part2 - MM203 Mechanics of Machines: Part 2 2:1 Kinetics of...

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2:1 MM203 Mechanics of Machines: Part 2

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2:2 Kinetics of systems of particles Extension of basic principles to general systems of particles Particles with light links Rigid bodies Rigid bodies with flexible links Non-rigid bodies Masses of fluid
2:3 Newton’s second law G – centre of mass F i – external force, f i – internal force ρ i – position of m i relative to G O G r i _ r ρ i m i m i F 1 F 2 F 3 f 1 f 2 f 3

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2:4 Newton’s second law By definition For particle i Adding equations for all particles = = i i i i i m m m m r r r r i i m r f f f F F F = + + + + + + + 3 2 1 3 2 1 = + i i m r f F
2:5 Newton’s second law Differentiating w.r.t. time gives Also so ( principle of motion of the mass centre ) = i i m m r r = i i m m r r 0 = f a r F m m = =

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2:6 Newton’s second law Note that is the acceleration of the instantaneous mass centre – which may vary over time if body not rigid. Note that the sum of forces is in the same direction as the acceleration of the mass centre but does not necessarily pass through the mass centre . , etc a m F x x =
2:7 Example Three people ( A , 60 kg, B , 90 kg, and C , 80 kg) are in a boat which glides through the water with negligible resistance with a speed of 1 knot. If the people change position as shown in the second figure, find the position of the boat relative to where it would be if they had not moved. Does the sequence or timing of the change in positions affect the final result? (Answer: x = 0.0947 m ). (Problem 4/15, M&K) B C A C B A 1 knot 0.6 m 1.8 m 2.4 m x 1.8 m 1.2 m 1.2 m

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2:8 Example The 1650 kg car has its mass centre at G . Calculate the normal forces at A and B between the road and the front and rear pairs of wheels under the conditions of maximum acceleration. The mass of the wheels is small compared with the total mass of the car. The coefficient of static friction between the road and the rear driving wheels is 0.8. (Answer: N A = 6.85 kN, N B = 9.34 kN ). (Problem 6/5, M&K)
2:9 Work-energy Work-energy relationship for mass i is where ( U 1-2 ) i is the work done on m i during a period of motion by the external and internal forces acting on it. Kinetic energy of mass i is ( 29 i i T U = - 2 1 2 2 1 2 2 1 i i i i i m v m T r = =

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2:10 Work-energy For entire system ( 29 2 2 1 1 2 1 2 1 T U T or T U or T U i i = + = = - - -
2:11 Work-energy Note that no net work is done by internal forces. If changes in potential energy possible (gravitational and elastic) then as for single particle 2 2 2 2 1 1 1 1 e g e g V V T U V V T + + = + + + -

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2:12 Work-energy For system Now and note that so = 2 2 1 i i v m T i i ρ v v + = i i i v v v = 2 ( 29 ( 29 i i i i i i i i m m v m m T ρ v ρ ρ v ρ v + + = + + = 2 2 1 2 2 1 2 1
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part2 - MM203 Mechanics of Machines: Part 2 2:1 Kinetics of...

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