Physics 207: Lecture 10, Pg 1 Lecture 10 l Goals:  Exploit Newton’s 3 rd Law in problems with friction  Employ Newton’s Laws in 2D problems with circular motion Assignment: HW5, (Chapter 7, due 2/24, Wednesday) For Tuesday: Finish reading Chapter 8, First four sections in Chapter 9

Physics 207: Lecture 10, Pg 2 Example: Friction and Motion l A box of mass m 1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (  k = 0.5) on top of a second box having mass m 2 = 2 kg, which in turn slides on a smooth (frictionless) surface.  What is the acceleration of the bottom box ? Key Question: What is the force on mass 2 from mass 1? m2m2m2m2 T m1m1m1m1  slides with friction (  k =0.5  ) slides without frictiona = ? v

Physics 207: Lecture 10, Pg 3 Example Solution l First draw FBD of the top box: m1m1 N1N1 m1gm1g T f k =  K N 1 =  K m 1 g v

Physics 207: Lecture 10, Pg 4 l Newtons 3 rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. m1m1 f f 1,2 =  K m 1 g = 5 N m2m2 f 2,1 = -f 1,2 l As we just saw, this force is due to friction: Example Solution Action Reaction

Physics 207: Lecture 10, Pg 5 l Now consider the FBD of box 2: m2m2 f 2,1 =  k m 1 g m2gm2g N2N2 N 1 = m 1 g Example Solution

Physics 207: Lecture 10, Pg 6 l Finally, solve F x = ma x in the horizontal direction: m2m2 f 2,1 =  K m 1 g  K m 1 g = m 2 a x Example Solution = 2.5 m/s 2

Physics 207: Lecture 10, Pg 7 Home Exercise Friction and Motion, Replay A box of mass m 1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m 2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are  s =1.5 and  k = 0.5. The second box can slide freely (frictionless) on an smooth surface. Question: Compare the acceleration of box 1 to the acceleration of box 2? m2m2 T m1m1  friction coefficients  s =1.5 and  k =0.5 slides without friction a2a2 a1a1

Physics 207: Lecture 10, Pg 8 Home Exercise Friction and Motion, Replay in the static case A box of mass m 1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m 2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are  s =1.5 and  k = 0.5. The second box can slide freely on an smooth surface (frictionless). In the case of “no slippage” what is the maximum frictional force between boxes 1 & 2? m2m2 T m1m1  friction coefficients  s =1.5 and  k =0.5 slides without friction a2a2 a1a1

Physics 207: Lecture 10, Pg 9 Home Exercise Friction and Motion f s = 10 N and the acceleration of box 1 is Acceleration of box 2 equals that of box 1, with |a| = |T| / (m 1 +m 2 ) and the frictional force f is m 2 a (Notice that if T were in excess of 15 N then it would break free) m2m2 T m1m1  friction coefficients  s =1.5 and  k =0.5 slides without friction a2a2 a1a1 T g m 1 g N fSfS f S   S N =  S m 1 g = 1.5 x 1 kg x 10 m/s 2 which is 15 N (so m 2 can’t break free)

Physics 207: Lecture 10, Pg 10 Exercise Tension example A. T a = ½ T b B. T a = 2 T b C. T a = T b D. Correct answer is not given Compare the strings below in settings (a) and (b) and their tensions.

Physics 207: Lecture 10, Pg 11 Problem 7.34 Hint (HW 6) Suggested Steps l Two independent free body diagrams are necessary l Draw in the forces on the top and bottom blocks l Top Block  Forces: 1. normal to bottom block 2. weight 3. rope tension and 4. friction with bottom block (model with sliding) l Bottom Block  Forces: 1. normal to bottom surface 2. normal to top block interface 3. rope tension (to the left) 4. weight (2 kg) 5. friction with top block 6. friction with surface N Use Newton's 3rd Law to deal with the force pairs (horizontal & vertical) between the top and bottom block.

Physics 207: Lecture 10, Pg 12 On to Chapter8 Reprisal of : Uniform Circular Motion For an object moving along a curved trajectory with constant speed a = a r (radial only) |a r |= vt2vt2 r arar v

Physics 207: Lecture 10, Pg 13 Non-uniform Circular Motion For an object moving along a curved trajectory, with non-uniform speed a = a r + a t (radial and tangential) arar atat dt d| | v |a t |= |a r |= vT2vT2 r

Physics 207: Lecture 10, Pg 14 Key steps l Identify forces (i.e., a FBD) l Identify axis of rotation l Apply conditions (position, velocity & acceleration)

Physics 207: Lecture 10, Pg 15 Example The pendulum Consider a person on a swing: When is the tension on the rope largest? And at that point is it : (A) greater than (B) the same as (C) less than the force due to gravity acting on the person? axis of rotation

Physics 207: Lecture 10, Pg 16 Example Gravity, Normal Forces etc. vTvT mgmg T at bottom of swing v T is max F r = m a c = m v T 2 / r = T - mg T = mg + m v T 2 / r T > mg mgmg T at top of swing v T = 0 F r = m 0 2 / r = 0 = T – mg cos  T = mg cos  T < mg  axis of rotation y x

Physics 207: Lecture 10, Pg 17 Conical Pendulum (very different) l Swinging a ball on a string of length L around your head (r = L sin  ) axis of rotation  F r = ma r = T sin   F z = 0 = T cos  – mg so T = mg / cos  (> mg) ma r = (mg / cos  ) (sin  ) a r = g tan  v T 2 /r  v T = (gr tan  ) ½ Period: t = 2  r / v T =2  (r cot  /g) ½ = 2  (L cos  /g) ½ r L

Physics 207: Lecture 10, Pg 18 Conical Pendulum (very different) l Swinging a ball on a string of length L around your head axis of rotation Period: t = 2  r / v T =2  (r cot  /g) ½ = 2  (L cos  / g ) ½ = 2  (5 cos  / 9.8 ) ½ = 4.38 s = 2  (5 cos  / 9.8 ) ½ = 4.36 s = 2  (5 cos 1  / 9.8 ) ½ = 4.32 s r L

Physics 207: Lecture 10, Pg 19 A match box car is going to do a loop-the-loop of radius r. What must be its minimum speed v t at the top so that it can manage the loop successfully ? Another example of circular motion Loop-the-loop 1

Physics 207: Lecture 10, Pg 20 To navigate the top of the circle its tangential velocity v T 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). Loop-the-loop 1 F r = ma r = mg = mv T 2 /r v T = (gr) 1/2 mgmg vTvT

Physics 207: Lecture 10, Pg 21 The match box car is going to do a loop-the-loop. If the speed at the bottom is v B, what is the normal force, N, at that point? Hint: The car is constrained to the track. Loop-the-loop 2 mgmg v N F r = ma r = mv B 2 /r = N - mg N = mv B 2 /r + mg

Physics 207: Lecture 10, Pg 22 Once again the car is going to execute a loop-the-loop. What must be its minimum speed at 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… Loop-the-loop 3

Physics 207: Lecture 10, Pg 23 Home Exercise Swinging around a ball on a rope in a “nearly” horizontal circle over your head. Eventually the rope breaks. If the rope breaks at 64 N, the ball’s mass is 0.10 kg and the rope is 0.10 m How fast is the ball going when the rope breaks? (neglect mg contribution, 1 N << 40 N) (mg = 1 N) T = 40 N F r = m v T 2 / r ≈ T v T = (r F r / m) 1/2 v T = (0.10 x 64 / 0.10) 1/2 m/s v T = 8 m/s

Physics 207: Lecture 10, Pg 24 Example, Circular Motion Forces with Friction (recall ma r = m |v T | 2 / r F f ≤  s N ) l How fast can the race car go? (How fast can it round a corner with this radius of curvature?) m car = 1600 kg  S = 0.5 for tire/road r = 80 m g = 10 m/s 2 r

Physics 207: Lecture 10, Pg 25 l Only one force is in the horizontal direction: static friction x-dir: F r = ma r = -m |v T | 2 / r = F s = -  s N (at maximum) y-dir: ma = 0 = N – mg N = mg v T = (  s m g r / m ) 1/2 v T = (  s g r ) 1/2 = (  x 10 x 80) 1/2 v T = 20 m/s Example N mgmg FsFs m car = 1600 kg  S = 0.5 for tire/road r = 80 m g = 10 m/s 2 y x

Physics 207: Lecture 10, Pg 26 A horizontal disk is initially at rest and very slowly undergoes constant angular acceleration. A 2 kg puck is located a point 0.5 m away from the axis. At what angular velocity does it slip (assuming a T << a r at that time) if  s =0.8 ? l Only one force is in the horizontal direction: static friction x-dir: F r = ma r = -m |v T | 2 / r = F s = -  s N (at  ) y-dir: ma = 0 = N – mg N = mg v T = (  s m g r / m ) 1/2 v T = (  s g r ) 1/2 = (  x 10 x 0.5) 1/2 v T = 2 m/s   = v T / r = 4 rad/s Another Example m puck = 2 kg  S = 0.8 r = 0.5 m g = 10 m/s 2 y x N mgmg FsFs

Physics 207: Lecture 10, Pg 27 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 mgmg FfFf What differs on a banked curve?

Physics 207: Lecture 10, Pg 28 Lecture 10 Assignment: HW5, (Chapters 7, due 2/24, Wednesday) For Tuesday: Finish reading Chapter 8 and start Chapter 9