Physics 211: Lecture 14 Today s Agenda Systems of Particles Center of mass Linear Momentum Example problems Momentum Conservation Inelastic collisions in one dimension Ballistic pendulum Physics 211: Lecture 14, Pg 1
System of Particles Until now, we have considered the behavior of very simple systems (one or two masses). But real life is usually much more interesting! For example, consider a simple rotating disk. An extended solid object (like a disk) can be thought of as a collection of parts. The motion of each little part depends on where it is in the object! Physics 211: Lecture 14, Pg 2
System of Particles: Center of Mass How do we describe the position of a system made up of many parts? Define the Center of Mass (average position): For a collection of N individual pointlike particles whose masses and positions we know: R CM N i 1 N i 1 m m r i i i m 1 r 1 m 4 y r 4 R CM x r m 2 2 r 3 (In this case, N = 4) m 3 Physics 211: Lecture 14, Pg 3
System of Particles: Center of Mass We can consider the components of R CM separately: m x m y i i i i i ( X CM, YCM, Z CM ),, M M m z i i i i M m 2 m 1 r 1 m 4 y r 4 R CM x r 2 r 3 (In this case, N = 4) m 3 Physics 211: Lecture 14, Pg 4
Example Calculation: Consider the following mass distribution: 2m (12,12) m (0,0) m (24,0) Physics 211: Lecture 14, Pg 5
System of Particles: Center of Mass Baton The center of mass is where the system is balanced! Building a mobile is an exercise in finding centers of mass. + m 1 m 2 + m m 1 2 Physics 211: Lecture 14, Pg 6
System of Particles: Center of Mass For a continuous solid, we have to do an integral. y x dm r R CM r dm dm r dm M where dm is an infinitesimal mass element. Physics 211: Lecture 14, Pg 7
System of Particles: Center of Mass We find that the Center of Mass is at the center of the object. y R CM x The location of the center of mass is an intrinsic property of the object!! (it does not depend on where you choose the origin or coordinates when calculating it). Physics 211: Lecture 14, Pg 8
System of Particles: Center of Mass We can use intuition to find the location of the center of mass for symmetric objects that have uniform density: It will simply be at the geometrical center! + CM + + + + + Physics 211: Lecture 14, Pg 9
Lecture 13, Act 1 Center of Mass The disk shown below (1) clearly has its CM at the center. Suppose the disk is cut in half and the pieces arranged as shown in (2): Where is the CM of (2) as compared to (1)? (a) higher (b) lower (c) same X CM (1) (2) Physics 211: Lecture 14, Pg 10
Lecture 13, Act 1 Solution The CM of each half-disk will be closer to the fat end than to the thin end (think of where it would balance). The CM of the compound object will be halfway between the CMs of the two halves. This is higher than the CM of the disk X X CM X X (1) (2) Physics 211: Lecture 14, Pg 11
+ CM System of Particles: Center of Mass Double cone The center of mass (CM) of an object is where we can freely pivot that object. pivot + Gravity acts on the CM of an object (show CM later) If we pivot the object somewhere else, it will orient itself so that the CM is directly below the pivot. pivot pivot This fact can be used to find the CM of odd-shaped objects. mg CM Physics 211: Lecture 14, Pg 12
System of Particles: Center of Mass Odd shapes Hang the object from several pivots and see where the vertical lines through each pivot intersect! pivot pivot pivot + CM The intersection point must be at the CM. Physics 211: Lecture 14, Pg 13
Lecture 13, Act 2 Center of Mass 3 pronged object An object with three prongs of equal mass is balanced on a wire (equal angles between prongs). What kind of equilibrium is this position? a) stable b) neutral c) unstable Physics 211: Lecture 14, Pg 14
Lecture 13, Act 2 Solution The center of mass of the object is at its center and is initially directly over the wire If the object is pushed slightly to the left or right, its center of mass will not be above the wire and gravity will make the object fall off CM mg CM mg (front view) Physics 211: Lecture 14, Pg 15
Lecture 13, Act 2 Solution Consider also the case in which the two lower prongs have balls of equal mass attached to them: CM mg In this case, the center of mass of the object is below the wire CM mg When the object is pushed slightly, gravity provides a restoring force, creating a stable equilibrium Physics 211: Lecture 14, Pg 16
Example: Astronauts & Rope Two astronauts at rest in outer space are connected by a light rope. They begin to pull towards each other. Where do they meet? M = 1.5m m Physics 211: Lecture 14, Pg 17
Example: Astronauts & Rope... They start at rest, so V CM = 0. V CM remains zero because there are no external forces. So, the CM does not move! They will meet at the CM. M = 1.5m m CM L x=0 x=l Finding the CM: If we take the astronaut on the left to be at x = 0: x cm M( 0) m(l) m(l) M m 2. 5m 2 5 L Physics 211: Lecture 14, Pg 18
Lecture 13, Act 3 Center of Mass Motion A man weighs exactly as much as his 20 foot long canoe. Initially he stands in the center of the motionless canoe, a distance of 20 feet from shore. Next he walks toward the shore until he gets to the end of the canoe. What is his new distance from the shore. (There no horizontal force on the canoe by the water). 20 ft (a) 10 ft before (b) 15 ft 20 ft (c) 16.7 ft? ft after Physics 211: Lecture 14, Pg 19
Lecture 13, Act 3 Solution Since the man and the canoe have the same mass, the CM of the man-canoe system will be halfway between the CM of the man and the CM of the canoe. Initially the CM of the system is 20 ft from shore. 20 ft X X CM of system x Physics 211: Lecture 14, Pg 20
Lecture 13, Act 3 Solution Since there is no force acting on the canoe in the x-direction, the location of the CM of the system can t change! 20 ft X X x CM of system Physics 211: Lecture 14, Pg 21
Linear Momentum: Definition: For a single particle, the momentum p is defined as: (p is a vector since v is a p = mv vector). So p x = mv x etc. Newton s 2nd Law: F = ma m dv dt d dt (mv) F dp dt Units of linear momentum are kg m/s. Physics 211: Lecture 14, Pg 22
Linear Momentum: So the total momentum of a system of particles is just the total mass times the velocity of the center of mass. P MV CM Observe: dp dt M d VCM MACM miai F dt, i i i net dp We are interested in so we need to figure out dt i F i, net Physics 211: Lecture 14, Pg 23
Linear Momentum: Only the total external force matters! dp dt i F, F i EXT NET, EXT m 3 Which is the same as: F NET, EXT dp dt MA CM m 1 F 1,EXT m 2 Newton s 2nd law applied to systems! Physics 211: Lecture 14, Pg 24
Center of Mass Motion: Recap We have the following law for CM motion: Pendulum F EXT dp MA dt This has several interesting implications: It tells us that the CM of an extended object behaves like a simple point mass under the influence of external forces: We can use it to relate F and A like we are used to doing. It tells us that if F EXT = 0, the total momentum of the system can not change. The total momentum of a system is conserved if there are no external forces acting. CM Physics 211: Lecture 14, Pg 25
Momentum Conservation F EXT dp dt dp 0 F EXT 0 dt The concept of momentum conservation is one of the most fundamental principles in physics. This is a component (vector) equation. We can apply it to any direction in which there is no external force applied. You will see that we often have momentum conservation even when energy is not conserved. Physics 211: Lecture 14, Pg 26
Elastic vs. Inelastic Collisions A collision is said to be elastic when kinetic energy as well as momentum is conserved before and after the collision. K before = K after Carts colliding with a spring in between, billiard balls, etc. v i A collision is said to be inelastic when kinetic energy is not conserved before and after the collision, but momentum is conserved. K before K after Car crashes, collisions where objects stick together, etc. Physics 211: Lecture 14, Pg 27
Inelastic collision in 1-D: Example 1 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 : What is the initial speed of the bullet v? What is the initial energy of the system? What is the final energy of the system? Is kinetic energy conserved? x v V before after Physics 211: Lecture 14, Pg 28
Example 1... Consider the bullet & block as a system. After the bullet is shot, there are no external forces acting on the system in the x-direction. Momentum is conserved in the x direction! x v V initial final Physics 211: Lecture 14, Pg 29
Physics 211: Lecture 14, Pg 30 Example 1... Now consider the kinetic energy of the system before and after: Before: After: So E mv m M m m V M m m M m V B 1 2 1 2 1 2 2 2 2 2 E M m V A 1 2 2 E m M m E A B Kinetic energy is NOT conserved! (friction stopped the bullet) However, momentum was conserved, and this was useful. v M m m V
Example 1... What if the bullet goes through? Physics 211: Lecture 14, Pg 31
Inelastic Collision in 1-D: Example 2 M m V v = 0 ice (no friction) M + m v =? Physics 211: Lecture 14, Pg 32
Example 2... Air track Physics 211: Lecture 14, Pg 33
Lecture 14, Act 4 Momentum Conservation Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. Which box ends up moving faster? (a) Box 1 (b) Box 2 (c) same 1 2 Physics 211: Lecture 14, Pg 34
Lecture 14, Act 4 Momentum Conservation Since the total external force in the x-direction is zero, momentum is conserved along the x-axis. In both cases the initial momentum is the same (mv of ball). In case 1 the ball has negative momentum after the collision, hence the box must have more positive momentum if the total is to be conserved. The speed of the box in case 1 is biggest! V 1 V 2 1 2 x Physics 211: Lecture 14, Pg 35
Lecture 14, Act 4 Momentum Conservation mv init = MV 1 - mv fin V 1 = (mv init + mv fin ) / M mv init = (M+m)V 2 V 2 = mv init / (M+m) V 1 numerator is bigger and its denominator is smaller than that of V 2. V 1 > V 2 x V 1 V 2 1 2 Physics 211: Lecture 14, Pg 36
Ballistic Pendulum L L L L V=0 m v M H V M + m A projectile of mass m moving horizontally with speed v strikes a stationary mass M suspended by strings of length L. Subsequently, m + M rise to a height of H. Given H, what is the initial speed v of the projectile? Physics 211: Lecture 14, Pg 37
Ballistic Pendulum... Two stage process: 1. m collides with M, inelastically. Both M and m then move together with a velocity V (before having risen significantly). 2. M and m rise a height H, conserving K+U energy E. (no non-conservative forces acting after collision) Physics 211: Lecture 14, Pg 38
Ballistic Pendulum... Stage 1: Momentum is conserved in x-direction: mv ( m M ) V V m m M v Stage 2: K+U Energy is conserved 1 2 2 ( E E ) I F ( m M ) V ( m M ) gh V 2gH 2 Eliminating V gives: v 1 M m 2gH Physics 211: Lecture 14, Pg 39
Ballistic Pendulum Demo L L L L m v H M + m M d In the demo we measure forward displacement d, not H: L-H H d L 2 2 2 L d L H H L L d 2 2 Physics 211: Lecture 14, Pg 40
H L L L L Ballistic Pendulum Demo... 2 1 d d L 2 2 2 L L 1 2 d 2L 2 L-H H 2 d 2L for d L d L 1 Ballistic pendulum v 1 M m 2gH v 1 M m d g L for d << L Let s see who can throw fast... Physics 211: Lecture 14, Pg 41
Recap of today s lecture Systems of particles (Text: 8-1) Center of mass (Text: 8-1 & 12-6) Linear Momentum (Text: 8-3 to 8-4) Inelastic collisions in one dimension (Text: 8-6) Ballistic pendulum (Ex. 8-14) Physics 211: Lecture 14, Pg 42