z_2) +........)/(m_1 + m_2 +..........) = (sum m_ia_(z_i))/(sum m_i)`

Note: if `veca_1 =veca_2 = ............=veca`

Then `veca_(cm) = veca`

Center of Mass Frame ( CM Frame)

(a) A frame of reference attached to centre of mass is called CENTRE or MASS frame.

(b) CM frame is also known as zero momentum frame.

(c) The kinetic energy of a system of particles can be expressed as a sum total of two kinetic energies

Kinetic energy of centre of mass `K_(cm)= 1/2MV_(cm)^2`

Kinetic energy of the system with respect to centre of mass `K = sum1/2 m_i v_i^2`

Where `v_i` is velocity of `i^(th)` particle with respect to centre of mass.

`K_(system) =K_(cm) + K`

In the absence of external force `K_(cm)` remains constant as velocity of centre of mass cannot be changed. In an inertial frame, in the absence of dissipative forces the mechanical energy of a system is conserved. i.e.

`triangle K_(system) + triangle U_(system) = 0`

`triangle(K_(cm) + K) + triangle U_(system) = 0`

In the absence of external force `V_(cm)` = constant , hence `triangle K_(cm) = 0`.

i.e. `triangleK + triangle U_(system) = 0`

In this situation the KE of the system with respect to CM may be converted into potential energy or vice-versa so that `(K + U)_(system)` remains constant .

Linear Momentum and Kinetic Energy

For a particle of mass m moving with speed v.

K = K.E of a body = `(1/2) mv^2`

P = magnitude of linear momentum = mv

`K = (mv^2)/2 = (m^2v^2)/2 = P^2/(2m)`

`K = P^2/(2m)` or `P = sqrt(2Km)`

(a) If P = constant

`Kprop1/m`

i.e if different bodies have same momentum, K.E will be maximum for the lightest one.

(b) If K = constant

`Pprop sqrtm`

i.e. If different bodies have same KE the heaviest body will have the maximum magnitude ofmomentum.

(c) If m = constant

`Pprop sqrtK`

i.e. If different bodies have same mass, the body having maximum KE will have maximum momentum.

Conservation of Linear Momentum

(a) If net external force acting on a particle or system is zero then net linear momentum of the particle or the system remains constant. This is called conservation of linear momentum.

(b) Newton's `2^(nd)` Law of motion.

`vecF_(ext) = (dvecP)/(dt)`

If `vecF_(ext) = 0 => (dvecP)/(dt) = 0 => vecP = constant`

i.e. Linear momentum vector is conserved

Note : `vecF_(ext) = F_(x^hati) + F_(y^hatj) + F_(z^hatk)`

`vecP = P_(x^hati) + P_(y^hatj) + P_(z^hatk)`

`F_(x^hati) + F_(y^hatj) + F_(z^hatk) = d/(dt) (P_(x^hati) + P_(y^hatj) + P_(z^hatk))`

= `(dP_x)/(dt) hati + (dP_y)/(dt) hatj + (dP_z)/(dt) hatk , F_x = (dP_x)/dt`

`F_y = (dP_y)/(dt) F_z = (dP_z)/(dt)`

If `F_x` = 0 then `P_x` = constt

If `F_x` = 0 `F_y` = 0 then `P_x` = constt ; `p_y` = constt

If `F_x` = 0, `F_y` = 0, `F_z` = 0 i.e `vecF_(ext)` = 0

`P_x` = constt, `P_y` = constt, `P_z` = constt i.e. `vecP` remains constant

Note:
Depending upon forces acting on a system, linear momentum might be conserved in one direction ( i.e. `P_x = 0`) or two directions ( i.e. `P_x = 0, P_y = 0`) or in all directions (i.e. `P_x = 0, P_y = 0, P_z = 0`). If a component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.