SYSTEMS OF PARTICLES

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Chapter 7
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Internal forces
External forces The weights of the particles
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In the absence of interaction among the
particles, the problem is rather simple.
One can solve the motion of each particle of the
system separately.
In the presence of interaction, the motion of the
system gets enormously complicated
With gravitational interaction, the motion of a
three-body system in unsolvable.
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Center of mass of a system of particles
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The three coordinates of the center of mass are
where, xi, yi zi are the coordinates of the ith
particle.
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Center of mass of a solid body
The body is divided into small masses ,
each of which is treated as a point particle.
So,
In the limit that each goes to zero,
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Similarly,
Thus,
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1. One dimensional body (along x-axis)
2. Two dimensional body (Lamina)
3. Three dimensional body (Solids)
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Prob. 7.6
Find the centre of mass of a homogeneous
semicircular plate of radius R.
Soln
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Overall Translational Motion of a System of
Particles
The Centre of Mass of a system of particles moves
according to Newtons law, as though the entire
mass of the system were concentrated at it and
the net external force were applied to it.
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Proof
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as, internal forces occur in pairs
and the sum of each pair is zero, from Newtons
third law.
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Principle of momentum conservation for a system
of particles
That is, the total linear momentum of a system of
particles is the total mass times the velocity of
center of mass.
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Differentiating w.r.t time,
That is, the rate of change of total momentum is
the net external force acting on the system
In the absence of external forces, the total
momentum of a system of particles is conserved.
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Double Integration in Cartesian Coordinates
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Summing along x-axis
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Integration along y-axis
y1
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D
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• Prob. 7.3
• A uniform flexible chain of length L, with mass
  per unit length ?, passes over a small,
  frictionless peg. It is released from a rest
  position with a length of chain x hanging from
  one side and a length L-x from the other side.
  Find the acceleration as a function of x.

b) Find the time taken by the chain to fall off
the peg
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y
x
Equations of motion
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Adding the two equations
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Integrating,
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Integrating,
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1.
2. L 5m, x0 3L/4
x0 2.51m, T 3.1s
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The net tangential force on the element is
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Equating this force to mass of the element times
the tangential acceleration,
Integrating from one end of the chain to the
other,
Where l is the length of the portion of the rope
that passes over the peg.
In the limit that l goes to zero, T1 T2
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Exercise
Show that the exact acceleration of the rope is
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SYSTEMS OF VARIABLE MASS 1. Rocket
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For the entire system S
Final momentum
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2. Conveyor Belt
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Rocket Motion
Dividing by M and integrating over time,
Or,
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Ex. 22
During a lunar mission, it is necessary to make a
mid-course correction of 22.6 m/s in the speed of
the spacecraft, which is moving at 388 m/s. The
exhaust speed of the rocket engine is 1230 m/s.
What fraction of the initial mass of the rocket
must be discarded as exhaust ?
Ans.
We have here,
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Prob. 12 A flexible in extensible string of
length L is threaded into a smooth tube. The tube
contains a right-angled bend and is positioned in
the vertical plane so that one arm is vertical
and the other is horizontal. Initially a length
y0 of the string hangs in the vertical arm. The
string is released and slides through the tube.
a) Show that in terms of the variable mass
problem vrel 0 so that the equation of motion
of the string is . b) Show that
is the solution
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Subsystem I
Observation Masses are added to each subsystem
with a relative velocity of zero
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Subsystem I
Subsystem II
Adding the above two equations
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b.
Moreover,
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c.
y(t)
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Prob. 11 Two long barges are floating in the
same direction in still water, one with speed of
9.65 km/hr and the other with a speed of 21.2
km/hr. While they are passing each other, coal is
shoveled from the slower to the faster barge at a
rate of 925 kg/min how much force must be
provided by the engine of each barge, if they
were to maintain their speeds.
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Equations of motion of the two barges are
Slower barge
Faster barge
Here, v1,rel, the relative velocity of shoveled
coal w.r.t. the slower barge, is zero, and v2,rel
is
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Moreover,
Since the speed of neither is to change,
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Falling Raindrop

Equation of Motion
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equating it to the given
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Claim The solution, with the condition that r
v 0 at t 0, is
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Full Solution
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Integrating both the sides,
Using the condition that at t 0, r v 0, we
get C 0.
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