2. Physics 151: Lecture 12, Pg 1 REVIEW-2 Phys-151 l Work & Energy l Momentum Conservation l Center of Mass l Torque, Angular Momentum, Moment of Inertia

3. Physics 151: Lecture 12, Pg 2 Definition of Work: Ingredients: Fr Ingredients: Force ( F ), displacement (  r ) F Work, W, of a constant force F r acting through a displacement  r is: F rrr W = F.  r = F  r cos  = F r  r  F r rr r displacement FrFr “Dot Product” See text: 7-1 Total Energy is Conserved

4. Physics 151: Lecture 12, Pg 3 xxxx vovo m toto F Example Work Kinetic-Energy Theorem  How much will the spring compress to bring the object to a stop if the object is moving initially at a constant velocity (v o ) on frictionless surface as shown below ? spring compressed spring at an equilibrium position V=0 t m

5. Physics 151: Lecture 12, Pg 4 Lecture 13 - ACT 3 Work of Springs l I have a spring with k = 20 N/m. Its compressed 30 cm by a 200 gram mass. How much work is done by the spring in this process ? A) 0.6 JB) 6 JC) 0.9 JD) –0.9 J

6. Physics 151: Lecture 12, Pg 5 Lecture 13, ACT 5 Work & Energy Two blocks having mass m 1 and m 2 where m 1 > m 2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e.  > 0) which slows them down to a stop. Which one will go farther before stopping ? (a) (b) (c) (a) m 1 (b) m 2 (c) they will go the same distance m1m1 m2m2

7. Physics 151: Lecture 12, Pg 6 Another Example A 3.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37° above the horizontal as shown. The speed of the block increases from 4.0 m/s to 6.0 m/s in a displacement of 5.0 m. What work was done by the friction force during this displacement? a.–34 J b. –64 J c. –30 J d. –94 J e. +64 J

8. Physics 151: Lecture 12, Pg 7 Work & Power: Example 3 : What is the power required for a car (m=1000 kg) to climb a hill (5%) at v=30m/s assuming the coefficient of friction  = 0.03 ? 100 5 Car 5% V = 30 m/s P tot = P horizontal + P vertical v=const. - > a = 0 P horizontal = F v =  mg v horizontal P vertical = F v = mg v vertical = (1000kg) (10m/s 2 ) (30m/s)(5/100) P vertical ~ 15 kW P horizontal ~ 0.03 (1000kg) (10m/s 2 ) (30m/s) ~ 10 kW P tot ~ 10 kW + 15 kW = 25 kW

9. Physics 151: Lecture 12, Pg 8 Example - 2 A 12-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kN/m). The block is initially at rest at its equilibrium position when a force (magnitude P = 80 N) acting parallel to the surface is applied to the block, as shown. What is the speed of the block when it is 13 cm from its equilibrium position? x F m k v 1 = 0 v 2 =? 0.78 m/s

10. Physics 151: Lecture 12, Pg 9 Lecture 16, ACT 1 The Roller Coaster l I have built a Roller Coaster. A motor tugs the cars to the top and then they are let go and are in the hands of gravity. To make the following loop, how high do I have to let the release the car ?? h ? R Car has mass m A) 2RB) 5RC) 5/2 RD) none of the above

11. Physics 151: Lecture 12, Pg 10 MOMENTUM pv p = mv Momentum Conservation l A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. l A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. K before  K after K before = K after

12. Physics 151: Lecture 12, Pg 11 ACT 4 l The value of the momentum of a system is the same at a later time as at an earlier time if there are no a. collisions between particles within the system. b.inelastic collisions between particles within the system. c. changes of momentum of individual particles within the system. d. internal forces acting between particles within the system. e. external forces acting on particles of the system.

13. Physics 151: Lecture 12, Pg 12 Lecture 17, ACT 3 Momentum Conservation l Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. l The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. çWhich box ends up moving fastest ? (a) (b) (c) (a) Box 1 (b) Box 2 (c) same 1 2

14. Physics 151: Lecture 12, Pg 13 Inelastic collision in 1-D: Example 1 l A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V : çWhat is the initial speed of the bullet v ? çWhat is the initial energy of the system ? çWhat is the final energy of the system ? çIs energy conserved ? v V beforeafter x See example 12-6

15. Physics 151: Lecture 12, Pg 14 Lecture 17 ACT 4 3.0-kg mass sliding on a frictionless surface has a velocity of 5.0 m/s east when it undergoes a one- dimensional inelastic collision with a 2.0-kg mass that has an initial velocity of 2.0 m/s west. After the collision the 3.0-kg mass has a velocity of 1.0 m/s east. How much kinetic energy does the two-mass system lose during the collision? a. 22 J b. 24 J c. 26 J d. 20 J e. 28

16. Physics 151: Lecture 12, Pg 15 Elastic Collision in 1-D l After some moderately tedious algebra, (see text book Chapter 9, section3) we can derive the following equations for the final velocities, See text: 9.4 1) m 1 v 1,b + m 2 v 2,b = m 1 v 1,a + m 2 v 2,a 2) v 1,b - v 2,b = - (v 1,a - v 2,a ) In general:

17. Physics 151: Lecture 12, Pg 16 Example - Elastic Collision l Suppose I have 2 identical bumper cars. One is motionless and the other is approaching it with velocity v 1. If they collide elastically, what is the final velocity of each car ? Note that this means, m 1 = m 2 = m v 2B = 0 See text: 9.4 Animation

18. Physics 151: Lecture 12, Pg 17 2-D Elastic Collisions Billiards l Consider the case where one ball is initially at rest. ppappa ppbppb F PPaPPa beforeafter the final direction of the red ball will depend on where the balls hit. v v cm See Figure 12- 14 See text: Ex. 9.11 Animation

19. Physics 151: Lecture 12, Pg 18 ACT ACT A 5.0-g particle moving 60 m/s collides with a 2.0-g particle initially at rest. After the collision each of the particles has a velocity that is directed 30° from the original direction of motion of the 5.0-g particle. What is the speed of the 2.0-g particle after the collision? a. 72 m/s b. 87 m/s c. 79 m/s d. 94 m/s e. 67 m/s

20. Physics 151: Lecture 12, Pg 19 Force and Impulse F t titi tftf tt I l The diagram shows the force vs time for a typical collision. The impulse, I, of the force is a vector defined as the integral of the force during the collision. I Impulse I = area under this curve ! See text: 9-2 See Figure 12- 2 Impulse has units of Ns.

21. Physics 151: Lecture 12, Pg 20 Force and Impulse F t tt l Using the impulse becomes: impulse = change in momentum ! See Figure 12- 2 See text: 9-2

22. Physics 151: Lecture 12, Pg 21 Lecture 19, ACT 2 Force & Impulse l What is the average force that wall exerts on ball (0.40kg) if duration of wall-ball contact is 0.01 s ? v i = 30m/s v f = 20m/s before collision after collision a) 20 N b) 200 N c) 2,000 N d) 20,000 N

23. Physics 151: Lecture 12, Pg 22 System of Particles: Center of Mass l How do we describe the “position” of a system made up of many parts ? Center of Mass l Define the Center of Mass (average position): çFor a collection of N individual pointlike particles whose masses and positions we know: (In this case, N = 2) y x rr2rr2 rr1rr1 m1m1 m2m2 R R CM

24. Physics 151: Lecture 12, Pg 23 Center of Mass Motion: Review l We have the following law for CM motion: l This has several interesting implications: l It tell us that the CM of an extended object behaves like a simple point mass under the influence of external forces: FA çWe can use it to relate F and A like we are used to doing. F l It tells us that if F EXT = 0, the total momentum of the system does not change. See text: 9.6

25. Physics 151: Lecture 12, Pg 24 Example: Astronauts & Rope l A male astronaut and a female astronaut are at rest in outer space and 20 meters apart. The male has 1.5 times the mass of the female. The female is right by the ship and the male is out in space a bit. The male wants to get back to the ship but his jet pack is broken. Conveniently, there is a rope connected between the two. So the guy starts pulling in the rope. l Does he get back to the ship? l Does he at least get to meet the woman? M = 1.5m m :-( :-)

26. Physics 151: Lecture 12, Pg 25 Lecture 19, ACT Center of Mass Motion l A woman weighs exactly as much as her 20 foot long boat. l Initially she stands in the center of the motionless boat, a distance of 20 feet from shore. Next she walks toward the shore until she gets to the end of the boat. çWhat is her new distance from the shore. (There is no horizontal force on the boat by the water). 20 ft ? ft 20 ft (a) (b) (c) (a) 10 ft (b) 15 ft (c) 16.7 ft before after

27. Physics 151: Lecture 12, Pg 26 Recall Kinematic of Circular Motion: R v s tt (x,y)  =  t s = v t s = R  = R  t v = Rv = R x = R cos(  )  = R cos(  t)  y = R sin(  )  = R sin(  t)  = tan -1 (y/x) For uniform circular motion:  is angular velocity x y Animation

28. Physics 151: Lecture 12, Pg 27 Summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R See text: 10.3

29. Physics 151: Lecture 12, Pg 28 Example: l The turntable of a record player has an angular velocity of 8.0 rad/s when it is turned off. The turntable comes to rest 2.5 s after being turned off. Through how many radians does the turntable rotate after being turned off ? Assume constant angular acceleration. a. 12 rad b. 8.0 rad c. 10 rad d. 16 rad e. 6.8 rad See text: 10.1

30. Physics 151: Lecture 12, Pg 29 Calculating Moment of Inertia... For a single object, I clearly depends on the rotation axis !! L I = 2mL 2 I = mL 2 mm mm I = 2mL 2 See text: 10.5 See example 10.4 (similar)

31. Physics 151: Lecture 12, Pg 30 Moments of Inertia Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center. Some examples of I for solid objects: R L r dr

32. Physics 151: Lecture 12, Pg 31 Moments of Inertia... Some examples of I for solid objects: Solid sphere of mass M and radius R, about an axis through its center. R See text: 10.5 See Table 10.2, Moments of Inertia Thin spherical shell of mass M and radius R, about an axis through its center. R

33. Physics 151: Lecture 12, Pg 32 Kinetic Energy of Rotation and Moment of Inertia See text: 10.4 l where I PARALLEL = I CM + MR 2 Parallel Axis Theorem

34. Physics 151: Lecture 12, Pg 33 Rotational Dynamics: What makes it spin? See text: 10.6 and 10.7  r F DIRECTION:  =  r X F   r F sin MAGNITUDE:  =  r F sin 

35. Physics 151: Lecture 12, Pg 34 How Much WORK is Done ? W =  l Analogue of W = F  r W will be negative if  and  have opposite sign ! See text: 10.8 Work & Kinetic Energy:

36. Physics 151: Lecture 12, Pg 35 Lecture 21, ACT 2 A uniform rod of mass M = 1.2kg and length L = 0.80 m, lying on a frictionless horizontal plane, is free to pivot about a vertical axis through one end, as shown. If a force (F = 5.0 N,  = 40°) acts as shown, what is the resulting angular acceleration about the pivot point ? a. 16 rad/s 2 b. 12 rad/s 2 c. 14 rad/s 2 d. 10 rad/s 2 e. 33 rad/s 2

37. Physics 151: Lecture 12, Pg 36 Angular Momentum: Definitions & Derivations l e have shown that for a system of particles Momentum is conserved if See text: 11.3 F  The rotational analogue of force F is torque  p l Define the rotational analogue of momentum p to be angular momentum p=mv Animation

38. Physics 151: Lecture 12, Pg 37 Definitions & Derivations... L l First consider the rate of change of L: So See text: 11.3

39. Physics 151: Lecture 12, Pg 38 Lecture 21, ACT 2 Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both are made of identical material (i.e. their density  = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest. Which disk has the biggest angular velocity after the pull ? (a) (a) disk 1 (b) (b) disk 2 (c) (c) same FF 11 22

40. Physics 151: Lecture 12, Pg 39 Example 2 l A rope is wrapped around the circumference of a solid disk (R=0.2m) of mass M=10kg and an object of mass m=10 kg is attached to the end of the rope 10m above the ground, as shown in the figure.  M m h =10 m T a)How long will it take for the object to hit the ground ? a)What will be the velocity of the object when it hits the ground ? a)What is the tension on the cord ? 1.7 s 11m/s 32 N

41. Physics 151: Lecture 12, Pg 40 Example: Rotating Road l A uniform rod of length L=0.5m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is a) angular speed when it reaches the lowest point ? b) initial angular acceleration ? c) initial linear acceleration of its free end ? See example 10.14 See text: 10.8 L m  = 7.67 rad/s a)  = 30 rad/s 2 b) c) a = 15 m/s 2

42. Physics 151: Lecture 12, Pg 41 Example: l A mass m = 4.0 kg is connected, as shown, by a light cord to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius R = 0.12 m and a moment of inertia I = 0.090 kg m 2. The cord does not slip on the pulley. What is the magnitude of the acceleration of m? a. 2.4 m/s 2 b. 2.8 m/s 2 c.3.2 m/s 2 d. 4.2 m/s 2 e. 1.7 m/s 2 See text: 10.1

43. Physics 151: Lecture 12, Pg 42 Example : Rolling Motion l A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? M  h M v ? Cylinder has radius R M M M M M

44. Physics 151: Lecture 12, Pg 43 How high do we have to start the ball ? 1 2 -> The rolling motion added an extra 2/10 R to the height) h h = 2.7 R = (2R + 1/2R) + 2/10 R