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Chapter 4 & 5 Dynamics: Newton's Laws and Its Application “The discovery of the laws of dynamics, or the laws of motion, was a dramatic moment in the history of science... after Newton there was complete understanding. —— R. Feynman Isaac Newton (1643-1727) “Nature and Nature's laws Mathematician and lay hid in night; physicist, the greatest God said, Let Newton be! scientist of all time. And all was light.” 2 “If I have been able to see a little farther than other man, it’s because I have stood on the shoulder of giants.” —— Isaac Newton Work and energy Newton’s laws Momentum Angular momentum 3 Force (action capable of accelerating an object) F m G F G 4 Newton’s First Law of Motion Aristotle (384-322 B.C. ) Force is necessary to keep a body in motion Galileo Galilei (1564-1642) Idealized experiment with no friction No forces → constant speed in a straight line Net force F2 F1 F 0 5 Newton’s first law of motion: Every body continues in its state of rest or of uniform speed in a straight line as long as no net force acts on it. Inertia: tendency to keep one’s state of motion Inertial reference frames Noninertial reference frames How to distinguish? By checking to see if Newton’s first law holds. 6 Mass Mass is a measure of the inertia of a body. More mass ←→ harder to change its velocity Unit of mass: kilogram (kg) ◆ Mass and weight 7 Newton’s Second Law of Motion No forces → constant speed in a straight line F 0 velocity will change Newton’s second law of motion: The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object. 8 F a m F ma Component form in rectangular coordinates F x max , F y may , F z maz. Unit of force is Newton (N) 1N=1kg·m/s2 Limitations: Macro world, low speed motion, inertial frames 9 Newton’s Third Law of Motion Where do forces come from? Action ←→ reaction F1 F2 Newton’s third law of motion: Whenever an object exerts a force on a second object, the second object exerts an equal and opposite force on the first. ◆ Action force and reaction force are acting on different objects! 10 Weight and normal force Weight: the magnitude of gravity FG mg g = 9.80m/s2 Direction: down toward the center of Earth Contact force: ∥ Friction ⊥ Normal force FN FG 11 Weight loss Example1: Someone reads his weight 600N in an elevator when it accelerates at 0.2g to the ground, what is the real weight? Solution: FN =600N FG FN m 0.2 g a=g FN 0.8mg 5 FG mg FN 750 N 4 What about this acceleration? a=0.2g FN FG 12 Friction Why there is a friction? (P105, Figure 5-1) Kinetic friction F k N k : coefficient of kinetic friction Static friction F s N s : coefficient of static friction k and s depend on the surfaces: materiel, smoothness, lubrication, … 13 How to solve problems 1. Identify all the bodies 2. Draw a free-body diagram, show all the forces 3. Choose a convenient x-y coordinate system 4. Component equations of Newton’s second law 5. Solve all the equations ▲ Be careful about limitations of the formulas! 14 Tensions in rope Example2: An object (m=10kg) is hanged on a rope as the figure. What are the tensions in the rope? (Ignoring the mass of rope) 60°30° Solution: 1) free-body diagram 2) coordinate system 3) component equations: FT1 m y o x FT2 mg 3 mg FT 1 cos 60 FT 2 cos 30 0 FT 1 2 1 FT 1 sin 60 FT 2 sin 30 mg FT 2 mg 2 FT 1 =? 15 *Mass of rope & Catenary Consider a rope with uniformly distributed mass, what is the shape of a hanging rope? x y cosh( ) C a Catenary Search and read literatures if you are interested 16 Pulling a box Example3: Someone is pulling a box, where µ=0.75, h=1.5m. If he wants to move with constant velocity and exert a minimum force, L=? Solution: x : F cos F 0 F y : F sin FN mg 0 FN y o mg F cos sin dF 0 tan d L=h/sin =2.5m L Fµ FN m µ mg F x h 17 Broken Atwood’s machine Example4: Two masses connected by a rope and a pulley (Atwood’s machine). The connection part of m2 is broken and m2 is moving with constant acceleration a0 relative to the rope. What are the accelerations of m1 and m2 relative to the ground? (Ignoring the mass of rope and pulley) Solution: T T m1: m1g- T = m1a1 m2:m2g- T= m2 a2 m1 m2 ao a2G a2 R aRG m1g m2g a2 a0 a1 or a1 a2 a0 a1 m1 m2 a2 18 m1: m1g- T = m1a1 m2:m2 g- T = m2 a2 a2G a2 R aRG or:a0 = a1 + a2 Solve these equations: m 2 a 0 ( m1 m 2 ) g a1 m1 m 2 m1a 0 ( m 2 m1 ) g a2 m1 m 2 ( 2 g a 0 ) m1m 2 T m1 m 2 a1 m1 T m2 a2 T m1 m2 ao m1g m2g 19 Uniform circular motion Radial acceleration Net force aR v 2 / r 2 F ma mv /r R Centripetal/radial force v aR F r o Not some new kind of force It can be tension, weight, friction or the net force of these kind of forces. Determined by aR 20 Conical pendulum Example5: conical pendulum. a) What is the direction of acceleration? b) Calculate the speed and period. Solution: a) direction of acceleration 2 F sin mv /r b) horizontal: T vertical: FT cos mg 0 r L sin v Lg sin 2 , cos L FT m r O aR G 2 L sin T v 21 Rotating hoop Example6: A thin circular horizontal hoop of mass m and radius R rotates at frequency f. Determine the tension within the hoop. T Solution: Consider a tiny section of the hoop with an angle d, d m it has a mass dm d 2 speed v 2 f R R T v2 d d 2T Centripetal force F dm 2T sin 2 2 R Tension: T 2 mRf 2 22 Highway curves Where does centripetal force come from? Unbanked curves v Centripetal force: friction F Banked curves Centripetal force: N Net force of N and G F (and friction) G 23 Skating in a bowl Example7: Someone is skating in a bowl shape track with constant , determine the height h. (no friction) horizontal: Nsin =m2Rsin Solution: R vertical: N mg h Ncos =mg g cos 2 R g h R R cos R 2 2 g / R h 0? Limitation of model 24 Nonuniform circular motion 2 a v /r Radial acceleration R dv Tangential acceleration atan dt Total acceleration a Ftan FR 2 a tan a R2 atan Fnet aR a Components of the net force: 2 v dv FR maR m , Ftan matan m r dt 25 Example8: An object (mass m) slides down along a bowl from position A with no friction, find the radial and tangential acceleration of the object. 2 v Solution: Radial: N mg cos maR m R dv Tangential: mg sin matan m dt So: atan = gsin o A N mg R Conservation 1 2 mv mgRcos of energy 2 v2 aR 2 gcos R What is the normal force N ? 26 Velocity-dependent force Question: An object falls from rest, under the action of gravity and the air friction F =- v, what is the speed at time t, and when t→∞? dv Solution: mg v ma m dt Separation of variables dv dt v mg / m t mg mg m ln(v ) t C v C1e m t mg m Using initial conditions v (1 e ) t→∞? 27 Example9: An object moves along a semi-circular wall on smooth horizontal plane. Known: v0 and µ, determine its speed v at time t. Solution: v2 Radial force N m R dv Tangential force N m dt 2 dv v dv 2 dt v R R dt 1 v t 0 v0 t C, v R v0 R Speed: v R v0 t µ v0 v . o R 28 *Shooting problem A cannon ball is fired with v0 at angle , and the air friction F =- v, how to describe the motion, and what is the range? F dv x m Fx v x y dt dv y o x m Fy mg v y mg G dt vx vx 0 ebt g bt g v y v y 0 b e b where b m 29 vx 0 bt vx vx 0 ebt x 1 e b g bt g 1 y v y 0 g 1 ebt gt v y v y 0 b e b b b b vx 0 vy 0 300m/s; a) b 0.05s1 b) b 0.3s 1 How about F =-’ v2 ? 30