MIT16_07F09_Lec14

# MIT16_07F09_Lec14 - J Peraire S Widnall 16.07 Dynamics Fall...

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Unformatted text preview: J. Peraire, S. Widnall 16.07 Dynamics Fall 2008 Version 2.0 Lecture L14- Variable Mass Systems: The Rocket Equation In this lecture, we consider the problem in which the mass of the body changes during the motion, that is, m is a function of t , i.e. m ( t ). Although there are many cases for which this particular model is applicable, one of obvious importance to us are rockets. We shall see that a significant fraction of the mass of a rocket is the fuel, which is expelled during ﬂight at a high velocity and thus, provides the propulsive force for the rocket. As a simple model for this process, consider the cases sketched in a) through f) of the figure. In a) we consider 2 children standing on a stationary ﬂat car. At t = 0, they both jump off with velocity relative to the ﬂat car of u . (While this may not be a good model for children jumping off a ﬂatcar, it is a good model for the expulsion of mass from a rocket where the mass ﬂow from the choked nozzle occurs at a relative velocity u from the moving rocket.) In b) we consider that they jump off in sequence, each jumping with a velocity u relative to the then velocity of the ﬂatcar, which will be different for the 2nd jumper since the car has begun to move as a result of jumper 1. For a), we have a final velocity of the ﬂat car as 2 mu V 2 R = (1) M + 2 m for b) we have mu mu V 22 = + (2) M + 2 m M + m where M is the mass of the ﬂatcar, and m is the mass of the jumper. We can see the case b) gives a higher final velocity. We now consider cases with more blocks; we introduce the notation V NN to be the final velocity with N blocks coming off one at a time. We use the notation V NR to denote the reference velocity if the N blocks come off together. If we consider 3 blocks, we can see by inspection that 3 mu V 3 R = (3) M + 3 m and V 33 = mu M + 3 m + mu M + 2 m + mu M + m (4) While for 4 blocks V 4 R = 4 mu M + 4 m (5) and V 44 = mu M + 4 m + mu M + 3 m + mu M + 2 m + mu M + m (6) 1 Generalizing this to arbitrary N, we can write Nmu V NR = (7) M + Nm and N mu V NN = (8) M + im i =1 The figure shows the results for V 100 , 100 /V 100 R , the increase in velocity as each block comes off. We choose 100 blocks with a mass m equal to .01 M; therefore the 100 blocks with m = . 01 equal the mass M, the propellant ratio of the ”rocket” is 50%. The final velocity of the cart is 40% higher than would be obtained if the masses come off together. Momentum is conserved. The sum of the momentum of the cart plus all the individual mass points is equal to zero. However, the momentum of all the particles left behind is of very little interest to us. We care only about the final velocity of the cart....
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## This note was uploaded on 11/06/2011 for the course AERO 112 taught by Professor Widnall during the Fall '09 term at MIT.

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MIT16_07F09_Lec14 - J Peraire S Widnall 16.07 Dynamics Fall...

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