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Lecture 11

Lecture 11. Goals Problem solving with Newton’s 1 st , 2 nd and 3 rd Laws Forces in circular motion Introduce concept of work Introduce dot product. Loop-the-loop 1.

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Lecture 11

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  1. Lecture 11 • Goals • Problem solving with Newton’s 1st, 2nd and 3rd Laws • Forces in circular motion • Introduce concept of work • Introduce dot product

  2. Loop-the-loop 1 The match box car is going to do a loop-the-loop. If the speed at the bottom is vB, what is the normal force, N, at that point? Hint: The car is constrained to the track. Fr = mar = mvB2/r = N - mg N = mvB2/r + mg N vB mg

  3. Another example of circular motionLoop-the-loop 2 A match box car is going to do a loop-the-loop of radius r. What must be its minimum speed vt at the top so that it can manage the loop successfully ?

  4. Loop-the-loop 2 vt mg To navigate the top of the circle its tangential velocity vt must be such that its centripetal acceleration at least equals the force due to gravity. At this point N, the normal force, goes to zero (just touching). Fr = mar = mg = mvt2/r vt = (gr)1/2

  5. Loop-the-loop 3 Once again the car is going to execute a loop-the-loop. What must be its minimum speedat the bottom so that it can make the loop successfully? This is a difficult problem to solve using just forces. We will skip it now and revisit it using energy considerations later on…

  6. Orbiting satellites Net Force: mar = mg = mvt2 / r gr = vt2 The only difference is that g is less because you are further from the Earth’s center!

  7. Example, Circular Motion Forces with Friction (mar = m vt 2 / rFf≤ms N ) • How fast can the race car go? (How fast can it round a corner with this radius of curvature?) mcar= 1600 kg mS = 0.5 for tire/road r = 80 m g = 10 m/s2 r

  8. Navigating a curve • Only one force is in the horizontal direction: static friction x-dir: Fr = mar = -m vt 2 / r = Fs= -ms N(at maximum) y-dir: ma = 0 = N – mg N = mg vt = (ms m g r / m )1/2 vt = (ms g r )1/2 = (0.5 x 10 x 80)1/2 vt = 20 m/s (or 45 mph) y N x Fs mg mcar= 1600 kg mS = 0.5 for tire/road r = 80 m g = 10 m/s2

  9. Banked Curves In the previous car scenario, we drew the following free body diagram for a race car going around a curve on a flat track. n Ff mg What differs on a banked curve?

  10. Banked Curves N mg Free Body Diagram for a banked curve. (rotated x-y coordinates) Resolve into components parallel and perpendicular to bank y x Ff No speed q

  11. Banked Curves y x mar N N Ff q mg mg mar Ff q Free Body Diagram for a banked curve. (rotated x-y coordinates) Resolve into components parallel and perpendicular to bank Low speed High speed

  12. Hanging Pink Fuzzy Dice Problem T mg • You are in a car going around a horizontal curve of radius 40.0 m and a speed of 10 m/s. There is a 0.10 kg pink fuzzy dice at the end of a 0.10 m string. What is the angle of the die with respect to vertical? (Little g is 10 m/s2) x-dir SFx = mar = -m vT2 / r = -Tx = -0.1 x 2.5= - 0.25 N y-dir SFy = may = 0 = Ty - mg  Ty = 0.1 x 10= 1.0 N tanq = -0.25  q = 14° But to you, if you are not paying attention, it may look like there is a mysterious force pushing the die out….a so-called centrifugal (center fleeing) or fictitious force. Observers in accelerated frames of reference do not see proper physics and so come to erroneous conclusions.

  13. Drag forces (forces that oppose motion) • Serway & Jewett describe three models: • The velocity dependent model give a general analytic solution. • Terminal velocity for velocity dependent drag forces occur where drag force equals applied force

  14. Drag at high velocities in a viscous medium • With a cross sectional area, A (in m2), D coefficient of drag (0.5 to 2.0),  density, and velocity, v(m/s), the drag force is: R = ½ D  Av2 Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m2 and drag coefficient 1 exerts a force of ~30 Newtons • At low speeds air drag is proportional to v but at high speeds it is v2 • Minimizing drag is often important

  15. Energy & Work • Work (Force over a distance) describes energy transfer • No “working” definition for energy….yet • Net forces result in acceleration • Velocity can change direction, magnitude or both • Only net force acting along the path changes the speed • Forces acting over a distance  work….energy changes • Forces acting over a time  impulse…..momentum changes

  16. Perpendicular Forces v Fc • I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle. • Is there a net force? • Does the force act over a distance? • What if there were a tangential force….what would happen? • Only parallel forces change speed

  17. Motion along a line Fy net = 0 Start Finish F F Fx mg x • The only net force is along the horizontal • A net force acting along the path, over a distance, induces changes in speed (magnitude of the velocity) • We call this action “work” • The vector “dot” product simplifies the notation

  18. Scalar Product (or Dot Product) A q Ay Ax î • Useful for finding parallel components A î= Ax î  î = 1 î j = 0 • Calculation can be made in terms of components. A B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz ) Calculation also in terms of magnitudes and relative angles. A B≡ | A | | B | cosq You choose the way that works best for you!

  19. Scalar Product (or Dot Product) Compare: A B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz ) with A as force F, B as displacement Dr Notice if force is constant: F Dr = (Fx )(Dx) + (Fy )(Dz ) + (Fz )(Dz) FxDx +FyDy + FzDz So here Fnet Dr≡ Wnet A Parallel Force acting Over a Distance does “Work”

  20. Units: Force x Distance = Work Newton x [M][L] / [T]2 Meter = Joule [L][M][L]2 / [T]2 mks cgs Other BTU = 1054 J calorie = 4.184 J foot-lb = 1.356 J eV = 1.6x10-19 J Dyne-cm or erg = 10-7 J N-m or Joule

  21. Circular Motion v • I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle. • How much work is done after the ball makes one full revolution? (A) W > 0 Fc (B) W = 0 (C) W < 0 (D) need more info

  22. Infinitesimal Work vs speed along a linear path Start Finish F F Fx x

  23. Work vs speed along a linear path • If F is constant • DK is defined to be the change in the kinetic energy

  24. Work in 3D…. • x, y and z with constant F:

  25. Examples of “Net” Work (Wnet) DK = Wnet • Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energy Examples of No “Net” Work • Pushing a box on a rough floor at constant speed • Driving at constant speed in a horizontal circle • Holding a book at constant height This last statement reflects what we call the “system” ( Dropping a book is more complicated because it involves changes in U and K, U is transferred to K. The answer depends on what we call the system. )

  26. Net Work: 1-D Example (constant force) Finish Start q = 0° F • Net Work is F x= 10 x 5 N m = 50 J • 1 Nm ≡ 1 Joule and this is a unit of energy • Work reflects energy transfer • A force F= 10 Npushes a box across a frictionless floor for a distance x= 5 m. x

  27. Net Work: 1-D 2nd Example (constant force) • Net Work is F x= -10 x 5 N m = -50 J • Work again reflects energy transfer • A forceF= 10 Nis opposite the motion of a box across a frictionless floor for a distance x = 5 m. Finish Start q = 180° F x

  28. Work: “2-D” Example (constant force) • (Net) Work is Fxx= F cos(-45°) x = 50 x 0.71 Nm = 35 J • Work reflects energy transfer • An angled force, F= 10 N,pushes a box across a frictionless floor for a distance x= 5 m and y= 0 m Finish Start F q = -45° Fx x

  29. Work and Varying Forces (1D) • Consider a varying force F(x) Area = Fx Dx F is increasing Here W = F·r becomes dW = Fdx Fx x Dx Finish Start F F q = 0° Dx Work has units of energy and is a scalar!

  30. Fini • Read all of Chapter 7

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