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C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 General Physics I (aka PHYS 2013) P ROF. VANCHURIN ( AKA V ITALY ) University of Minnesota, Duluth (aka UMD) R EVIEW C HAPTER 5 C HAPTER 6 O UTLINE C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 C HAPTER 7 C HAPTER 8 R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 5.1: U SING N EWTON ’ S F IRST L AW First Law. A body acted on by no net force, i.e. X ~Fi = 0 i has a constant velocity (which may be zero) and zero acceleration. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 5.1: U SING N EWTON ’ S F IRST L AW Example 5.1. A gymnast with mass mG = 50 kg suspends herself from the lower end of a hanging rope of negligible mass. The upper end of the rope is attached to the gymnasium ceiling. (a) What is the gymnasts’s weight? (b) What force (magnitude and direction) does the rope exert on her? (c) What is the tension at the top of the rope? R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 5.1: U SING N EWTON ’ S F IRST L AW Example 5.2. Find the tension at each end of rope in Example 5.1 if the weight of the rope is 120 N. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 5.1: U SING N EWTON ’ S F IRST L AW Example 5.4. A car of weight w rests on a slanted ramp attached to a trailer. (See figure below. Angle α is given. ) Only a cable running from the trailer to the car prevents the car from rolling off the ramp. (The car brakes are off and its transmission is neutral.) Find the tension in the cable and the force that the ramp exerts on the car’s tires. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 5.2: U SING N EWTON ’ S S ECOND L AW 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 mass of the body times the acceleration vector of the body equals to the net force vector, i.e. X ~Fi = m~a (1) i C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 5.2: U SING N EWTON ’ S S ECOND L AW Example 5.6. An iceboat is at rest on a frictionless horizontal surface. Due to the blowing wind, 4.0 s after the iceboat is released, it is moving to the right at 6.0 m/s. What constant horizontal force FW does the wind exert on the iceboat? The combined mass of iceboat and rider is 200 kg. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 5.2: U SING N EWTON ’ S S ECOND L AW Example 5.10. A toboggan loaded with physics students (total weight w) slides down a snow-covered hill that slopes at a constant angle α. The toboggan is well waxed, so there is virtually no friction. (a) What is its acceleration? (b) What is normal force? R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 5.3: F RICTION F ORCES Friction force. There two types of contact forces between macroscopic objects: ~n − normal force always perpendicular to the contact surface ~f − friction force is always parallel to the contact surface . Both forces arise due to microscopic (electromagnetic) interaction between molecules, but we shall only study their macroscopic properties. Kinetic Friction. If there is a motion along the surface of contact, then these two forces are related to each other by the so-called coefficient of kinetic friction fk µk = (2) n or fk = µk n. (3) Note that the friction coefficients are dimensionless, i.e. have no units. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 5.3: F RICTION F ORCES Static Friction If there is no motion along the surface of contact, then the friction force is bounded from above fs ≤ (fs )max = µs n. (4) It turns out that µs > µk and thus fk < fsmax . (5) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 5.3: F RICTION F ORCES Table of Friction Coefficients Exact values of both coefficients depend on the materials in constant, e.g. Material steel on steel ice on steel dry rubber on dry concrete wet rubber on wet concrete µs 0.74 0.03 1.0 0.3 µk 0.57 0.015 0.8 0.25 There is also rolling friction which is typically much smaller. For steel wheels on steel rails it is ∼ 0.002 − 0.003. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 5.3: F RICTION F ORCES Example 5.13. You want to move a 500 N crate across a level floor. To start the crate moving, you have to pull with a 230 N horizontal force. Once the crate starts to move, you can keep it moving at constant velocity with only 200 N force. What are the coefficient of static and kinetic friction? R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 5.3: F RICTION F ORCES Example 5.16. A toboggan loaded with physics students (from Example 5.10) slides down a snow-covered hill. The wax has worn off, so there is a nonzero coefficient of kinetic friction µk . The slope has just the right angle α to make the toboggan slide with constant velocity. Find the angle in terms of w and µk . C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 5.3: F RICTION F ORCES Fluid (air) resistance. The (magnitude of) force of fluid resistance depends on the velocity, ( kv for “small” velocities f = (6) 2 Dv for “large” velocities where the coefficients depend on: type of fluid, shape of object, etc. Terminal velocity. For small velocities the second law implies mg − (kv) = ma (7) This acceleration must stop when the terminal velocity is reached mg (8) vt = k Similarly in the regimes of large velocities the second law implies mg − (Dv2 ) = ma. (9) and thus the terminal velocity is r vt = mg . D (10) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 5.3: F RICTION F ORCES Example 5.18. For a human body falling through air in a spread-eagle position, the numerical value of the constant D in Eq. (5.6) is about 0.25 kg/m. Find the terminal speed for a 50 kg skydiver. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 5.4: D YNAMICS OF CIRCULAR MOTION In Sec. 3.4 we derived the following eqs. for uniform circular motion v2 R (11) 4π 2 R T2 (12) a⊥ = and a⊥ = where a⊥ − v − R − T − magnitude of acceleration constant speed radius of circular path period of motion. (13) An object in such a motion experiences a constant (in magnitude) acceleration and thus according to second law Fnet = ma⊥ = m v2 4π 2 R =m 2 . R T (14) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 5.4: D YNAMICS OF CIRCULAR MOTION Example 5.19. A sled with a mass of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00 m rope to a post set in the ice. Once given a push, the sled revolves uniformly in a circle around the post . If the sled makes five complete revolutions every minute, find the force F exerted on it by the rope. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.1: W ORK Constant speed Speed of an object is unchanged if I the sum of all forces is acting on it is zero. I the sum of all forces is acting in the direction orthogonal to the direction of motion. An important observation is that in both cases any function of speed would also not change, e.g. f (v(t)) ∝ v(t)2 = const. (15) Changing speed Speed of an object is changing on if the net force acting on the body has a component parallel to the direction of motion. Then for example f (v(t)) ∝ v(t)2 6= const. (16) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.1: W ORK Work Let us define a physical quantity, we shall call work , which will quantify whether there is a component of constant force ~F parallel to a straight-line displacement, ~s, W = ~F · ~s = Fs cos φ. (17) Although the work is done by each and every force acting on a body, according to above equation it may be positive, negative or even zero. Units of work [Work] = [Force] × [Distance]. (18) In SI units 1 J = 1 N×1 m. (19) Note that the dimensions of work can be also written as [Work] = [Mass] × [Acceleration] × [Distance] (20) [Work] = [Mass] × [Velocity]2 . (21) or C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 6.1: W ORK Example 6.2. A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance of 20 m along ground. The total weight of sled and load is 147000 N. The tractor exerts a constant 5000 N force at an angle of 36.9◦ above the horizontal. A 3500 N friction force opposes the sled’s motion. Find the work done by each force acting on the sled and the total work done by the forces. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.2: K INETIC E NERGY Back in second chapter we argued that 1D motion with constant acceleration is described by following equations 1 = x0 + v0x t + ax t2 2 v(t) = v0x + ax t x(t) a(t) = ax . (22) Then we derived a useful relation 2ax (x(t) − x0 ) = vx (t)2 − v20x (23) which can be written as max (x − x0 ) = 1 2 1 2 mv − mv . 2 x 2 0x (24) Note that this equation does not depend on time explicitly, but only through time-dependence of x and vx . C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.2: K INETIC E NERGY Kinetic energy. In 3D the equation can be written as m~a · ~s = 1 2 1 2 mv − mv0 2 2 (25) where the left hand side takes the form of the equation for work. What about the right hand side of (25)? If we define kinetic energy by K(v) = 1 2 mv 2 (26) then equation (25) tells us that the work acting on a system changes its kinetic energy by the amount equal to work W = K(t) − K(t0 ) = ∆K. (27) In fact what is relevant is the total work of all forces, i.e. Wtot = ∆K. (28) This is known as the work-energy theorem or as we will see later an equation representing conservation of energy. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 6.2: K INETIC E NERGY Example 6.3. Let’s come beck to Example 6.2. What is the speed of the sled after it moves 20 m? Suppose sled’s initial speed v1 = 2.0 m/s. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.2: K INETIC E NERGY Example 6.4. The 200 − kg steel hammerhead of a pile driver is lifted 3.00 m above the top of a vertical I-beam being driven into the ground. The hammerhead is then dropped, driving the I-beam 7.4 cm deeper into the ground. The vertical guide rails exert a constant 60 − N friction force on the hammerhead. Use the work-energy theorem to find (a) the speed of the hammerhead just as it heats the I-beam and (b) the average force the hammerhead exerts on the I-beam. Ignore effects of air. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.2: K INETIC E NERGY Example 6.5. Two iceboats hold a race on a frictionless horizontal lake. The two iceboats have masses m and 2m. The iceboats have identical sails, so the wind exerts the same constant force ~F on each iceboat. They start from rest and cross the finish line a distance s away. Which iceboat crosses the finish line with greater kinetic energy. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.3: VARYING F ORCES Up to now we were dealing with work done by constant forces W = Fx xf − xi . (29) This can generalized in two different ways: I we can consider a motion along a curve and/or I we can consider forces which change with position. In the latter case the motion can be broken into small segments W1 W2 = F1 ∆x = F2 ∆x ... Wn = Fn ∆x (30) and by adding all these small pieces of work we get n X i=1 Wi = n X i=1 Fi ∆x (31) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.3: VARYING F ORCES In the limit of infinite n this sum reduces to integral W ≡ lim n→∞ n X Z xf Fx (x)dx Fi ∆x = (32) xi i=1 where Fx (x) is x-component of ~F(x) which is a function of position. For example, force required to stretch an ideal spring is Fx (x) = kx. (33) where k is the so-called spring constant measured [k] = [Force] . [Distance] From (32) and (33) the work need to stretch an ideal spring is Z W= 0 X 1 2 Fx (x)dx = kx 2 X = 0 1 2 kX . 2 (34) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.3: VARYING F ORCES Work done on an ideal spring to stretch it X Z X 1 2 1 W= Fx (x)dx = kx = kX2 . 2 2 0 0 (35) Note that the position for non-stretched spring was set to x = 0, but if the origin is displaced then the equations would have to be modified Fx (x) W = k(x − x0 ) 1 2 = k (X − x0 ) . 2 (36) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 6.3: VARYING F ORCES Example 6.6. A woman weighing 600 N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring is compressed 1.0 cm under her weight. Find the force constant of the spring and the total work done on it during compression. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.3: VARYING F ORCES Example 6.7. An air-track glider of mass 0.100 kg is attached to the end of a horizontal air track by a spring with force constant 20.0 N/m. Initially the spring is unstretched and the glider is moving at 1.50 m/s to the right. Find the maximum distance d that the glider moves to the right (a) is the air track is turned on, so that there is no friction, and (b) if the air is turned off, so that there is a kinetic friction coefficient µk = 0.47 C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.4: P OWER I So far we did not pay attention to time it takes to do the work. I The physical quantity which represents the rate at which the work is done is called power. I This suggest that dimensions of power should be [Power] = I (37) In SI units power is measures in watts (W) which is defined as 1W = I [Work] . [Time] 1J 1s (38) Average power is defined from total work done in the same way as average velocity was defined from total displacement vavg = Pavg = ∆x ∆t ∆W . ∆t (39) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.4: P OWER Similarly the (instantaneous) power is defined as (instantaneous) velocity by taking limit of interval to zero v ≡ P ≡ ∆x dx = ∆t dt dW ∆W = . lim ∆t→0 ∆t dt lim ∆t→0 (40) Since work is given by W = ~F · ~s the power can also be expressed as ~F · ~s d dW d~F d~s P= = = · ~s + ~F · dt dt dt dt (41) (42) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 6.4: P OWER But for constant (time-independent) forces acting on a particle d~F =0 dt (43) the power is P= d~F d~s ~ · ~s + ~F · = F · ~v dt dt (44) d~s . dt (45) where ~v = is the velocity of particle. Example 6.9. Each of the four jet engines on an Airbus A380 airliner develops a thrust (a forward force on the airliner) of 322000 N (72000 lb). When the airplane is flying at 250 m/s (900 km/h, or roughly 560 mi/h) what is horsepower does each engine develop. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY Potential Energy I So far the only form of energy that we defined was kinetic energy which is a function of velocity K(v) = I 1 2 mv . 2 (46) It turns out that it is also useful to define another form of energy - potential energy - which is a function of position, e.g. U(x) = mgx or U(x) = 1 2 kx . 2 (47) I The idea is that it might be possible to store energy in the form of potential energy by placing an object at certain position. I The process of storing energy require work to be done on the object, but later on the potential energy can be released. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY Gravitational Potential Energy I The most familiar example of potential energy is the gravitational potential energy, when the work must be done on an object to lift it up (or down). I For example, if on object has mass m and is lifted from height x1 to height x2 , then the work done by gravitational force is Wgrav = −mg î (x2 − x1 ) î = mgx1 − mgx2 (48) I Then it is convenient to define a gravitational potential energy Ugrav ≡ mgx (49) and then the work done by gravitational force can be written as Wgrav = −∆Ugrav = Ugrav,1 − Ugrav,2 . (50) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY Gravitational Potential Energy I Now if we assume that there are no other forces acting on a body, then according to work-energy theorem Wgrav = Kf − Ki (51) and by combining (50) and (51) we get Ugrav,i − Ugrav,f = Kf − Ki (52) Ugrav,i + Ki = Ugrav,f + Kf . (53) or I The latter equation suggest that the total energy is unchanged E ≡ K + Ugrav = constant. I More generally there might be other forces acting on a given system and then the work energy theorem would imply Wother = Ef − Ei I (54) (55) Conservation of energy is the first of many conservation laws. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY Example 7.1. You throw a 0.145 − kg baseball straight up, giving it an initial velocity of magnitude 20.0 m/s. How high it goes? C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY Example 7.2. In Example 7.1 suppose your hand moves upward by 0.50 m while you are throwing the ball. The ball leaves your hand with an upward velocity of 20 m/s. (a) Find the magnitude of the force (assumed constant) that your hand exerts on the ball. (b) Find the speed of the ball at a point 15.0 m above the point where it leaves your hand. Ignore air resistance. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY Example 7.3. A batter hits two identical baseballs with the same initial speed and from the same initial height but at different initial angles. Prove that both balls have the same speed at any height h is air resistance can be neglected. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.1: G RAVITATIONAL P OTENTIAL E NERGY Example 7.6. We want to slide a 12 − kg crate up a 2.5 − m-long ramp inclined at 30◦ angle. A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5.0 m/s at the bottom and letting it go. But, friction is not negligible; the crate slides only 1.6 m up the ramp, stops, and slides back down. (a) Find the magnitude of friction acting on crate, assuming that it is constant. (b) How fast is the crate moving when it reaches the bottom of the ramp? C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.2: E LASTIC P OTENTIAL E NERGY I Remember that the work done on a spring is given by W= 1 2 1 2 kx − kxi 2 f 2 (56) but so the work done by a spring is Wel = 1 2 1 2 kx − kxf 2 i 2 (57) Similarly to work done by gravity, Wgrav = −∆Ugrav = mgyi − mgyf (58) the work done by spring can be expressed as Wel = −∆Uel = 1 2 1 2 kx − kxf 2 i 2 (59) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.2: E LASTIC P OTENTIAL E NERGY I The main difference is that Ugrav = mgy (60) but 1 2 kx 2 where the origin corresponds to unstretched spring. Uel = I (61) It is sometime useful to define the total potential energy 1 U = Ugrav + Uel = mgy + kx2 2 (62) and the total mechanical energy as E=U+K I (63) We can also modify the work-energy theorem further Wother = Ef − Ei . (64) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 7.2: E LASTIC P OTENTIAL E NERGY Example 7.7. A glider with mass m = 0.200 kg sits on a frictionless horizontal air track, connected to a spring with constant k = 5.00 N/m. You pull on the glider, stretching the spring 0.100 m, and releasing it from rest. The glider moves back towards its equilibrium position (x = 0). What is its x-velocity when x = 0.080 m? R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 7.2: E LASTIC P OTENTIAL E NERGY Example 7.9. A 2000 − kg (or 19600 − N) elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of the shaft. The spring is intended to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17000 − N frictional force to the elevator. What is the necessary force constant k for the spring. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 7.3: C ONSERVATIVE AND N ONCONSERVATIVE FORCES I We were able to define potential energy associated with work done by gravitational and elastic forces. All such forces are called conservative forces. I Work done by conservative forces: I can always be expressed as difference between initial and final values of a suitably defined potential energy I it is reversible and is independent on the trajectory of the body, but only on initial and final points I One might wonder if it is possible to do the same for all macroscopic forces which would allow to rewrite the work-energy theorem as a simple law of conservation of energy. I It turns out that there are other or non-conservative forces for which it is not possible to define potential energy. I For example, frictional force or air resistance forces are nonconservative. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.2: C ONSERVATIVE / N ONCONSERVATIVE Example 7.11. In a region of space the force of an electron is ~F = C x ĵ, where C is a positive constant. The electron moves around a square loop in the xy-plane. Calculate the work done on the electron by the force ~F during a counterclockwise trip around square. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.3: C ONSERVATIVE / N ONCONSERVATIVE I The non-conservative forces cannot be described in terms of mechanical potential energies, but one can still associate with the other energies such as internal energy. I For example the frictional force is non-conservative, but when friction is applied to objects in contact the internal properties of objects change. In particular friction leads to increase in temperature or internal energy. I On the microscopic level it corresponds to the change of kinetic energy of individual molecules. But from a macroscopic point of view one can think of work done by all non-conservative forces as a measure of change of internal energies Wother = −∆Uint (65) and the work energy theorem implies the conservation law ∆K + ∆U + ∆Uint = 0. (66) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.3: C ONSERVATIVE / N ONCONSERVATIVE I The main difference is that one cannot use internal energy to do any useful work. I This follows from the so-called second law of thermodynamics that we will see later in the course. I The conservation law of energy and the second law of thermodynamics had passed a very large number of tests, but this does not stop people from trying to build a perpetual motion machine. I Perpetual motion machines: I I first kind (do work without input of energy) second kind (do work using internal energy). C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.4: F ORCE AND POTENTIAL ENERGY I Work done on a particle when displacement is small W = Fx (x)∆x (67) which can also be expressed as a change in potential energy W = −∆U. (68) By equating (67) and (68) we get Fx (x)∆x = −∆U (69) or in the limit of small infinitesimal displacement Fx (x) = − lim ∆x→0 ∆U dU =− . ∆x dx (70) For example U= 1 2 kx 2 ⇒ Fx (x) = −kx (71) Fx (x) = −mg. (72) or U = mgx ⇒ C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.4: F ORCE AND POTENTIAL ENERGY I This can be easily generalized to 3D ~F = − ∂U(x, y, z) , ∂U(x, y, z) , ∂U(x, y, z) . ∂x ∂y ∂z (73) It is convenient to define a gradient operator (pronounced nabla) ~ ≡ ∂ , ∂ , ∂ (74) ∇ ∂x ∂y ∂z such that ~F = −∇U(x, ~ y, z). (75) For example the gravitational potential energy is U = mgy (76) and so ∂ ∂ ∂ ~F = −∇(mgy) ~ = − î + ĵ + k̂ (mgy) = −mgĵ. ∂x ∂y ∂z (77) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.4: F ORCE AND POTENTIAL ENERGY Example 7.14. A puck with coordinates x and y slides on a level, frictionless air-hockey table. It is acted on by a conservative force described by the potential-energy function U(x, y) = 1 k x2 + y2 . 2 Find a vector expression for the force acting on the puck, and find an expression for the magnitude of the force. (78) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.5: E NERGY D IAGRAM I Physical systems can be described in terms of energy diagrams. I For example harmonic oscillator whose potential energy is U= 1 2 kx 2 (79) and total energy is E = K + U. I (80) The main point is to study the possible solutions qualitatively without having to solve equations of motion quantitatively. C HAPTER 5 C HAPTER 6 C HAPTER 7 S ECTION 7.5: E NERGY D IAGRAM C HAPTER 8 R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 7.5: E NERGY D IAGRAM Which statement correctly describes what happens to particle when it is at the maximum (of red line) between x2 and x3 ? 1. The particle’s acceleration is zero. 2. The particle accelerates in the positive x-direction; the magnitude of the acceleration is less than at any other point between x2 and x3 . 3. The particle accelerates in the positive x-direction; the magnitude of the acceleration is greater than at any other point between x2 and x3 . 4. The particle accelerates in the negative x-direction; the magnitude of the acceleration is less at any other point between x2 and x3 . 5. The particle accelerates in the negative x-direction; the magnitude of the acceleration is greater than at any other point between x2 and x3 . R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 7.5: E NERGY D IAGRAM I The total energy was defined as a sum 2 dx 1 dx 1 E x, = m + kx2 . dt 2 dt 2 (81) I However, a more fundamental quantity (called Lagrangian) is 2 1 dx 1 dx = m − kx.2 (82) L x, dt 2 dt 2 I For any given trajectory x(t) Lagrangian L is a function of time and thus can be integrated over some interval, i.e. Z ti tf dx L x, dt dt. (83) I Clearly, for any trajectory x(t) the integral would be some real number and one might wonder what trajectory would produce the smallest (or largest) number? Answer: Classical trajectories. I Note, however, that all other trajectories are also possible! C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.1: M OMENTUM AND I MPULSE I Newton’s second law can be expressed as X ~Fi = d~p dt (84) i where the new quantity (known as momentum) is defined as: ~p ≡ m~v = m d~x . dt (85) I In context of the so-called Hamiltonian mechanics it is defined as ∂E x, dx dt ~p = . (86) ∂( dx dt ) I For example if d~x 2 ~ ∂E x , dt d~x 1 d~x 1 d~x E ~x, = m + kx2 ⇒ ~p = =m x dt 2 dt 2 dt ∂( d~ ) dt (87) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.1: M OMENTUM AND I MPULSE I The units of momentum are given by [Momentum] = [Mass] × [Velocity] (88) and for example in SI units 1 N · 1 s = 1 kg · m/s. I Alternatively one can write units of momentum as [Momentum] I (89) = [Mass] × [Velocity] = [Mass] × [Acceleration] × [Time] = [Force] × [Time] The latter form suggests that when some force is applied to a system for some time then, it may effect the momentum. (90) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.1: M OMENTUM AND I MPULSE I If we define impulse (of the time-independent net force) as X ~J = ~Fi ∆t (91) i then from (84) we get ~J = ~pf − ~pi . I When the net force is time dependent the impulse is defined as Z t X f ~J = ~Fi dt ti (92) (93) i and from (84) we get ~J = Z ti tf X i ~Fi dt = tf Z ti d~p dt = ~pf − ~pi dt I Note also that if the average force is defined as R tf P ~ i Fi dt ~Favg ≡ ti tf − ti (94) (95) then the impulse is just ~J = ~Favg (tf − ti ) (96) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.1: M OMENTUM AND I MPULSE I Similarly to the energy conservation which is fundamentally due to time-shift symmetry of physics laws, the momentum conservation is due to space-shift symmetry. I Conservation of energy expresses changes in time X ~Fi dt = ~pf − ~pi ~J = (97) i I Conservation of momentum expresses changes in space X ~Fi · d~l = Kf − Ki . W= (98) i I The fact that the two expressions look so much alike might be surprising at first but this is what led people to eventually discover a more fundamental and unified conserved quantity the energy-momentum tensor as well as other conserved quantities such as electric charge. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.1: M OMENTUM AND I MPULSE Example 8.1. We can now go back to the example 6.5 where we considered a race of two iceboats on a frictionless frozen lake. The boats have masses m and 2m, and the wind exerts the same constant horizontal force ~F on each boat. The boats start from rest and cross the finish like a distance s away. Which boat crosses the final line with greater momentum. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.1: M OMENTUM AND I MPULSE Example 8.2. You throw a ball with a mass of 0.40 kg against a brick wall. It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20 m/s. (a) Find the impulse of the net force on the ball during its collision with the wall. (b) If the ball is in contact with the wall for 0.010 s, find the average horizontal force that the wall exerts on the ball during impact. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.2: C ONSERVATION OF M OMENTUM I Similarly to the energy conservation which is fundamentally due to time-shift symmetry of physics laws, the momentum conservation is due to space-shift symmetry. I Conservation of energy expresses changes in time X ~Fi dt = ~pf − ~pi ~J = (99) i I Conservation of momentum expresses changes in space X ~Fi · d~l = Kf − Ki . W= (100) i I The fact that the two expressions look so much alike might be surprising at first but this is what led people to eventually discover a more fundamental and unified conserved quantity the energy-momentum tensor as well as other conserved quantities such as electric charge. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.2: C ONSERVATION OF M OMENTUM It is useful to distinguish two types of forces: I internal forces (forces exerted by objects inside the system) I external forces (forces exerted by objects outside of the system) When there are no external forces the systems is said to be closed or isolated. For isolated systems one can write down the impulse-momentum theorem for each object separately ~FA on B = ~FB on A = d~pB dt d~pA dt (101) but because of the Newton’s third law ~FA on B = −~FB on A (102) d~pB d~pA =− . dt dt (103) we get C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.2: C ONSERVATION OF M OMENTUM If we now define the total momentum of all particles as ~P = ~pA + ~pB (104) then we get the law of conservation of total momentum d~P = 0. dt This is true for an arbitrary collection of particles, i.e. ~P = ~pA + ~pB + ~pC + ... = mA~vA + mB~vB + mC~vC + ... (105) (106) given that there are no external (but only internal) forces. Note that since (105) is a vector equation we have three equatiions dPx dt dPy dt dPz dt = d(mA VAx +mB VBx +mC VCx +...) dt =0 = d(mA VAy +mB VBy +mC VCy +...) dt =0 = d(mA VAz +mB VBz +mC VCz +...) dt =0 (107) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 8.2: C ONSERVATION OF M OMENTUM Example 8.4. A marksman holds a rifle of mass mR = 3.00 kg loosely, so it can recoil freely. He fires a bullet of mass mB = 5.00 g horizontally with a velocity relative to the ground of vBx = 300 m/s. What is the recoil velocity vRx of the rifle? What are the final momentum and kinetic energy of the bullet and rifle? R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 8.2: C ONSERVATION OF M OMENTUM Example 8.6. Consider two batting robots on a frictionless surface. Robot A, with mass 20 kg, initially moves at 2.0 m/s parallel to the x-axis. It collides with robot B, which has mass 12 kg and is initially at rest. After the collision, robot A moves at 1.0 m/s in a direction that makes an angle α = 30◦ with its initial direction. What is the final velocity of robot B. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.3: I NELASTIC C OLLISIONS In this section we will discuss instantaneous events - we call collisions - which suddenly change the kinetic energies of objects. I If forces between colliding objects are conservative, then the total kinetic energy right before and right after collision is the same and the the collision is called elastic. ∆K = 0 I (108) If the forces are non-conservative, then the total kinetic energy is not conserved and the collision is called inelastic (or completely inelastic is objects stick together). ∆K 6= 0 (109) The key point is that although the (kinetic) energy might not be conserved, the momentum is still conserved for both elastic and inelastic collisions: ∆~P = 0 (110) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.3: I NELASTIC C OLLISIONS Consider a completely inelastic two-bodies collision, i.e. ~vA2 = ~vB2 = ~v2 (111) then conservation of momentum implies mA~vA1 + mB~vB1 = (mA + mB ) ~v2 (112) or mA~vA1 + mB~vB1 . (mA + mB ) For example if object B was originally at rest, then mA v2 = vA1 (mA + mB ) ~v2 = (113) (114) and then after the (completely inelastic) collision K2 mA = <1 K1 (mA + mB ) (115) or the kinetic energy after the collision is lower than before collision K2 < K1 . (116) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.3: I NELASTIC C OLLISIONS Example 8.8. Consider a ballistic pendulum, a simple system of measuring speed of a bullet. A bullet of mass mB makes a completely inelastic collision with a block of wood of mass mW , which is suspended like a pendulum. After impact, the block swings up to a maximum height y. In terms of y, mB , and mW , what is the initial speed v1 of the bullet. C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.3: I NELASTIC C OLLISIONS Example 8.9. A 1000 − kg car traveling north at 15 m/s collides with a 2000 − kg truck traveling east at 10 m/s. The occupants, wearing seat belts, are uninjured, but the two vehicles move away from the impact point as one. The insurance adjustor asks you to find the velocity of the wreckage just after the impact. What is your answer? C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.4: E LASTIC C OLLISIONS In elastic collisions the total kinetic energy before and after collisions is unchanged and so one can use both conservation of energy and momenta ∆K ∆~P = 0 = 0 (117) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 8.4: E LASTIC C OLLISIONS Example 8.12. Consider an elastic collision of two pucks (masses mA = 0.500 kg and mB = 0.300 kg) on a frictionless air-hockey table. Puck A has an initial velocity of 4.00 m/s in the positive x-direction and a final velocity of 2.00 m/s in an unknown direction α. Puck B is initially at rest. Find the final speed vB2 of puck B and the angles α and β. R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW S ECTION 8.5: C ENTER OF M ASS The conservation of momentum can be restated in terms of center of mass defined as the mass-weighted average position of particle P mi~ri m1~r1 + m2~r2 + m3~r3 + ... ~rcm ≡ = Pi . m1 + m2 + m3 + ... i mi Then the total momentum can be written as ~P = (m1 + m2 + m3 + ...) d~rcm dt or simply ~P = M~vcm where M= X mi i is the total mass and ~vcm = is the velocity of the center of mass. d~rcm dt (118) C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 8.5: C ENTER OF M ASS Example 8.14. James (mJ = 90.0 kg) and Ramon (mR = 60.0 kg) are 20 m apart on a frozen pond. Midway between them is a mug of their favorite beverage. They pull on the ends of a light rope stretched between them. When James has moved 6.0 m towards the mug, how far and in what direction has Ramon moved? R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 S ECTION 8.5: C ENTER OF M ASS If there are external forces acting on the objects, then the total momentum changes as X ~ ~Fext = dP = d (M~vcm ) = M~acm . dt dt Example. Will the center of mass continue on the same parabolic trajectory even after one of the fragments hits the ground? Why or why not? R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 C HAPTER 0: M ATHEMATICS I Algebra I I I I Trigonometry I I I I I Solving linear equations (numerically/algebraically) Solving quadratic equations (numerically/algebraically) Solving system of two equations with two unknowns (numerically/algebraically) Hypotenuse from two sides (numerically/algebraically) Side from hypotenuse and another side (numerically/algebraically) Angle from hypothenuse and one side (numerically/algebraically) Angle from two sides, etc. (numerically/algebraically) Calculus I I I I Differentiation of standard functions Product rule, quotient rule, chain rule. Indefinite integrals (i.e. antiderivitives) Definite integrals over interval R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 C HAPTER 5: A PPLYING N EWTON L AWS I I I I I I Coordinate system I Choosing 1D, 2D or 3D coordinate system I Drawing a free-body diagram for each object Using Newton’s First Law I Identifying equilibrium or motion without acceleration I Applying First Law to every object in equilibrium Using Newton’s Second Law I Identifying non-equilibrium or motion with acceleration I Applying Second Law to every object out of equilibrium Using Newton’s Third Law I Identifying action-reaction pairs for objects in contact I Applying Third Law to every action-reaction pair Frictional Force I Static and kinetic frictional forces and frictional coefficients I Relation between normal force and frictional forces Circular motion I Relation between radius, velocity and acceleration I Relation between radius, period and acceleration R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 C HAPTER 6: W ORK AND K INETIC E NERGY I Work I I I Kinetic Energy I I I Definition of kinetic energy Applying the work-energy theorem to every object Varying forces I I I I Work done by constant force for straight-line displacement Average acceleration and instantaneous acceleration Work done by varying forces and along straight line Work done by varying forces and along curved line Applying the work-energy theorem to varying forces Power I I Definition of average and instantaneous power Power from velocity of a particle and force acting on it R EVIEW C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 R EVIEW C HAPTER 7: P OTENTIAL E NERGY I Potential Energy I I I I I Conservative and Nonconservative forces I I I I Distinguishing conservative and non-conservative forces Applying energy conservation for only conservative forces Applying work-energy theorem with nonconservative force Force and potential energy I I I Definition of gravitational potential energy Definition of elastic potential energy Understanding conservation of total mechanical energy Applying the Work-energy theorem for every object Expressing force as a derivative of potential energy in 1D Expressing force as a gradient of potential energy in 3D Energy diagrams I I Drawing energy diagrams Determining stable/unstable equilibriums C HAPTER 5 C HAPTER 6 C HAPTER 7 C HAPTER 8 C HAPTER 8: M OMENTUM , I MPULSE , C OLLISIONS I Momentum and Impulse I Definition of momentum I Definition of impulse I Applying the impulse-momentum theorem to every object I Conservation of momentum I Conservation of the total momentum of all objects I Applying the conservation of total momentum I Inelastic collisions I Distinguishing inelastic and elastic collisions I Applying momentum conservation to inelastic collisions I Elastic collisions I Applying momentum conservation to elastic collisions I Applying energy conservation to elastic collisions I Center of mass I Definition of the center of mass I Applying second law for motion of center of mass R EVIEW