#### Transcript Center of Mass and Momentum

Center of Mass and Momentum Lecture 08 Monday: 9 February 2004 A special point… •If the net external force on a system of particles is zero, then (even if the velocity of individual objects changes), there is a point associated with the distribution of objects that moves with zero acceleration (constant velocity). •This point is called the “center of mass” of the system. It is the balancing point for the mass distribution. CENTER OF MASS Three component equations : 1 1 1 xcm mn xn ycm mn yn zcm M M M M = m1 + m2 + ··· M = S mn m z n n Position, Velocity and Acceleration of CENTER OF MASS xcm 1 M m xn v cm 1 M m vn acm 1 M m an n n n MOMENTUM • For a system of particles, total momentum is: P = p1 + p2 + p3 ··· or P = S mnvn along with v cm 1 M m v n • P = Mvcm n MOMENTUM dv cm dP M Ma cm dt dt Fext dP dt CENTER OF MASS (CONT.) •The overall translational motion of a system of particles can be analyzed using Newton's laws as if all the mass were concentrated at the center of mass and the total external force were applied at that point. MOMENTUM • For a particle, translational momentum is: p mv • Momentum is a vector • Its direction is the same as the object’s velocity • Momenta must be added as vectors MOMENTUM (Review) • For a particle, translational momentum is: p mv • Momentum is a vector • Its direction is the same as the object’s velocity • Momenta must be added as vectors MOMENTUM of a SYSTEM (Review) • A system is any group of objects that we wish to consider as a group. • For a system of particles, momentum is: ptotal p1 p2 p3 ... pn i n i n ptotal pi mi vi i 1 i 1 • This is a vector sum!! We have to take direction into account. dp F dtimply? What does (Review) 1. Forces cause changes in an object’s momentum. That is, forces cause the object’s velocity to change over time. 2. We can determine the change in an object’s velocity over time using this expression in the form: dp Pf Pi F dt t t f o Internal and External Forces •There are two classes of forces that act on and within systems. •INTERNAL FORCES are forces between an object within the system and another object within the system. •EXTERNAL FORCES are forces between an object within the system and an object outside the system. Only External Forces Change the Total Momentum of the System • Recall Newton’s third law. For every force there is an equal and opposite force. These “paired” forces are called Newton’s third law pairs. • Two forces are a Newton’s third law pair if 1. The two different forces are between the same two objects 2. The object exerting the force and the object being acted on switch roles in the two forces. • Third Law force pairs cancel out. So only forces external to a system matter. ALL internal forces are third law pairs. Conservation of Momentum dp Recall that F dt dp For a system this is FEXTERNAL dt If the sum of external forces is zero, then dp 0 dt What does this imply? Conservation of Momentum • dP 0 dt states that the momentum does not change over time. That is, it is constant. • If you know the momentum at any one time, then you know it for all other times. EXAMPLE: FExt = 0, FInt 0 Two masses, initially at rest. PBefore = PAfter 0 = m1v1 + m2v2 m1v1 = – m2v2 If, m1 = m2 v1 = – v2 Relationship between Force and Momentum In one dimension (so dropping the vector symbol) p mv if we take the derivative of both sides with respect to time dp d (mv) dt dt for a constant mass dp dv m ma dt dt and F ma so F dp dt What does dp F dtimply? 1. Forces cause changes in an object’s momentum over time. For constant mass objects, that means forces cause the object’s velocity to change over time. 2. We can determine the change in an object’s velocity over time using this expression in the form: dp Pf Pi F dt t t f o Reworking dp F dt dp dp or Fnet dt dt we can rewrite this as Fnet (dt ) dp F integratin g both sides of the equation tf ti tf ti pf Fnet dt dp pif Fnet dt p f pi p IMPULSE tf J Fdt p f p i ti In many cases, the integral is easily evaluated as the area under a curve. 4 F (N) 0 5 8 Time (sec) MOMENTUM of a SYSTEM • A system is any group of objects that we wish to consider as a group. • For a system of particles, the total momentum is: ptotal p1 p2 p3 ... pn i n i n ptotal pi mi vi i 1 1 • This is a vector sum!! We ihave to take direction into account. MOMENTUM (Review) • For a particle, translational momentum is: p mv • Momentum is a vector • Its direction is the same as the object’s velocity • Momenta must be added as vectors MOMENTUM of a SYSTEM (Review) • A system is any group of objects that we wish to consider as a group. • For a system of particles, momentum is: ptotal p1 p2 p3 ... pn i n i n ptotal pi mi vi i 1 i 1 • This is a vector sum!! We have to take direction into account. dp F dtimply? What does (Review) 1. Forces cause changes in an object’s momentum. That is, forces cause the object’s velocity to change over time. 2. We can determine the change in an object’s velocity over time using this expression in the form: dp Pf Pi F dt t t f o Internal and External Forces •There are two classes of forces that act on and within systems. •INTERNAL FORCES are forces between an object within the system and another object within the system. •EXTERNAL FORCES are forces between an object within the system and an object outside the system. Only External Forces Change the Total Momentum of the System • Recall Newton’s third law. For every force there is an equal and opposite force. These “paired” forces are called Newton’s third law pairs. • Two forces are a Newton’s third law pair if 1. The two different forces are between the same two objects 2. The object exerting the force and the object being acted on switch roles in the two forces. • Third Law force pairs cancel out. So only forces external to a system matter. ALL internal forces are third law pairs. Conservation of Momentum dp Recall that F dt dp For a system this is FEXTERNAL dt If the sum of external forces is zero, then dp 0 dt What does this imply? Conservation of Momentum • dP 0 dt states that the momentum does not change over time. That is, it is constant. • If you know the momentum at any one time, then you know it for all other times. Conversation of Momentum In one dimension, m v m v .. m dp v m v ... Recall that F dt 1 1i 2 2i 1 1f 2 2f EXTERNAL dp If the sum of external forces is zero, then 0 dt (That is, the momentum is constant.) This does not mean that the momentum of any one object in the system stays the same. It means that if you add up all of the momenta for all of the objects in the system that this total doesn’t change as time passes. This is only true for systems on which no net external force acts. When motion occurs in two dimensions, TWO conservation of momentum equations are required. mv m v 2 iy mv m v 2 fy mv m v 2 ix mv m v 2 fx 1 1 1 iy 1 ix 2 2 1 1 1 fy 1 fx 2 2 Momentum is conserved or not conserved in each direction SEPERATELY. When motion occurs in two or more directions, we need to consider each direction separately.. dp x FEXT , x dt dp y FEXT , y dt dp z FEXT , z dt If the sum of external forces is zero in any one or all of the directions , then the momentum in that (or those) direction( s) is constant. Conversation of Momentum in Two Dimensions m v m v ... m v m v 2 fy ... m v m v ... m v m v 2 fx ... 1 1 1 iy 1 ix 2 2 2 iy 2 ix 1 1 1 fy 1 fx 2 2 Remember, you know how to get the x and y components of velocity….. f V Vy= VCos f Vx=VSinf Momentum Transfers In systems made up of several objects, momentum can be transferred from one object to another and still be conserved. However, momentum in the x direction can not be transferred into the y direction and visa-versa. Collisions Momentum is conserved only if there is no net external force acting on the system. •Remember, if a system is composed of two or more objects and these objects undergo a collision, then all forces between the colliding objects are INTERNAL forces and so sum to zero. •There can be external forces acting on a system in which a collision occurs. An example is friction between the objects and the surface on which they move. EXAMPLE: FExt = 0, FInt 0 Two masses, initially at rest. PBefore = PAfter 0 = m1v1 + m2v2 m1v1 = – m2v2 If, m1 = m2 v1 = – v2