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# Variable mass system

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### Variable mass system

1. 1. Lecture 17: More on Center of Mass, and Variable-Mass Systems• A Note on Center of Mass Location: – The center of mass is of a solid object is not required to be within the volume of the material• Examples: – Ship: – Hollow shell: Center of Mass
2. 2. Applications of Center of Mass Motion• Some basketball players are said to “hang” in the air• How can that be, given the their center of mass must move as a projectile – that is, parabolically?• Consider how the player configures his body as he flies through the air
3. 3. • Mid-jump: Dunk: Center of Mass• The center-of-mass moves parabolically, but the distance between the center-of-mass and the ball varies throughout the jump (less in the middle, greatest at the end) – Ball appears to “hang”, or move in a straight line
4. 4. Another Application: High Jump• High-jumpers contort their bodies in a peculiar way when going over the bar:• This keeps the jumper’s center of mass below any part of his body – Means he might clear the bar even though his center of mass goes below it
5. 5. Variable-Mass Systems• So far, we’ve considered the motion of systems of particles with constant mass• Not too much of a restriction, since we know that mass is never created nor destroyed• However, in some cases it’s more convenient to draw our system boundary such that mass can leave (or enter) the system• A rocket is the best example – It expels gas at high velocity – since the rocket applies a force to the gas, the gas in turn applies a force to the rocket (Newton’s Third Law again!); this force propels the rocket forward – While we care about the motion of the rocket, we don’t care about how the gas moves after it’s exhausted
6. 6. • In other words, we want to draw our system boundary as:
7. 7. • At some time t, our system has mass M and is moving at velocity v • At a later time t + dt both the mass and velocity of the system have changed • Newton’s Second Law tells us that: dp Fext,net = dt • Here p is the momentum of everything that was within the system at time t – including the mass that was ejected during dt Velocity of ejected mass • So: p i = Mv pf = ( M + dM )( v + dv ) + u ( −dM )Note the sign: If rocket is ejecting mass, dm is a negative number!
8. 8. dp = pf − p i = Mv + Mdv + vdM + dvdM − udM − Mv Product of two small numbers – can be ignored!• So, our original equation becomes: Mdv + vdM − udM dv dM Fext,net = =M + ( v − u) dt dt dt dv dM =M − v rel dt dt• vrel is the velocity of the ejected mass with respect to the rocket
9. 9. • Consider the case where no external forces act on the rocket: dv dM M − v rel =0 dt dt dv dM Thrust of rocket M = + v rel dt dt• So the rocket accelerates even though no external forces act on it – However, momentum is conserved for the rocket + gas system as a whole• Our equation works equally well for cases in which a system is gaining mass – Sand being poured into a moving rail car, for example
10. 10. Example: Saturn V Rocket• The first stage of a Saturn V rocket (used to launch astronauts to the moon) burns 15 tons of fuel per second, and ejects the gasses at a velocity of 2700m/s. The rocket, when fully loaded, has a mass of 2.8 x 106 kg.• Can the rocket lift off the pad, and if so, what is its initial acceleration? T• The force diagram looks like: mg
11. 11. • The mass of the fuel ejected per second is: dM 1 dW 1 = = ⋅ 1.3 × 105 N/s dt g dt 9.8m/s2 = 1.3 × 104 kg/s = − ( −2.7 × 103 m/s )(1.3 × 104 kg/s ) dM T = − vrel dt = 3.5 × 107 N• The net force is the thrust minus the weight of the rocket, or: Fnet = T − mg = 3.5 × 107 N − 2.8 × 106 kg ⋅ 9.8m/s 2 = 7.6 × 106 N• So the rocket does lift off, with initial acceleration of: Fnet 7.6 × 106 N a= = = 2.7m/s2 M 2.8 × 106 kg