Centre Of Mass

Centre of mass is defined as the point at which the total mass of a system is supposed to be concentrated

On the application of external force Centre of mass of system of particles move in the same way as a point having mass equal to that of the whole mass move

Position Of Centre of Mass

1) For Two Particles R_cm = \(\frac{m_1r_1+m_2r_2}{m_1+m_2}\) = \(\frac{m_1r_1 + m_2r_2}{M}\)

2) For Three Dimensional Axis X_cm = \(\frac{m_1x_1 + m_2x_2}{m_1 + m_2}\) = \(\frac{m_1x_1 + m_2x_2}{M}\) = ∑\(\frac{m_ix_i}{ M}\)

\(Y_{cm} =\frac{m_1y_1 + m_2y_2}{m_1 + m_2} = \frac{m_1y_1 + m_2y_2}{M} \)= ∑\(\frac{m_iy_i}{M}\)

Zcm = \(\frac{m_1z_1 + m_2z_2} {m_1 + m_2} = \frac{m_1z_1 + m_2z_2}{M} = ∑\frac{m_iz_i}{M}\)


For Large Number of Particles R_cm = ∑\(\frac{m_ir_i}{M}\)

3)For Continuous Bodies R_cm = \(\frac{∫r^2 dm}{M}\)

In the Coordinates X_cm = \(\frac {∫x^2 dm}{M}\)

Y_cm = \(\frac {∫ y^2 dm}{M}\)

Z_cm = \(\frac{∫ z^2 dm}{M}\)

Motion of Centre of Mass

1) V_cm = \(\frac{m_1v_1 + m_2v_2 + …………. + m_nv_n}{M}\) = ∑\(\frac{m_iv_i}{M}\)

2) a_cm = \(\frac{m_1a_1 + m_2a_2 + …………. + m_na_n}{M}\) = ∑\(\frac{m_ia_i}{M}\)

3) F_cm = \(F_1 + F_2 + …………… + F_n\) = ∑ \(F_i\)

If System contains more than one particle and Fext in a particular direction is zero then centre of mass is also zero even though some particles may move along the direction

Motion of center of mass is not affected by internal force because it always makes an action reaction pairs and their contribution to acceleration is zero

If m₁ and m₂ are placed by separation r and move towards each other by their mutual attraction force then 1) A_cm = zero but not individual velocities 2) V_cm = Zero

If larger number of particles are projected under the effect of gravity only in different directions then acceleration of center of mass is equal to the acceleration due to gravity irrespective of directions of projection of particles

The total momentum of a system of particles is equal to product of the total mass and velocity of its centre of mass i.e.. \(P = MV_{cm}\) and we can also define total linear momentum as the sum of linear momenta of individual particles is

\(P = P_1 + P_2 + P_3 + ………+ P_n\)

The relation between KE and Momentum is \(\sqrt{2mk}\)

Conservation of linear Momentum for System of particles :

If there is no external force the sum of total linear momentum of system remains constant or conserved

\(P = MV_{cm}\)

By differentiating P and \(M_{Vcm}\) we get \(\frac{dp}{dt}\) = \(F_{ext}\)

If \(F_{ext}\) = 0 \(\frac{dp}{dt}\) = 0 or P = 0

\(MV_{cm}\) = constant and \(V_{cm}\) = Constant

Collision

It is an isolated event in which two or more colliding bodies exert strong forces on each other for a relatively short time

In all types of collision Total linear momentum is conserved i.e.. initial and final momentum of the system i.e.. Total momentum before collision = Total momentum after collision \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\)

If there are absence of any dissipative forces the mechanical system of the system will also remain conserved i.e\(\frac{1}{2}mv_1^2\)+ \(\frac{1}{2}mv_2^2\)= \(\frac{1}{2}mu_1^2\)+ \(\frac{1}{2}mu_2^2\)

Types of collisions

1) Elastic and inelastic collisions

Elastic: Both Kinetic Energy and Momentum are conserved and constant

Inelastic: Only Momentum are conserved and constant and not Kinetic Energy

Perfectly Inelastic Collision: If the approaching particles permanently stick and move together with common velocity

Head on and oblique Collisions

Head on Collisions: The velocity vectors of colliding particles are directed along the line of impact

Oblique Collisions: The velocity vectors of colliding particles are not along the line of impact

Head on Elastic collisions

If two balls of masses m₁ and m₂ are moving with velocities u1 and u2 in same direction and after collisions the velocities become < v₁ and v₂then momentum and Kinetic Energy will become

\(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\)


\(\frac{1}{2}mv_1^2\)+ \(\frac{1}{2}mv_2^2\)= \(\frac{1}{2}mu_1^2\)+ \(\frac{1}{2}mu_2^2\)

Solving Above two equations we get

v₁ = \(\frac{(m_1-m_2)u_1}{m_1 + m_2}\) + \(\frac{2m_2u_2}{ m_1 + m_2}\)

v₂ = \(\frac{2m_1u_1}{m_1 + m_2}\) +\(\frac{(m_2-m_1)u_2}{m_1+m_2}\)

Cases

1) If m₁ = m₂ and subtitling in above equations we get

V₁ = U₂

V₂ = U₁

2) If m₁ >> m₂and subtitling in above equations we get

V₁ = 0

V₂ = -U₂


3) If m₁ <<<< m₂, u₁ = 0 and subtitling in above equations we get

V₁ = 2u₂

V₂ = u₂

Head on Inelastic collisions

If two balls of masses m₁ and m₂ are moving with velocities u₁ and u₂ in same direction and after collisions the velocities become Same (V)
(m₁u₁ + m₂u₂) = (m₁ + m₂)V

Loss in Kinetic Energy (ΔK)

\(\frac{1}{2}mu_1^2\)+ \(\frac{1}{2}mu_2^2\)= \(\frac{1}{2} (\frac{m_1m_2}{m_1+m_2}) (u_1^2-u_2^2\))
and by solving this we get

ΔK = \(\frac{1}{2}\) \(\frac{m_1m_2}{m_1+m_2}\) (V₁ ² – V₂ ²) and we see that there is loss in KE in inelastic collision

Newton Law of Restitution

When two bodies are in direct impact the speed with which they can separate after impact is usually less than or equal to speed of approach before impact

E = \(\frac{Velocity of separation}{Velocity of approach}\)

E = \(\frac{V_1 – V_2}{U_2 – U_1}\)

V₁ = \(\frac{(m_2-em_1)u_2}{(m_1+m_2)}\) +\(\frac{2m_2u_2}{m_1 + m_2}\)

V₂ = \(\frac{2m_1u_1}{m_1 + m_2}\) + \(\frac{(m_2-em_1)u_2}{ m_1 + m_2}\)

For Elastic Collision e =1 and for inelastic collision e = 0 and for perfectly inelastic collision e = 0

If M₂ >> M₁ then after collision velocity of M₂ does not change appreciably

If a ball is freely fallen and e is the coefficient of restitution b/w ball and ground then then

V_n = eⁿ V_o = eⁿ = \(\sqrt{2gh}\)

2)h_n = e²ⁿ h

3) H = h (\(\frac{1+e^2}{1-e^2}\))

4) T = \(\frac{1+e}{1-e}\) = \(\sqrt\frac{2h}{g}\)

Loss in KE in inelastic collision

\(\frac{1}{2}mu_1^2\)+\(\frac{1}{2}mu_2^2\)=\(\frac{1}{2}\)\(\frac{m_1m_2}{m_1+m_2}\)(V₁–V₂²)(1- e²)