PHY183-Lecture21pre - Fire Hose What is the magnitude of...

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1 February 22, 2012 Physics for Scientists&Engineers 1 1 Physics for Scientists & Engineers 1 Spring Semester 2012 Lecture 22 Rocket Motion and Calculating the Center of Mass Fire Hose February 22, 2012 Physics for Scientists&Engineers 1 2 Δ m = Δ V ρ = 360 L () 1 kg/L = 360 kg Δ p = v Δ m ( ) 39.0 m/s 360 kg 234 N 60 s pvm F tt ΔΔ == = = The weight of a fireman is 880 N What is the magnitude of the force that acts on a firefighter holding a fire hose that ejects 360 L of water per minute with a muzzle speed of 39.0 m/s? SOLUTION The total mass of water that is being ejected per minute is The change in momentum of the water is The force is then February 22, 2012 Physics for Scientists&Engineers 1 3 Rocket Motion Here we consider the case where the object in motion changes mass For example, the space shuttle in flight burning its fuel A rocket works by ejecting material A rocket does not “push against” anything This material is usually the result of a chemical reaction NY Times Did Not Believe It Jan. 13, 1920, Topics of the Times Editorial "That Professor Goddard, with his 'chair' in Clark College and the countenancing of the Smithsonian Institution, does not know the relation of action to reaction, and of the need to have something better than a vacuum against which to react -- to say that would be absurd. Of course he only seems to lack the knowledge ladled out daily in high schools." July 17, 1969, NY Times “Further investigation and experimentation have confirmed the findings of Isaac Newton in the 17th century, and it is now definitely established that a rocket can function in a vacuum as well as in an atmosphere. The Times regrets the error.” February 22, 2012 Physics for Scientists&Engineers 1 4 February 22, 2012 Physics for Scientists&Engineers 1 5 Rocket Motion Let’s imagine a spaceship that powers itself by shooting cannonballs out its nozzle Each cannonball has mass Δ m and the initial mass of the rocket including all cannonballs is 0 Each cannonball is fired with velocity v c relative to the rocket producing a cannonball momentum of c Δ February 22, 2012 Physics for Scientists&Engineers 1 6 After firing the first cannonball, the mass of the rocket is reduced to 0 - Δ Firing the cannonball does not change the center of mass of the system (rocket plus cannonball) The momentum of the cannonball is The rocket then receives a recoil momentum opposite to that of the cannonball so that the momentum of the rocket is Momentum is conserved so Rocket Motion p r = m 0 −Δ m v 1 p c = v c Δ m p r + p c = 0 m 0 m v 1 + v c Δ m = 0
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2 Rocket Motion We define the change in velocity of the rocket after firing one cannonball to be The recoil velocity of the rocket after firing one cannonball is Now we fire a second cannonball The mass of the rocket is reduced from m 0 Δ to 0 –2 Δ
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PHY183-Lecture21pre - Fire Hose What is the magnitude of...

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