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Physics 2211: Lecture 07 A Gallery of Forces Newton’s 2nd Law of Motion Newton’s 1st Law of Motion Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 1 Force A force is a push or a pull. agent A force is caused by an agent and acts on an object. More precisely, the object and agent INTERACT. A force is a vector. F1 Physics 2211 Spring 2005 Dr. Bill Holm object Lecture 07, Page 2 Review: Newton's Laws Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, Fnet F ma . Law 3: Forces occur in action-reaction pairs: Fab Fba where Fab is the force due to object a acting on object b. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 3 Weight The Earth is the agent (on this planet) The “weight force” pulls the box down (toward the center of the Earth) long-range force W Physics 2211 Spring 2005 Dr. Bill Holm no contact needed! Lecture 07, Page 4 Gravity What is the force of gravity exerted by the earth on a typical physics student? Typical student mass m = 55kg g = 9.8 m/s2. Fg = mg = (55 kg)x(9.8 m/s2 ) Fg = 539 N (weight) FES Fg mg FSE mg Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 5 Example Mass vs. Weight An astronaut on Earth kicks a bowling ball and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon with the same force. Ouch! His foot hurts... (1) (2) (3) more less the same The masses of both the bowling ball and the astronaut remain the same, so his foot will feel the same resistance and hurt the same as before. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 6 Example Mass vs. Weight Wow! However the weights of the bowling ball and the astronaut are less: W = mgMoon That’s light. gMoon < gEarth Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 7 The Spring Force The spring is the agent A compressed spring pushes on the object. Physics 2211 Spring 2005 Dr. Bill Holm A stretched spring pulls on the object. Lecture 07, Page 8 Springs Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position (linear restoring force). FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position FX = 0 x=0 Physics 2211 Spring 2005 Dr. Bill Holm x Lecture 07, Page 9 Springs Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position (linear restoring force). FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position FX = -kx > 0 x0 Physics 2211 Spring 2005 Dr. Bill Holm x Lecture 07, Page 10 Springs Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position (linear restoring force). FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality. relaxed position FX = - kx < 0 x x>0 Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 11 Tension The rope is the agent The rope force (tension) is a pull along the rope, away from the object Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 12 Tools: Ropes & Strings Ropes & strings can be used to pull from a distance. Tension (T) at a certain position in a rope is the magnitude of the force acting across a cross-section of the rope at that position. The force you would feel if you cut the rope and grabbed the ends. An action-reaction pair. T cut T T Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 13 Tools: Ropes & Strings Consider a horizontal segment of rope having mass m: Draw a free-body diagram (ignore gravity). m T1 a T2 Using Newton’s 2nd law (in x direction): FNET = T2 - T1 = ma So if m = 0 (i.e., the rope is light) then T1 = T2 Physics 2211 Spring 2005 Dr. Bill Holm x Lecture 07, Page 14 Tools: Ropes & Strings An ideal (massless) rope has constant tension along the rope. T T If a rope has mass, the tension can vary along the rope For example, a heavy rope hanging from the ceiling... T = Tg T=0 We will deal mostly with ideal massless ropes. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 15 Tools: Ropes & Strings The direction of the force provided by a rope is along the direction of the rope: T Since ay = 0 (box not moving), m T = mg mg Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 16 The Surface Contact Force (one force — two components) Component perpendicular Component parallel (normal) to surface to surface n f NORMAL FORCE—the surface pushes outward against the object. Physics 2211 Spring 2005 Dr. Bill Holm FRICTION FORCE pulls in a direction to (try to) prevent slipping of the object Lecture 07, Page 17 Surface Contact Force (microscopic view) FRICTION is caused by the making and breaking of molecular bonds between the two surfaces. The NORMAL force is caused by the molecular-scale repulsion between the object and the surface. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 18 There are two types of Friction Kinetic Friction Static Friction v fs fk FRICTION doesn’t prevent slipping of object along surface. Physics 2211 Spring 2005 Dr. Bill Holm FRICTION successfully prevents slipping of the object along surface Lecture 07, Page 19 Rolling friction | f r | r | N | coefficient of rolling friction Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 20 Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 21 Drag A force experienced by a body that moves through air. Experiment shows that to a good approximation, 1 2 D Av vˆ 4 Unit vector that points in the direction of the velocity of the body Cross sectional area of the body measured in m2 Square of the body's velocity measured in m2 /s2 . Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 22 Drag Force DRAG occurs when an object moves in a gas or liquid. Like friction, the force of drag always points opposite to the direction of motion The fluid is the agent. Skin Drag (like friction) Form Drag (rowboat) Drag is very small for most of the objects we will discuss. We will always neglect drag in our models unless we explicitly state otherwise. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 23 Throw a ball upward vertically Drag decreases as the ball slows down Drag adds to the weight Physics 2211 Springas 2005it rises Dr. Bill Holm Drag increases as the ball speeds up slows Drag opposes the weight as it falls Lecture 07, Page 24 Terminal velocity At the terminal velocity, the drag force balances the force of gravity. F y Dw0 1 4 AvT2 mg 0 4mg vT A Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 25 Drop a particle from rest (v = 0) As the speed (and thus drag) increases, the slope decreases vT Slope approaches zero as v approaches terminal velocity Without drag, v = - at Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 26 The dimensions of a 1500 kg car, as seen from the front, are 1.6 m wide by 1.4m high. At what speed does the drag force equal the force of rolling friction? Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 27 Empirical (Contact) Forces (examples) Linear Restoring Force: Friction Force: Fluid Force: F kx F N Fs sN; Fk k N F bv n n 1 (v small); n 2, (v large) Characterized by variable, experimentally determined “constants”: k, , b, etc. These forces are all electromagnetic in origin. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 28 Thrust THRUST is a contact force exerted on rockets and jets (and leaky balloons) by exhaust gases (the agent). Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 29 Electric and Magnetic Forces (Non-Contact) E E E q E E -q E E E Electric Field Physics 2211 Spring 2005 Dr. Bill Holm Magnetic Field Lecture 07, Page 30 Identifying Forces---The Skier Tension T 1. Identify SYSTEM Normal force N 3. Find contact points between SYSTEM & ENVIRONMENT— Name and label the CONTACT FORCES Friction force f 2. Draw PICTURE with a closed curve around the Weight W SYSTEM. (Everything outside the curve is the environment. Physics 2211 Spring 2005 Dr. Bill Holm 4. Name and label LONG-RANGE FORCES Lecture 07, Page 31 Newton’s 2nd Law of Motion Newton’s 1st Law of Motion Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 32 Newton’s 2nd Law the connection between force and motion Fnet a m Fnet ma The acceleration of an object a is proportional to the net force F net that acts upon it. The constant of proportionality is called the "mass". The unit of mass m is the kilogram (kg). Mass is an intrinsic property of matter. The unit of force is [F] = [m][a] = kg- m s2 = N (Newton) Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 33 Newton’s First Law An object moves with constant velocity v if and only if the total force F net acting on the object is zero. dynamics: a F net m a 0 kinematics: v v0 at v v0 Rest is just a special case where v0 0. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 34 Mechanical Equilibrium: (Fnet 0) Dynamic Equilibrium (object moving with constant velocity) Static Equilibrium (object at rest) N W Physics 2211 Spring 2005 Dr. Bill Holm An object moving in a straight line at constant Lecture 07, Page 35 velocity is in dynamic equilibrium Problem: Accelerometer A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g. a iˆ Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 36 Accelerometer Draw a free body diagram (FBD) for the mass: What are all of the forces acting? a T (string tension) y m x mg (gravitational force) Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 37 Accelerometer Resolve forces into components: Sum forces in each dimension separately: x : SFx = Tx = T sin = ma y: SFy = Ty - mg T = T cos - mg = 0 Tx iˆ Eliminate T : T sin = ma T cos = mg Physics 2211 Spring 2005 Dr. Bill Holm => a tan g Ty ˆj y x a mg mg ˆj Lecture 07, Page 38 Accelerometer tan a g Let’s put in some numbers: Say the car goes from 0 to 60 mph uniformly in 10 seconds: 60 mph = (60 x 0.45) m/s = 27 m/s. Acceleration a = Δv/Δt = 2.7 m/s2. So a/g = 2.7 / 9.8 = 0.28 . = arctan (a/g) = 15.6 deg a Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 39 Problem: Inclined plane A block of mass m slides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration a ? m a Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 40 Inclined plane Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only. y m a x Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 41 Inclined plane Draw a FBD. Resolve forces into components & sum forces in x and y directions separately: F F y : N - mg cos = may= 0 x : mg sin = max = ma a ĵ mg Physics 2211 Spring 2005 Dr. Bill Holm a = g sin N Assume forces are acting at “center of mass” of block. - mg cos N = mg cos y mg sin iˆ x Lecture 07, Page 42 Angles of an Inclined plane Lines are perpendicular, so the angles are the same! - mg cos ĵ mg Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 43 Vocabulary A Reference Frame is the (x,y,z) coordinate system you choose for making measurements. An Inertial Reference Frame is a frame where Newton’s Laws are valid. USE NEWTON’S LAWS TO TEST FOR INERTIAL REFERENCE FRAMES Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 44 Test for Inertial Reference Frame EXAMPLE: Airplane parked on runway A ball is placed on the floor of the plane; no net forces act on the ball. Observation : The ball's motion doesn't change (no acceleration) Conclusion : The plane is an inertial reference frame. Less obvious: a plane cruising at constant velocity is also an inertial reference frame! Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 45 Test for Inertial Reference Frame EXAMPLE: Airplane taking off. Ball placed on floor of plane; no net forces act on ball. Observation: The ball rolls back (it accelerates) Conclusion : Newton's 1st Law is violated. The accelerating plane is NOT an inertial frame. . Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 46 We live in a World of Approximations . . . Strictly speaking, an Inertial Reference Frame has zero acceleration with respect to the “distant stars”. The Earth accelerates a little (compared to the distant stars) due to its daily rotation and its yearly revolution around the Sun. Nevertheless, to a good approximation, the Earth is an Inertial Reference Frame. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 47 Is Atlanta a good IRF? Is Atlanta accelerating? YES! Atlanta is on the Earth. The Earth is rotating. What is the acceleration (centripetal) of Atlanta? 2 T = 1 day = 8.64 x 104 sec, 2 v 2 R 1 aatl R ~ RE = 6.4 x 106 meters . R T R Plug this in: aatl = 0.034 m/s2 ( ~ 1/300 g) Close enough to zero that we will ignore it. Atlanta is a pretty good IRF. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 48 Physics 2211: Lecture 12 2-D, 3-D Kinematics and Projectile Motion Independence of x and y components *Georgia Tech track and field example *Football example *Shoot the monkey Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 49 3-D Kinematics The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as: r xiˆ yjˆ zkˆ v vxiˆ v y ˆj vz kˆ a axiˆ a y ˆj az kˆ We have already seen the 1-D kinematics equations: x x(t ) Physics 2211 Spring 2005 Dr. Bill Holm dx v dt dv d 2 x a 2 dt dt Lecture 07, Page 50 3-D Kinematics For 3-D, we simply apply the 1-D equations to each of the component equations. x x(t ) vx ax dx dt d2x dt 2 y y( t ) vy ay dy dt vz d2y dt z z( t ) 2 az dz dt d2z dt 2 Which can be combined into the vector equations: r r t Physics 2211 Spring 2005 Dr. Bill Holm dr v dt dv d 2 r a 2 dt dt Lecture 07, Page 51 3-D Kinematics So for constant acceleration we can integrate to get: a constant v v0 at r r0 v0t 12 at 2 Aside: the “4th” kinematics equation can be written as v v0 2a r 2 Physics 2211 Spring 2005 Dr. Bill Holm 2 (more on this later) Lecture 07, Page 52 2-D Kinematics Most 3-D problems can be reduced to 2-D problems when acceleration is constant: Choose y axis to be along direction of acceleration Choose x axis to be along the “other” direction of motion Example: Throwing a baseball (neglecting air resistance) Acceleration is constant (gravity) Choose y axis up: ay = -g Choose x axis along the ground in the direction of the throw Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 53 Uniform Circular Motion Physics 2211 Spring 2005 Dr. Bill Holm What does it mean? How do we describe it? What can we learn about it? Lecture 07, Page 54 What is Uniform Circular Motion? v is constant Motion with Constant Radius R (Circular) y v (x,y) R Constant Speed (Uniform) Physics 2211 Spring 2005 Dr. Bill Holm x Lecture 07, Page 55 How can we describe Uniform Circular Motion? In general, one coordinate system is as good as any other: Cartesian: y » (x,y) [position] » (vx ,vy) [velocity] v Polar: (x,y) R » (R, ) [position] » (vR , ) [velocity] In uniform circular motion: R is constant (hence vR = 0). (angular velocity) is constant. Polar coordinates are a natural way to describe Uniform Circular Motion! Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 56 x Polar Coordinates Conversion from polar to Cartesian coordinates: y And back: R x R cos y R sin 1 (x,y) R x2 y 2 x sin 0 /2 arctan y x cos 3/2 2 -1 Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 57 Angular Motion The arc length s (distance along the circumference) is related to the angle in a simple way: s = R, where is the angular displacement. units of are called radians. y For one complete revolution: 2R = Rc c = 2 has a period 2. v (x,y) R s x 1 revolution = 2radians Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 58 Angular Motion In Cartesian coordinates, we say velocity vx = dx/dt. x = vx t (vx constant) In polar coordinates, angular velocity d/dt = . =t ( constant) y has units of radians/second. Displacement s = v t. but s = R = R t, so: v=R Physics 2211 Spring 2005 Dr. Bill Holm v (x,y) R t s x x Lecture 07, Page 59 Aside: Period and Frequency Recall that 1 revolution = 2 radians (a) frequency ( f ) = revolutions / second (b) angular velocity ( ) = radians / second By combining (a) and (b) (rad/s) = [2 rad/rev] x f (rev/s) = 2f Realize that: period (T) = seconds / revolution So T = 1 / f = 2/ y v (x,y) R s x x = 2 / T = 2f Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 60 Summary Relationship between Cartesian and Polar coordinates: x R cos y R sin R x2 y 2 arctan y x y Angular motion: displacement: velocity: constant velocity: Linear Angular Links s s R ds v dt s vt d dt t Physics 2211 Spring 2005 Dr. Bill Holm v R v (x,y) R t s x x Lecture 07, Page 61 Polar Unit Vectors We are familiar with the Cartesian unit vectors: iˆ, ˆj , kˆ Now introduce “polar unit-vectors” rˆ and ˆ : rˆ points in radial direction ˆ points in tangential direction (counterclockwise) ˆ rˆ y R ĵ iˆ Physics 2211 Spring 2005 Dr. Bill Holm x Lecture 07, Page 62 Acceleration in Uniform Circular Motion Even though the speed is constant, velocity is not constant since the direction is changing: acceleration is not zero! Consider average acceleration in time t v v2 R t Physics 2211 Spring 2005 Dr. Bill Holm v2 v1 v1 aav v / t Notice that v (hence v / t ) points at the origin! In the limit t 0 , a dv / dt and a points in the rˆ direction. Lecture 07, Page 63 Acceleration in Uniform Circular Motion This is called Centripetal Acceleration. Now let’s calculate the magnitude: v v2 R R v2 v1 Similar triangles: v1 v R v R But R = v t for small t v vt So: v R v v 2 t R R v2 a R Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 64 Centripetal Acceleration Uniform Circular Motion results in acceleration: Magnitude: a = v2 / R Direction: -^ r (toward center of circle) a R Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 65 Centripetal Acceration (in terms of ) We know that v2 a R v = R and Substituting for v we find that: R a 2 R a = 2R Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 66 Example: Uniform Circular Motion A fighter pilot flying in a circular turn will pass out if the centripetal acceleration he experiences is more than about 9 times the acceleration of gravity g. If his F18 is moving with a speed of 300 m/s, what is the approximate diameter of the tightest turn this pilot can make and survive to tell about it ? (1) 500 m (2) 1000 m (3) 2000 m Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 67 Example: Solution v2 a 9g R R 10000 R m 1000 m 9.81 2 2 90000 ms2 v 9g 9 9.81 sm2 D 2R 2000m 2km Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 68 Example: Propeller Tip The propeller on a stunt plane spins with frequency f = 3500 rpm. The length of each propeller blade is L = 80 cm. What centripetal acceleration does a point at the tip of a propeller blade feel? what is a here? f L Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 69 Example First calculate the angular velocity of the propeller: 1 rpm 1 rev x min 1 min 60 s x rad rad -1 2 0.105 0.105 s rev s so 3500 rpm means = 367 s-1 Now calculate the acceleration. a = 2R = (367s-1)2 x (0.8m) = 1.1 x 105 m/s2 = 11,000 g direction of acceleration points towards the propeller hub. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 70 Example: Newton & the Moon What is the acceleration of the Moon due to its motion around the Earth? What we know (Newton knew this also): T = 27.3 days = 2.36 x 106 s (period ~ 1 month) R = 3.84 x 108 m (distance to moon) RE = 6.35 x 106 m (radius of earth) R Physics 2211 Spring 2005 Dr. Bill Holm RE Lecture 07, Page 71 Moon Calculate angular velocity: 1 rev 1 day rad x x 2 2.66 x106 s -1 27.3 day 86400 s rev So = 2.66 x 10-6 s-1. Now calculate the acceleration. a = 2R = 0.00272 m/s2 = 0.000278 g direction of acceleration points is towards the center of the Earth. Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 72 Moon So we find that amoon / g = 0.000278 Newton noticed that RE2 / R2 = 0.000273 amoon g R RE This inspired him to propose that FMm 1 / R2 Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 73 Example Centripetal Acceleration The Space Shuttle is in Low Earth Orbit (LEO) about 300 km above the surface. The period of the orbit is about 91 min. What is the acceleration of an astronaut in the Shuttle in the reference frame of the Earth? (The radius of the Earth is 6.4 x 106 m.) (1) (2) (3) 0 m/s2 8.9 m/s2 9.8 m/s2 Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 74 Example Centripetal Acceleration First calculate the angular frequency : 1 rev 1 min rad x x 2 0.00115 s -1 91min 60 s rev Realize that: RO RO = RE + 300 km = 6.4 x 106 m + 0.3 x 106 m = 6.7 x 106 m 300 km RE Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 75 Example Centripetal Acceleration Now calculate the acceleration: a = 2R a = (0.00115 s-1)2 (6.7 x 106 m) RO a = 8.9 m/s2 300 km RE Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 76 Physics 2211: Lecture 16 Physics 2211 Spring 2005 Dr. Bill Holm Circular Orbits Fictitious forces Lecture 07, Page 77 SEEM to be very different, BUT they have the same free body diagram . . . Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 78 Orbital Motion is Projectile Motion! Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 79 Orbiting projectile is in free fall w mg (downward ) w mg (center ) 2 vorbit ar g r Fnet a g (center ) m Flat earth approximation Spherical Earth, near Earth h vorbit rg RE g Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 80 Fictitious forces in Non-Inertial Frames of Reference A car and driver move a constant speed. Suddenly, the driver brakes Outside observer: if the seat is frictionless, the driver continues forward at constant speed and collides with the front window. F Physics 2211 Spring 2005 Dr. Bill Holm Inside observer: a force F “throws” the driver against the front window. F is fictitious; the observer is in a noninertial (accelerated) frame where Newton’s laws do not apply. There is no true “push” or “pull” from Lecture anything. 07, Page 81 Fictitious forces in Non-Inertial Frames of Reference path of book (Newton’s 1st Law) ● Car turns ● Book continues on straight line ● In driver’s reference frame, apparent (fictitious) force moves book across the bench seat book v car frictionless bench seat Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 82 A car is passing over the top of a hill at (non-zero) speed v. At this instant, 1 2 3 4 n>w n=w n<w Can’t tell without knowing v Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 83 A ball on a string swings in a vertical circle. The 3 2 string breaks when the string is horizontal and the ball is moving straight up. Which trajectory 1 does the ball follow thereafter? Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 84 4 Roller-Coaster Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 85 Bottom: mv 2 Fr n w mar r mv 2 n w wapp r Top: mv 2 Fr n w mar r mv 2 n w wapp r Lose contact : n 0 vcritical Physics 2211 Spring 2005 Dr. Bill Holm rw rg m Lecture 07, Page 86 Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 87 Full Disclosure vtop vbottom and force is purely radial (centripetal). mg Except at the top and bottom, N N mg aT 0 since there is a non-zero N a component of the weight in the N a mg tangential direction. Since v is entirely tangential, the speed v must speed up or slow down on the sidewalls mg Physics 2211 Spring 2005 Dr. Bill Holm acceleration is not purely 07, tangential Page 88 radial or Lecture purely Example Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 89 Example Treat horizontal motion and vertical motion separately Then just add the results: Principle of Superposition Horizontal Motion: Velocity at highest point: ax dv x 0 dt v x constant v x v 0 x 10 m/s in +x direction Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 90 Example ay g v y v 0 y gt y y 0 v 0 y t 21 gt 2 Vertical Motion: Need to determine magnitude of take-off velocity: v 0 v 0 v 02x v 02y v 0y ? We know when y y max h, v y 0 v y v 0 y gt 0 v 0 y gth th v 0 y g y y 0 v 0 y t 21 gt 2 h 0 v 0 y th 21 gth2 v h 0 v 0 y goy 21 g OR, use v y2 v 02y 2g y . When y h, v y 0 Physics 2211 Spring 2005 Dr. Bill Holm v oy g 2 v 02y 2gh v 02y 2gh Lecture 07, Page 91 Example Magnitude of take-off velocity: v 0 v 0 v 02x v 02y v 02x 2gh 10.2 Physics 2211 Spring 2005 Dr. Bill Holm 10 ms 2 9.81 sm2 0.25 m 2 m s Lecture 07, Page 92 Example Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it? (1) D2 = 2D1 Physics 2211 Spring 2005 Dr. Bill Holm (2) D2 = 4D1 (3) D2 = 8D1 Lecture 07, Page 93 Example The horizontal distance a ball will go is simply x = (horizontal speed) x (time in air) = v0x t To figure out “time in air”, consider the y y 0 v 0 y t 21 gt 2 equation for the height of the ball: When the ball is caught, y = y0 v 0y t gt t 0 1 2 two solutions Physics 2211 Spring 2005 Dr. Bill Holm v 0 y t 21 gt 2 0 2v 0 y g t 0 (time of catch) (time of throw) Lecture 07, Page 94 Example So the time spent in the air is: tR 2v 0 y g The range, R, is thus : R x tR v 0 x tR v0 2v 0 xv 0 y g 2v 02 sin cos v 02 sin 2 g g v oy v 0 sin v ox v 0 cos Ball 2 will go 4 times as far as ball 1! Notice: For maximum range, sin 2 1 45o for max. range Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 95 Shooting the Monkey (tranquilizer gun) Where does the zookeeper aim if he wants to hit the monkey? ( He knows the monkey will let go as soon as he shoots ! ) Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 96 Shooting the Monkey If there were no gravity, simply aim at the monkey r r0 r v0t Physics 2211 Spring 2005 Dr. Bill Holm Lecture 07, Page 97 Shooting the Monkey With gravity, still aim at the monkey! 1 2 r v0t gt 2 Physics 2211 Spring 2005 Dr. Bill Holm 1 2 r r0 gt 2 Dart hits the monkey! Lecture 07, Page 98 Shooting the Monkey x = v0 t y = -1/2 g t2 This may be easier to think about. It’s exactly the same idea!! Physics 2211 Spring 2005 Dr. Bill Holm x = x0 y = -1/2 g t2 Lecture 07, Page 99