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Tutorial 12 Linear Momentum Angular Momentum. Zhengjian, XU Nov 26 th , 2008. Conservation of Linear momentum. Newton’s third law:. Central impact. Direct impact:. Coefficient of restitution. Oblique impact:. Question 16.76.
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Tutorial 12 Linear Momentum Angular Momentum Zhengjian, XU Nov 26th, 2008
Conservation of Linear momentum Newton’s third law:
Central impact Direct impact: Coefficient of restitution Oblique impact:
Question 16.76 • Two small balls, each of mass m = 0.12 kg, hang from strings of length L = 1 m. The left ball is released from rest with θ= 30o. As a result of the initial collision, the right ball swings through a maximum angle of 25o. Determine the coefficient of restitution.
Question 16.76 Since the right ball can reach a maximum angle of 25o, After impact, the velocity of B is: A The impact velocity of Ball A and B: B Recalling:
Augular Momentum • Principle of Angular Impulse and momentum Angular momentum:
Central-Force Motion • If F goes through the center: • In a polar coordinate system: • For central motion:
Question 16.87/16.88 • A satellite is in the elliptic earth orbit shown. Its velocity at perigee A is 8640 m/s. The radius of the earth is 6370 km. • (a) Use conservation of angular momentum to determine the magnitude of the satellite’s velocity at apogee C. • (b) Use conservation of energy to determine the magnitude of the velocity at C. • (c) To determine the magnitudes of the radial velocityvrand transverse velocity vθ. at B.
(a) : conservation of angular momentum to determine the magnitude of the satellite’s velocity at apogee C. For a central-force motion (b): Use conservation of energy to determine the magnitude of the velocity at C.
(c) To determine the magnitudes of the radial velocity vrand transverse velocity vθ.at B.
Question 16.95 • Two gravity research satellites (mA= 250 kg, mB= 50 kg)are tethered by a cable. The satellite and cable rotate with angular velocity ,ω0 = 0.25 revolution per minute. Ground controllers order satellite A to slowly unreel 6 m of additional cable. What is the angular velocity afterward?
rA rB A O B Solution Consider these two objects as a system, then the angular momentum of this system relative to its center of mass is conserative. The total angular momentum:
rA rB A O B Solution • After the string isunreeled to 18m The new angular velocity can be obtained by:
Mass flows • Use the conservation of momentum
Question 104 • A nozzle ejects a stream of water horizontally at 40 m/s with a mass flow rate of 30 kg/s, and the stream is deflected in the horizontal plane by a plate. Assume that the magnitude of the velocity of the water when it leaves the plate is approximately equal to the initial velocity. • Determine the force exerted on the plate by the stream in cases (a), (b), and (c).
Consider a time interval ∆t, the change of momentum: • The average force in ∆t:
Solution for Question 16.104 Case (a): Case (b): Case (c):
Question 16.110/16.111 • The rocket consists of a 1000-kg payload and a 9000-kg booster. Eighty percent of the booster’s mass is fuel, and its exhaust velocity is 1200 m/s. If the rocket starts from rest and external forces are neglected, what velocity will it attain? • The booster has two stages whose total mass is 9000 kg. Eighty percent of the mass of each stage is fuel, and the exhaust velocity of each stage is 1200 m/s. When the fuel of stage 1 is expended, it is discarded and the motor of stage 2 is ignited. Determine the velocity attained by the rocket if the masses of the stages are m1 = 6000 kg and m2 = 3000 kg.
Payload: Mp=1000kg; Booster: Mb=9000kg 80% of the booster is fuel; Mf=7200kg Exhaust velocity: 1200m/s • The rocket has just one booster After the fuel is exhausted:
Question(b) • The rocket has two booster Payload: Mp=1000kg; Booster 1: M1=6000kg; Booster 2: M2=3000kg 80% of the booster is fuel; Exhaust velocity: 1200m/s (1) After the first booster is discarded: (2) After the second booster is discarded: