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Physics 111: Lecture 4 Today’s Agenda

Physics 111: Lecture 4 Today’s Agenda. Recap of centripetal acceleration Newton’s 3 laws How and why do objects move? Dynamics. Review: Centripetal Acceleration. UCM results in acceleration: Magnitude : a = v 2 / R =   R Direction : - r (toward center of circle). ^. v =  R.

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Physics 111: Lecture 4 Today’s Agenda

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  1. Physics 111: Lecture 4Today’s Agenda • Recap of centripetal acceleration • Newton’s 3 laws • How and why do objects move? • Dynamics

  2. Review: Centripetal Acceleration • UCM results in acceleration: • Magnitude: a= v2 / R =R • Direction: - r (toward center of circle) ^ v =R Useful stuff: f = rotations / sec T = 1 / f ω = 2 / T = 2 f = rad/sec a R 

  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.)

  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 aninertial 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!

  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? • T = 1 day = 8.64 x 104 sec, • R ~ RE = 6.4 x 106 meters . • 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.

  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 thatobject, and is independent of external influences. • Force has units of [M]x[L / T2] = kg m/s2= N (Newton)

  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 FNET = ma F1 F1 FNET F2 F2

  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:

  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

  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

  11. Example: Pushing a Box on Ice... • Start with F= ma. • a= F / m. • Recall that v2 - v02 = 2a(x - x0 ) (Lecture 1) • So v2 = 2Fd / m v F m a i d

  12. Example: Pushing a Box on Ice... • Plug in F = 50 N, d = 10 m, m = 100 kg: • Find v = 3.2 m/s. v F m a i d

  13. Lecture 4, Act 1Force and acceleration • A force F acting on a mass m1results in an acceleration a1.The same force acting on a different mass m2 results in an acceleration a2= 2a1. m1 m2 F a1 F a2 = 2a1 • If m1 and m2 are glued together and the same force F acts on this combination, what is the resulting acceleration? m1 m2 F a = ? (a)2/3 a1(b)3/2 a1(c)3/4 a1

  14. but F/m = a a = 2/3 a1 Lecture 4, Act 1Force and acceleration • Since a2 = (1/2) a1 for the same applied force, m2 = (1/2)m1 ! • m1+ m2= 3m2 /2 m1 m2 F a = F/(m1+ m2) • So a = (2/3)F / m1 (a)2/3 a1(b)3/2 a1(c)3/4 a1

  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

  16. Contact forces: • Objects in contact exert forces. • Convention: Fa,bmeans “the force acting on adue to b”. • So Fhead,thumb means “the force on the head due to the thumb”. Fhead,thumb

  17. Action at a distance • Gravity:

  18. Gravitation(Courtesy of Newton) • Newton found that amoon/g= 0.000278 • and noticed that RE2 / R2= 0.000273 • This inspired him to propose the Universal Law of Gravitation:|FMm|= GMm / R2 amoon g R RE where G = 6.67 x 10 -11 m3 kg-1 s-2

  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: • The direction of F12 is attractive, and lies along the line connecting the centers of the masses. m1 m2 F12 F21 R12

  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. m Fg h M RE

  21. Gravity... Leaky Cup • Near the Earth’s surface...  =g • So |Fg|= mg=ma • a = g All objects accelerate with acceleration g, regardless of their mass! Where:

  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 Fg • The force that gravity exerts on any object is called its Weight W= 539 N

  23. Lecture 4, Act 2Force 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)0.75W (b)1.5 W(c)2.25 W E X

  24. Ratio of weights = ratio of forces: Lecture 4, Act 2Solution • The gravitational force on a person of mass m by another object (for instance a planet) having mass M is given by:

  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 R12

  26. Fm,w Fw,m Ff,m Fm,f Newton's Third Law... 2 Skateboards • FA ,B = - FB ,A. is true for contact forces as well:

  27. Example of Bad Thinking • Since Fm,b = -Fb,m, why isn’t Fnet = 0 and a = 0 ? Fm,b Fb,m a ?? ice

  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

  29. Lecture 4, Act 3Newton’s 3rd Law • Two blocks are stacked on the ground. How many action-reaction pairs of forces are present in this system? (a) 2 (b) 3 (c) 4 a b

  30. a a a a Fa,E Fb,a b Fa,b b b b Fb,E Fb,g Fg,b FE,b FE,a Lecture 4, Act 3Solution: (c) 4

  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 problemsChapter 4: # 3, 5, 7, 19

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