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| Lami's Theorem |
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| If three concurrent forces acting on a body keep it in equilibrium, then each force is proportional to the sine of angle between the other two forces. |
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If three concurrent forces keep a body A in
equilibrium, the 3 forces must be coplanar. |
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| We know by Newton's laws that |
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| Consider the case of sand falling vertically on a horizontal conveyor belt moving at a velocity 'v'. |
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 The external force on the belt is in the direction of the velocity v. |
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| The mass of the belt is not considered as it is moving at constant velocity v. |
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| The power supplied by the external force F is |
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| Rocket propulsion |
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| The motion of a rocket is an application of Newton's third law of motion. A jet of hot gases is forced downwards (action) thereby pushing the rocket upward (reaction). |
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| Let the mass of the rocket be m at time t. |
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| Let a mass Dm of gas be ejected from the rocket in a very short interval of time Dt. |
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| Let |
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| vr = downward velocity of gases relative to the rocket |
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| v = velocity of rocket is upward direction |
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 velocity of gas relative to earth i.e., |
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| The mass of the rocket and unburnt fuel will decrease to (m-Dm) at the end of the interval Dt, and let the velocity increase v+Dv Momentum of rocket and unburnt fuel at time t + Dt = (m-Dm) (v + Dv) + Dm (v-vr). |
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| If air resistance is neglected, the external force F on the rocket is its weight -mg. [The upward direction is taken as positive.] |
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| We know that impulse = change in momentum |
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| The change in momentum in time Dt is the difference between the momentum of the system at the end and at the beginning of the time interval Dt. |
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| Therefore, -mgDt = mDv - DmDv - vDm |
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| The term Dm
Dv can be dropped. This is because it is a
product of two small quantities and thus is much smaller than the other
terms. Dropping this term, we obtain |
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 resultant force on the rocket = upward thrust on the rocket - weight of rocket |
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Thus, the resultant force equals the difference between the thrust and the weight of the rocket. The upward thrust is proportional both to the relative velocity (vr) of the ejected gas and the mass of the
gas ejected per unit time  |
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| Again, |
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As the rocket goes higher and higher, the value of 'g' continues to
decrease [Newton's law of gravitation]. The values of vr
and  remain practically constant while the fuel is being consumed but the remaining mass m continually decreases. So, acceleration continues to increase until all the fuel is burnt up. |
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| Equation (1) can be rewritten in terms of time t. Let mo be the mass of the rocket at t = 0 i.e., the initial mass of the rocket when no gas has escaped. Then the mass of the rocket at time t is given by |
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| If the effect of gravity is neglected i.e., if we consider the situation where the rocket moves in outer space (where no external forces are supposed to act), then |
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The left hand side is just the acceleration of the rocket. This should be equal to a force divided by a mass according to Newton's second law.
The force term is Vr ,
that is a constant velocity multplied by the rate of change in mass. The
mass in the denominator 
is a variable mass, since the rocket keeps ejecting mass in the form of exhaust gases. Even though the force is constant, the acceleration increases because of the decrease in mass. |
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| Let us now find a relation between the velocity at any time and the remaining mass. |
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| In equation (1), taking limit when Dt tends to zero, we get |
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| Here, dm is a positive quantity. It represents the mass ejected in time dt. So, the change in mass of the rocket in that time is '-dm'. Thus, for calculating the total mass change in the rocket, we must change the sign of the term containing dm. |
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| Let, at time t = 0, the mass and velocity be m0 and v0 respectively. |
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| Integrating equation (2) under necessary limits, we get |
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| Or |
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| Or |
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| Burn out speed of the rocket |
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| The speed acquired by the rocket when the whole of the fuel gets burnt is called the burn-out speed of the rocket. It is the maximum speed acquired by the rocket. When the rocket acquires burn-out speed vb, the mass of the rocket becomes equal to the mass of the empty container, say, mc. When the rocket acquires burn-out speed, it moves in outer space where the external forces can be neglected. |
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| From equation (3), |
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