Rutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 7 JJ II. Home Page. Title Page.

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1 Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics Fall 2015 Lecture 7 Page 1 of 31

2 Midterm 1: Monday October 10th Motion in one, two and three dimensions Forces and Motion No energy and work. 1:55-2:50pm in the Physics Lecture Hall Page 2 of 31

3 Gradebook Page 3 of 31

4 Friction v 0 6. Forces and Motion Suppose an object is launched with velocity ~v 0 along a surface. Newton s 1st law: if ~F net =0it should move forever with constant velocity ~v 0. Does it really happen in practice? Page 4 of 31

5 Experiment: a toy truck is quickly accelerated to some velocity ~v 0 and then released. Page 5 of 31 t stops in finite time; x vs time parabola v x vs time linear once it starts deccelerating ) constant negative acceleration

6 What force is necessary to set a static object in motion and then maintain constant velocity? Page 6 of 31 Static n motion

7 Page 7 of 31

8 Static friction ~ f s acts when no relative motion between object and contact surface. The magnitude f s increases as the applied force increases until it reaches a maximum value (f s ) max. static dynamic Kinetic friction ~ f k acts when there is relative motion between object and contact surface. Usually f k < (f s ) max. Page 8 of 31

9 F > f k F = ( f s ) > f max k Static Dynamic Transition F = f s F= f k Page 9 of 31

10 Properties of friction 1. For a stationary object ~ F + ~ f s =0i.e. the applied force and the static friction balance each other out. 2. The magnitude of static friction has a maximum value depending on the magnitude of the normal force F N : (f s ) max = µ s F N where µ s = coe cient of static friction. When F (f s ) max the object begins to slide. 3. When sliding begins, the magnitude of friction rapidly decreases to a value f k = µ k F N, µ k = coe cient of kinetic friction Page 10 of 31

11 i-clicker An object is placed on an inclined plane of angle. The coe cient of static friction between at the contact surface is µ s. For which values of will the object stay on the plane. q F g A) For all values of. B) There is no such value of. C) For tan apple µ s. D) For sin apple µ s. E) For tan >µ s. Page 11 of 31

12 Answer An object is placed on an inclined plane of angle. The coe cient of static friction between at the contact surface is µ s. For which values of will the object stay on the plane. q F g A) For all values of. B) There is no such value of. C) For tan apple µ s. D) For sin apple µ s. E) For tan >µ s. Page 12 of 31

13 y q FN F g f s F gx F N F g q fs F gy x The object will stay on the plane if Page 13 of 31 F gx = f s F gx = mgsin f s apple (f s ) max = µ s F N F N = F gy = mgcos mg sin apple µ s mg cos ) tan apple µ s.

14 Example: emergency braking Page 14 of 31 A car slides 290 m on pavement with wheels locked. Assuming µ k =0.6 and constant acceleration, what is v 0?

15 Page 15 of 31 Newton s 2nd law: ~F net = m~a = ma x bi (no vertical motion) x axis : ma x = µ k F N y axis : 0 = F N mg

16 x axis : ma x = µ k F N y axis : 0 = F N mg a x = µ k g Constant acceleration model: v 2 = v0 2 2µ k g(x x 0 ) v =0 ) v0 2 =2µ k g(x x 0 )=58m/s =210km/h. Page 16 of 31

17 Example: A horizontal tension force ~ T is appplied to an object of mass m on an inclined plane. a The object moves upwards along the plane with constant acceleration ~a. q T Find the kinetic friction coefficient µ k between the object and the plane. For which values of T,a is this motion possible? Page 17 of 31

18 y fk FN q F g T Newton s 2nd law: ~F net = m~a = mabi ~ Fnet = ~ T + ~ F g + ~ F N + ~ f k F gx f k F N F g T F T y x g y T x Page 18 of 31 (F net ) x = T cos mgsin f k (F net ) y = T sin mgcos + F N

19 y (F net ) x = ma (F net ) y =0 F N T x x F gx f k F g T F y g y T (F net ) x = T cos mgsin f k (F net ) y = T sin mgcos + F N Page 19 of 31 f k = T cos mgsin ma, f k = µ k F N F N = T sin + mgcos µ k = T cos mgsin ma T sin + mgcos

20 For which values of T,a is the motion possible? µ k = T cos mgsin ma T sin + mgcos 0 T cos m(gsin + a) Page 20 of 31

21 Drag force and terminal speed Page 21 of 31 Drag force: force caused by relative motion of an object and a fluid opposed the relative motion and points in the direction the fluid flows relative to the object

22 For a fast blunt object moving through air such that the flow becomes turbulent D = 1 2 C Av2 A = e ective crossection: area of crossection? ~v C = drag coe cient = air density Page 22 of 31

23 Terminal velocity: 1 2 CA v2 y mg =0 ) v y = Body falling from rest through air: The drag force increases gradually until it balances the gravitational force. The body reaches terminal velocity. ~D + F ~ g =0 s 2mg C A Page 23 of 31

24 i-clicker Theoretically, how long does it take to reach terminal velocity? (A) t = (B) t =0 q 2m gc A (C) nfinite amount of time. Page 24 of 31 (D) None of the above.

25 Answer Theoretically, how long does it take to reach terminal velocity? (A) t = (B) t =0 q 2m gc A (C) nfinite amount of time. Page 25 of 31 (D) None of the above. Why?

26 ~F net = m~a ~ Fnet = ~ D + ~ F g F net y = 1 2 CA v2 y mg dv y dt = applev2 y mg, apple = 1 2 CA + Page 26 of 31 v y = r mg apple tanh p mgapplet = r mg e p mgapplet e p mgapplet apple e p mgapplet + e p mgapplet

27 Terminal velocity: r mg lim v y = t!1 apple = s 2mg lim t!1 a y =lim t!1 dv y dt =0 C A Page 27 of 31

28 Page 28 of 31

29 i-clicker A sky diver jumps from a flying airplane and falls for several seconds before she reaches terminal velocity. She then opens her parachute, reaches a new terminal velocity, and continues her descent to the ground. Which one of the following graphs of the drag force versus time best represents this situation? Page 29 of 31

30 Answer A sky diver jumps from a flying airplane and falls for several seconds before she reaches terminal velocity. She then opens her parachute, reaches a new terminal velocity, and continues her descent to the ground. Which one of the following graphs of the drag force versus time best represents this situation? Page 30 of 31

31 Before opening the parachute terminal velocity is reached when ~D + ~ F g =0 D = F g After opening the parachute terminal velocity is again reached when ~D + ~ F g =0 D = F g Page 31 of 31 Asu ciently long time after opening the magnitude of the drag force must equal its value before opening. At the same time it should spike upwards during opening.

Rutgers University Department of Physics & Astronomy. 01:750:271 Honors Physics I Fall Lecture 8. Home Page. Title Page. Page 1 of 35.

Rutgers University Department of Physics & Astronomy 01:750:271 Honors Physics I Fall 2015 Lecture 8 Page 1 of 35 Midterm 1: Monday October 5th 2014 Motion in one, two and three dimensions Forces and Motion