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Lecture 17: More on Center of Mass, and Variable-Mass Systems

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Lecture 17: More on Center of Mass, and Variable-Mass Systems 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 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 • 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 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 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 • In other words, we want to draw our system boundary as: • 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 F = ext,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: = pi Mv = ( + )( + ) + (− ) pf M dM v dv u dM Note the sign: If rocket is ejecting mass, dm is a negative number! = − = + + + − − dp pf pi Mv Mdv vdM dvdM udM Mv Product of two small numbers – can be ignored! • So, our original equation becomes: Mdv + vdM − udM dv dM F = = M + (v − u) ext,net dt dt dt dv dM = M − v dt rel dt • vrel is the velocity of the ejected mass with respect to the rocket • Consider the case where no external forces act on the rocket: dv dM M − v = 0 dt rel dt dv dM Thrust of rocket M = +v dt rel 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 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 • 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 dM T = −v = −(−2.7 ×103 m/s)(1.3×104 kg/s) rel dt = 3.5×107 N • The net force is the thrust minus the weight of the rocket, or: = − = × 7 − × 6 ⋅ 2 Fnet T mg 3.5 10 N 2.8 10 kg 9.8m/s = 7.6 ×106 N • So the rocket does lift off, with initial acceleration of: F 7.6 ×106 N a = net = = 2.7m/s2 M 2.8×106 kg.
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