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Tutorial 12 Linear Momentum Angular Momentum Zhengjian, XU Nov 26 th, 2008.

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Presentation on theme: "Tutorial 12 Linear Momentum Angular Momentum Zhengjian, XU Nov 26 th, 2008."— Presentation transcript:

1 Tutorial 12 Linear Momentum Angular Momentum Zhengjian, XU Nov 26 th, 2008

2 Conservation of Linear momentum Newton’s third law:

3 Central impact Direct impact: Oblique impact: Coefficient of restitution

4 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 = 30 o. As a result of the initial collision, the right ball swings through a maximum angle of 25 o. Determine the coefficient of restitution.

5 Question 16.76 Since the right ball can reach a maximum angle of 25 o, After impact, the velocity of B is: A B The impact velocity of Ball A and B: Recalling:

6 Augular Momentum Principle of Angular Impulse and momentum Angular momentum:

7 Central-Force Motion If F goes through the center: In a polar coordinate system: For central motion:

8 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 velocity v r and transverse velocity v θ. at B.

9 (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.

10 (c) To determine the magnitudes of the radial velocity v r and transverse velocity v θ. at B.

11 Question 16.95 Two gravity research satellites (m A = 250 kg, m B = 50 kg) are tethered by a cable. The satellite and cable rotate with angular velocity,ω 0 = 0.25 revolution per minute. Ground controllers order satellite A to slowly unreel 6 m of additional cable. What is the angular velocity afterward?

12 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: rArA rBrB O A B

13 Solution After the string isunreeled to 18m rArA rBrB O A B The new angular velocity can be obtained by:

14 Mass flows Use the conservation of momentum

15 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).

16 The average force in ∆t: Consider a time interval ∆t, the change of momentum:

17 Solution for Question 16.104 Case (a): Case (b): Case (c):

18 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.

19 The rocket has just one booster Payload: Mp=1000kg; Booster: Mb=9000kg 80% of the booster is fuel; Mf=7200kg Exhaust velocity: 1200m/s After the fuel is exhausted :

20 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 (2) After the second booster is discarded : (1) After the first booster is discarded :


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