| Tutorial On Centre of Mass
- Centre of Mass Tutorial & Sample Questions:
a.) Two particle system
- Centre of mass of a body or a system of particles is the point at which the whole mass of the system or body is supposed to be concentrated and moves as if the whole external force is applied at that point.
- The motion of centre of mass of a body represents the motion of the whole body.
- Position of centre of mass :
Two particles of masses m1and m2are separated by a distance 'd'. If x1and x2are the distances of their centre of mass from m1and m2, then
b.) The centre of mass of heavier and lighter mass system lies nearer to the heavier mass .
c.) Number of particles lying along x-axis.
Particles of masses m1, m2, m3------- are at distances x1x2x3------ from the origin, the distance of centre of mass from the origin.
d.) Number of particles lying in a plane
Particles of masses m1, m2, m3, --------- are lying in xy plane at positions (x1,y1), (x2,y2), (x3,y3) -------, then the position co-ordinates of their centre of mass.
e.) Particles distributed in space.
If (x1y1z1), (x2y2z2) - are the position co-ordinates of particles of masses m1, m2- the position co-ordinates of their centre of mass are
f.) In Vector notation.
If r1, r2, r3. . . are the position vectors of particles of masses m1, m2, m3...... then the position vector of their centre of mass is
g.) Relative to centre of mass :
, where m1, m2.... mr have position vectors relative to centre of mass i.e. Algebraic sum of the moments of masses of a system about its centre of mass is always zero.
a.) depends on shape of the body.
- Position of centre of mass of a body.
b.) depends on distribution of mass for a given shape of the body.
c.) coincides with geometric centre of the body if the body is in uniform gravitational field.
If ....... are the velocities of particles of masses
- There may or may not be any mass at centre of mass.
- Centre of mass may be within or outside the body.
- For symmetrical bodies with uniform distribution of mass it coincides with geometric centre.
- Velocity of centre of mass
m1, m2, m3, ......, mn the velocity of their centre of mass.
- i.e total momentum of the system is the product mass of the whole system and the velocity of the centre of mass.
- LAW OF CONSERVATION OF LINEAR MOMENTUM :
In the absence of net external force, the total linear momentum of a system remains constant. i.e. if = constant
Effect of Internal Forces- The linear momentum of particles remains constant under the influence of internal forces.
a.) The linear momentum is conserved in all types of collisions (elastic and inelastic)
where u1and u2and v1and v2are the velocities of two particles with masses m1and m2 before and after the collision.
b.) In the absence of external forces, the linear momenta of individual particles can change but the total linear momentum of the whole system remains constant.
c.) The law of conservation of linear momentum is based on the Newton's laws of motion. This is the fundamental law of nature and there is no exception to it.
d.) Examples of laws of conservation of linear momentum
i.) Motion of a Rocket
(mv)gases= - (MV)rocket
ii.) Firing of a bullet from a gun
iii.) Explosion of a shell fired from a cannon
iv.) Two masses m1and m2, attached to the two ends of a spring, when stretched in opposite directions and released, then the linear momentum of the system is conserved.
e.) This law is valid only for linear motion.
f.) Rocket propulsion, motion of jet aeroplane and sailing of a boat all depend upon the law of conservation of momentum.
- If two particles of masses m1and m2are moving with velocities 1and 2at right angles to each other, then the velocity of their centre of mass is given by
If , ........ are the accelerations of particles of masses m1, m2, m3.......mnthen the acceleration of their centre of mass is
- Acceleration of centre of mass.
m1+ m2+ m3+ .....+mn= M total mass of the system, then m1
a.) the centre of mass of a system is at rest if the centre of mass is initially at rest.
- Centre of mass can be accelerated only by a net external force.
- Internal forces cannot accelerate the centre of mass or change the state of centre of mass.
- In the absence of external forces,
b.) if the centre of mass of a system is moving with constant velocity, it continues to move with the same velocity.
a.) The acceleration of centre of mass before and immediately after explosion is acm= g downward.
- For a ring the centre of mass is its centre where there is no mass.
- For a circular disc the centre of mass is at its Geometric centre where there is mass.
- For a triangular plane lamina, the centre of mass is the point of intersection of the medians of the triangle.
- The centre of mass of an uniform square plate lies at the intersection of the diagonals.
- Out of a uniform circular disc of radius R, if a circular sheet of r is removed; the Centre of mass of remaining part shits by a distance . d is the distance of the centre of the smaller part from the original disc.
- Out of a uniform solid sphere of radius R, if a sphere of radius r is removed, the centre of mass of the remaining part, shifts by . d is the distance of the smaller sphere from the centre of the original sphere.
- When shell in flight explodes
b.) The centre of mass of all the fragments will continue to move along the same trajectory as long as all the fragments are still in space.
c.) If all the fragments reach the ground simultaneously, the centre of mass will complete the original trajectory.
d.) If some of the fragments reach the ground earlier than the other fragments, the acceleration of centre of mass changes and its tragectory will change.
a.) If the man walks a distance L on the boat, the boat is displaced in the opposite direction relative to shore or water by a distance
- When a person walks on a boat in still water, centre of mass of person, boat system is not displaced.
(m = mass of man, M = mass of boat)
b.) distance walked by the mass relative to shore or water is (L-x)
a.) in the above case Vcm and a cm= 0
- Two masses starting from rest move under mutual force of attraction towards each other, they meet at their centre of mass.
b.) If the two particles are m1and m2and their velocities are v1and v2, then m1v1= -m2v2
c.) If the two particles have accelerations a1and a2.
d.) If s1and s2are the distances travelled before they meet
i) The particles come closer before collision and after collision they either stick together or move away from each other.
- The event or the process in which two bodies, either coming in contact with each other or due to mutual interaction at a distance apart, affect each others motion (velocity, momentum, energy or the direction of motion) is defined as a collision between those two bodies. In short, the mutual interaction between two bodies or particles is defined as a collision.
ii) The particles need not come in contact with each other for a collision.
iii) The law of conservation of linear momentum is necessarily conserved in all types of collisions whereas the law of conservation of mechanical energy is not
(i) Elastic collision or perfect elastic collision
(ii) Semi elastic collision
(iii) Perfectly inelastic collision or plastic collision
i) One dimensional collision: The collision, in which the particles move along the same straight line before and after the collision, is defined as one dimensional collision.
ii) According to the law of conservation of kinetic energy
iii) According to the law of conservation of momentum
(i) The velocity of first body after collision
- Newton's law of elastic collision - The relative velocity of two particles before collision is equal to the negative of relative velocity after collision i.e
(v1- v2) = -(u1- u2)
- Important formulae and features for one dimensional elastic collision.
(ii) The velocity of second body after collision
(iii) If the body with mass m2 is initially at rest, and u2= 0 then and
iv) When a particle of mass m1moving with velocity u1collides with another particle with mass m2at rest and
v) m1= m2 then v1= 0 and v2= u1. Under this condition the first particle come to rest and the second particle moves with the velocity of first particle before collision. In this state there occurs maximum transfer of energy.
vi) If m1>> m2then v1= u1and v2= 2u1under this condition the velocity of first particle remains unchanged and velocity of second particle becomes double that of first.
vii) If m1<< m2then v1= -u1and v2= under this condition the second particle remains at rest while the first particle moves with the same velocity in the opposite direction
viii) When m1= m2= m but u20 then v1= u2 and v2= u1i.e the particles mutually exchange their velocities.
ix) Exchange of energy is maximum when m1= m2. This fact is utilised in atomic reactor in slowing down the neutrons. To slow down the neutrons these are made to collide with nuclei of almost similar mass. For this hydrogen nuclei are most appropriate.
x.) Target Particle at rest : If m2is at rest, before collision
xi) If m2is at rest and kinetic energy of m1before collision with m2is E. The kinetic energy of m1and m2 after collision is
xii) In the above case fraction of KE retained by m1is
Fraction of KE transferred by m1to m2is
i) The collision, in which the kinetic energy of the system decreases as a result of collision, is defined as inelastic collision
- One dimensional inelastic collision:
ii) According to law of conservation of momentum
iii) According to law of conservation of energy
Q = other forms of energy like heat energy, sound energy etc. Q0
iv) According to Newton's law of inelastic collision (v1- v2) = -e(u1- u2)
e = Coefficient of restitution
- Coefficient of Restitution (e) -
iii) e is dimensionless and carries no limit.
iv) Limits of e 0 < e < 1
v) For plastic bodies and for perfectly inelastic collision e = 0. e = 1 for perfect elastic collision.
vi) The value of e depends upon the material of colliding bodies.
- SEMI - ELASTIC COLLISIONS
- The velocity of first body after collision
and velocity of second body
- Loss of energy in inelastic collision
- In case of inelastic collision the body gets strained and its temperature changes
- If a body falls from a height h and strikes the ground level with velocity and rebounds with velocity v up to a height h1then the coefficient of restitution is given by
If the body rebounds again and again to heights h1, h2, h3.... then
Thus the total time taken by the body in coming to rest
- The total distance covered by the body for infinite number of collisions
- Time taken by the body in falling through height h is
Perfectly inelastic collision:
- For a semi elastic collision 0 < e < 1
i) The collision, in which the two particles stick together after the collision, is defined as the perfectly inelastic collision.
ii) For a perfectly inelastic collision e = 0
iii) According to law of conservation of momentum
iv) Loss in Kinetic energy of system = (u1-u2)2.
v) Loss of kinetic energy is maximum when the colliding particles have equal momentum in opposite directions.
38. In an explosion, linear momentum is conserved, but kinetic energy is not conserved. The kinetic energy of the system after explosion increases. The internal energy of the system is used for the above purpose.
39. If a stationary shell breaks into two fragments, they will move in opposite directions, with velocities in the inverse ratio of their masses.
40. In the above, the two fragments have the same magnitude of linear momentum.
41. In the above case, the Kinetic energy of the two fragments is inversely propotional to their masses.
42. If the two fragments have equal masses, the two fragments have equal speeds in opposite directions.
43. If a shell breaks into three fragments, the total momentum of two of the fragments must be equal and opposite to the momentum of the third fragment is
44. When a stationary shell explodes, its total momentum is zero, before or after explosion.
Recoil of Gun
45. If a stationary gun fires a bullet horizontally, the total momentum of the gun + bullet is zero before and after firing.
46. If M and are the mass of the gun and velocity of recoil of the gun, m and is the mass of the bullet and velocity of the bullet, then
M + m = 0 i.e. |MV| = |mv|
or magnitude of momentum of the gun is equal to magnitude of momentum of the bullet.
- The bullet has greater Kinetic energy than the gun.