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Physics 111: Lecture 4 Today’s Agenda Recap of centripetal acceleration Newton’s 3 laws How and why do objects move? Dynamics Physics 111: Lecture 4, Pg 1 Review: Centripetal Acceleration UCM results in acceleration: Magnitude: a = v2 / R = R Direction: - r^ (toward center of circle) v=R Useful stuff: R f = rotations / sec T=1/f ω = 2 / T = 2 f = rad/sec a Physics 111: Lecture 4, Pg 2 Physics 111: Lecture 4, Pg 3 Dynamics Isaac Newton (1643 - 1727) published Principia Mathematica in 1687. In this work, he proposed three “laws” of motion: 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 pairs: FA ,B = - FB ,A (For every action there is an equal and opposite reaction.) Physics 111: Lecture 4, Pg 4 Newton’s First Law 1. Dishes 2. Monkey An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. If no forces act, there is no acceleration. The following statements can be thought of as the definition of inertial reference frames. An IRF is a reference frame that is not accelerating (or rotating) with respect to the “fixed stars”. If one IRF exists, infinitely many exist since they are related by any arbitrary constant velocity vector! Physics 111: Lecture 4, Pg 5 Is Urbana a good IRF? Ice puck Is Urbana accelerating? YES! Urbana is on the Earth. The Earth is rotating. What is the centripetal acceleration of Urbana? 4 sec, 2 T = 1 day = 8.64 x 10 2 v 2 aU 2R R R ~ RE = 6.4 x 106 meters . T R Plug this in: aU = .034 m/s2 ( ~ 1/300 g) Close enough to 0 that we will ignore it. Urbana is a pretty good IRF. Physics 111: Lecture 4, Pg 6 Newton’s Second Law For any object, FNET = F = ma. The acceleration a of an object is proportional to the net force FNET acting on it. The constant of proportionality is called “mass”, denoted m. » This is the definition of mass. » The mass of an object is a constant property of that object, and is independent of external influences. Force has units of [M]x[L / T2] = kg m/s2 = N (Newton) Physics 111: Lecture 4, Pg 7 Newton’s Second Law... What is a force? A Force is a push or a pull. A Force has magnitude & direction (vector). Adding forces is like adding vectors. a a F1 F1 FNET = ma FNET F2 F2 Physics 111: Lecture 4, Pg 8 Newton’s Second Law... Components of F = ma : FX = maX FY = maY FZ = maZ Suppose we know m and FX , we can solve for aX and apply the things we learned about kinematics over the last few weeks: 1 x = x0 + v 0 x t + ax t 2 2 v x = v 0 x + ax t Physics 111: Lecture 4, Pg 9 Example: Pushing a Box on Ice. A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10 m? v=0 F m a i Physics 111: Lecture 4, Pg 10 Example: Pushing a Box on Ice. A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the i direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ? v F m a i d Physics 111: Lecture 4, Pg 11 Example: Pushing a Box on Ice... Start with F = ma. a = F / m. Recall that v2 - v02 = 2a(x - x0 ) So v2 v = 2Fd / m (Lecture 1) 2 Fd m v F m a i d Physics 111: Lecture 4, Pg 12 Example: Pushing a Box on Ice... v 2 Fd m Plug in F = 50 N, d = 10 m, m = 100 kg: Find v = 3.2 m/s. v F m a i d Physics 111: Lecture 4, Pg 13 Lecture 4, Act 1 Force and acceleration A force F acting on a mass m1 results in an acceleration a1. The same force acting on a different mass m2 results in an acceleration a2 = 2a1. m1 F a1 F m2 a2 = 2a1 If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? F (a) 2/3 a1 m1 m2 (b) 3/2 a1 a=? (c) 3/4 a1 Physics 111: Lecture 4, Pg 14 Lecture 4, Act 1 Force and acceleration m1 F m2 a = F / (m1+ m2) Since a2 = (1/2) a1 for the same applied force, m2 = (1/2)m1 ! m1+ m2 = 3m2 /2 So a = (2/3)F / m1 but F/m = a a = 2/3 a1 (a) 2/3 a1 (b) 3/2 a1 (c) 3/4 a1 Physics 111: Lecture 4, Pg 15 Forces We will consider two kinds of forces: Contact force: » This is the most familiar kind. I push on the desk. The ground pushes on the chair... Action at a distance: » Gravity » Electricity Physics 111: Lecture 4, Pg 16 Contact forces: Objects in contact exert forces. Convention: Fa,b means “the force acting on a due to b”. So Fhead,thumb means “the force on the head due to the thumb”. Fhead,thumb Physics 111: Lecture 4, Pg 17 Action at a distance Gravity: Physics 111: Lecture 4, Pg 18 Gravitation (Courtesy of Newton) Newton found that amoon / g = 0.000278 and noticed that RE2 / R2 = 0.000273 amoon g R RE This inspired him to propose the Universal Law of Gravitation: |FMm |= GMm / R2 where G = 6.67 x 10 -11 m3 kg-1 s-2 Physics 111: Lecture 4, Pg 19 Gravity... The magnitude of the gravitational force F12 exerted on an object having mass m1 by another object having mass m2 a distance R12 away is: F12 G m1 m2 2 R12 The direction of F12 is attractive, and lies along the line connecting the centers of the masses. m1 F12 F21 m2 R12 Physics 111: Lecture 4, Pg 20 Gravity... Near the Earth’s surface: R12 = RE » Won’t change much if we stay near the Earth's surface. » i.e. since RE >> h, RE + h ~ RE. h m Fg Fg G ME m RE2 M RE Physics 111: Lecture 4, Pg 21 Gravity... Near the Earth’s surface... So |Fg| = mg = ma a=g Where: g G Leaky Cup ME ME m Fg G 2 m G 2 RE RE =g All objects accelerate with acceleration g, regardless of their mass! ME 2 9 . 81 m / s RE2 Physics 111: Lecture 4, Pg 22 Example gravity problem: 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 The force that gravity exerts on any object is called its Weight W = 539 N Fg Physics 111: Lecture 4, Pg 23 Lecture 4, Act 2 Force and acceleration Suppose you are standing on a bathroom scale in 141 Loomis and it says that your weight is W. What will the same scale say your weight is on the surface of the mysterious Planet X ? You are told that RX ~ 20 REarth and MX ~ 300 MEarth. (a) 0.75 W (b) 1.5 W (c) 2.25 W E X Physics 111: Lecture 4, Pg 24 Lecture 4, Act 2 Solution The gravitational force on a person of mass m by another object (for instance a planet) having mass M is given by: Ratio of weights = ratio of forces: MX m R X2 M m G E2 RE F G Mm R2 W X FX WE FE G M X ME R E RX 2 2 WX 1 300 .75 20 WE Physics 111: Lecture 4, Pg 25 Newton’s Third Law: Newton’s Sailboard Forces occur in pairs: FA ,B = - FB ,A. For every “action” there is an equal and opposite “reaction”. We have already seen this in the case of gravity: m1 m2 F12 F21 F12 G m1 m2 F21 2 R12 R12 Physics 111: Lecture 4, Pg 26 Newton's Third Law... 2 Skateboards FA ,B = - FB ,A. is true for contact forces as well: Fm,w Fw,m Ff,m Fm,f Physics 111: Lecture 4, Pg 27 Example of Bad Thinking Since Fm,b = -Fb,m, why isn’t Fnet = 0 and a = 0 ? Fm,b Fb,m a ?? ice Physics 111: Lecture 4, Pg 28 Example of Good Thinking Consider only the box as the system! Fon box = mabox = Fb,m Free Body Diagram (next time). Fm,b Fb,m abox ice Physics 111: Lecture 4, Pg 29 Lecture 4, Act 3 Newton’s 3rd Law Two blocks are stacked on the ground. How many action-reaction pairs of forces are present in this system? (a) 2 a (b) 3 b (c) 4 Physics 111: Lecture 4, Pg 30 Lecture 4, Act 3 Solution: a Fa,E b a b Fb,E a a b b Fb,a Fa,b Fb,g Fg,b FE,a FE,b (c) 4 Physics 111: Lecture 4, Pg 31 Recap of today’s lecture Extinguisher Cart Newton’s 3 Laws: (Text: 4-1 to 4-5) 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. (Text: 4-1) Law 2: For any object, FNET = F = ma (Text: 4-2 & 4-3) Law 3: Forces occur in pairs: FA ,B = - FB ,A. (Text: 4-4 & 4-5) Look at Textbook problems Chapter 4: # 3, 5, 7, 19 Physics 111: Lecture 4, Pg 32