1. Classic Mechanics : Dynamics

2. Dynamics Newton’s laws Work and energy Momentum & angular momentum Momentum & angular momentum Newton’s laws Newton’s laws application Work and energy Work and energy Momentum and angular momentum Momentum and angular momentum Conservation of energy Conservation of energy Conservation of Momentum and angular momentum Conservation of Momentum and angular momentum

3. "Nature and Nature's laws lay hid in night; God said, Let Newton be! and all was light." Newton, Sir Isaac (1642-1727), mathematician and physicist, one of the foremost scientific intellects of all time.

4. a profound impact: astronomy, physics, and mathematics achievements: reflecting telescope three laws of motion; the law of universal gravitation the invention of calculus

5. Key terms: dynamics force mass superposition net force Newton’s law of motion Inertia equilibrium action-reaction pair elasticity tension friction force gravitational interaction free-body diagram Chapter4-5 Newton’s Laws of Motion

6. 1. Newton’s law of motion 1.1 Newton’s first law A body acted on by no net force moves with constant velocity and zero acceleration inertial force 1.2 Newton’s second law If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. The net force vector is equal to the mass of the body times the acceleration of the body. Chapter4-5 Newton’s Laws of Motion

7. If m is a constant, then: Caution: This is a vector equation used for a instant. We will use it in component form. (net force) In the natural coordinate axis: In the rectangular coordinate axis: m

8. 1.3 Newton’s third law Whenever two bodies interact, the two forces that they exert on each other are always equal in magnitude and opposite indirection. 1.4 Application area of Newton’s law low speed macroscopic practicality inertial frame

9. 2. What is a force? Forces are the interactions between two or more objects.

10. unified field theory:gravity and electromagnetic interaction Einstain : S.L.Glashow physicists Grand Unification Theory weak and electromagnetic interaction 2.1 Fundamental Forces:

11. x 2.2 the Common forces in mechanics 1) Gravitation o k: force constant x: elongation 2) elasticity 3) Frictional force Static friction Kinetic friction

12. 3. Applications of Newton’s law * Identify the body  Examine the force, draw a free body diagram  Construct a coordinate * Write the Newton’s law in component form * Calculate the equation Problem-solving strategy

13. Example: A wedge with mass M rests on a frictionless horizontal table top. A block with mass m is placed on the wedge, and a horizontal force F is applied to the wedge. What must be the magnitude of F if the block is to remain at a constant height above the table top? Solution: x y F M  m N mg For m: x:Nsin  =ma y:Ncos  =mg a=g.tg  For (M+m): F=(M+m)a =(M+m)g.tg 

14. a m m N mg FsFs Discussion:What must be the magnitude of a if the block does not slide down the wedge? Draw a free body diagram of m. horizontal: N=ma perpendicular:  N=mg So: a=g/  。 If a ,then N , then  N>mg, will the m go up?

15.  m R m N mg FsFs Discussion:What must be the  if the block does not slide down the cylinder? Draw a free body diagram of m. horizontal: N=mR  2 perpendicular:  N=mg

16. Example: A man pulls a box by a rope with constant speed along a straight line, known:  k =0.6, h=1.5m. Find how long the rope is when F=F min.  L m h F fkfk N mg Solution: horizontal ： Fcos  - f k =0 perpendicular ： Fsin  + N - mg=0 f k = µN So: tg  = µ 。 that is: when L=h/sin  =2.92m F=F min F=F min ：

17. Example: A parachute man drop into air, the resisting force is approximately proportional to the man’s speed v, find the velocity in any instantaneous time and the final velocity? Solution: o y Draw a free body diagram of the man, establish Newton’s law in y direction: Suppose: mg f

18. o y f final velocity: Can you describe the man’s motion? Suppose:

19. Example: Try to find the path of the particle according to the picture. x y m F=f 0 t i v0v0 O Solution:

20. x y m F=f 0 t i v0v0 O

21. Example: A small bead can slide without friction on a Circular hoop that is in a vertical plane and has a radius of R=0.1m. The hoop rotates at a constant rate of  =4.0rev/s about a vertical diameter. a) Find the angle  at which the bead is in vertical equilibrium. b) Is it possible for the bead to “ride” at the same elevation as the center of the loop? c) What will happen if the hoop rotates at 1.00rev/s p. 161, 5-103 Solution: a) For the bead normal ： Nsin  =m  2 Rsin  perpendicular ： Ncos  =mg cos  =g/ (  2 R)  =80.9 0  =4.0rev/s R=0.1m  N mg

22. b) N can not balance mg, so… c) When  =1.0rev/s=2  rad/s, cos  =2.5, so the bead stay at the bottom of the loop Example: A small bead can slide without friction on a Circular hoop that is in a vertical plane and has a radius of R=0.01m. The hoop rotates at a constant rate of  =4.0rev/s about a vertical diameter. a) Find the angle  at which the bead is in vertical equilibrium. b) Is it possible for the bead to “ride” at the same elevation as the center of the loop? c) What will happen if the hoop rotates at 1.00rev/s  =4.0rev/s R=0.1m  N mg

23. Example: A small bead at rest slide down a frictionless bowl of radius R from point A. Find N, a n, a t at this position.  R o A N mg  Solution: tangential ： mgsin  = ma t = m normal ： N-mgcos  = ma n = m so ： a t =gsin  f n =ma n, f t =ma t

24. 4. Noninertial Frame of reference B A a m k B: mass m is accelerating. A: mass m is at rest. Which one is right? The bus is not a inertial frame of reference. A frame of reference in which Newton’s first law is valid is called an inertial frame of reference. Any frame of reference will also be inertial if it moves relative to earth with constant velocity. Newton’s laws of motion become valid in non-inertial system by applying a inertial force on the object.

25. B A a m k Transposition: : real force : inertial force : acceleration of A relative to B Then:

26. Example: A wedge rests on the floor of a elevator. A block with mass m is placed on the wedge. There is no friction between the block and the wedge. The elevator is accelerating upward. The block slides along the wedge. Try to find the acceleration of the block with respect to the elevator. a  m a  mg ma N m a Solution: The elevator is the frame of reference, there are three forces acts on the block. m(g+a)sin  =ma a=(g+a)sin 

27. Example: A wedge with mass M rests on a frictionless Horizontal table top. A block with mass m is placed on The wedge. There is on friction between the block and the wedge. The system is released from rest. a) Calculate the acceleration of the wedge and the Horizontal and vertical components of the acceleration of the block. b) Do your answer to part (a) reduce to the correct results When M is very large? c) As seen by a stationary observer, what is the shape of the trajectory of the block? (see page 182, 5-108) M  m x y

28. Tracing problem Plane: x=x 0 +vt, y=h Missile: dY/dX=(y-Y)/(x-X) h v x y O u So: dY/dt=k(y-Y) dX/dt=k(x-X) And: k 2 [(y-Y) 2 +(x-X) 2 ]=u 2 k=u/[(y-Y) 2 +(x-X) 2 ] -1/2 So: Y(n+1)=Y(n)+k(y-Y)  t X(n+1)=X(n)+k(x-X)  t Y(0)=0, X(0)=0