Physics Principles of Physics. Lecture 12 (Serway & Jewett Vol.1) - Chapter 9 October 9, 2007 Dr. Sung-Won Lee

Transcription

1 Physics Principles of Physics Lecture 1 (Serway & Jewett Vol.1) - Chapter 9 October 9, 007 Dr. Sung-Won Lee Sungwon.Lee@ttu.edu

2 Announcements Lecture notes SI Session (Andrew Wells) Schedule: M/W 5:00-6:30 SC 7 HW Assignment #5 has been placed on WebAssign, and is due by 11:30pm on Tuesday, 10/9 The solutions for Exam 1 will be posted on the wall; Next to my office (Sci 117) The grades for Exam 1 is done!!

3 Announcement II Exam 1 Average = 60. % Congrat! 99% Phuong Hoang, 93% Chukwuemeka Egbuna 91% Jason Roers, Jeff Macha

4 Chapter 9 Linear Momentum & Collisions Energy 1. Linear Momentum and its Conservation. Impulse and Momentum 3. Collisions in 1-dimension 4. -dimensional Collisions 5. The Center of Mass 6. Motion of a System of Particles 7. Rocket Propulsion

5 r r Definition: "momentum"! p = mv p = mv ; p = mv ; p = mv ; x x y y z z Momentum The product of a particle s mass and velocity is called the momentum Momentum is a vector quantity In terms of components: Newton formulated his nd law in terms of momentum: r r r r nd r dv d( mv) dp Law: F = ma = m = = dt dt dt! px = px " p1 x = # Fx ( t) dt t t 1 Force changes momentum

6 Conservation of Linear Momentum Whenever or more particles interact, the total momentum of the system remains constant (See below) Conservation of momentum can be expressed mathematically in various ways p total = p 1 + p = constant p 1i + p i = p 1f + p f In component form, the total momenta in each direction are independently conserved p ix = p fx p iy = p fy p iz = p fz Conservation of momentum can be applied to systems with any number of particles

7 9. Impulse and Momentum From Newton s nd-law, F = dp/dt Solving for dp gives dp = Fdt Integrating to find the change in momentum over some time interval The integral is called the impulse, I, of the force F acting on an object over t Impulse-Momentum Theorem This equation expresses the impulse-momentum theorem : The impulse of the force F acting on a particle equals the change in the momentum of the particle This is equivalent to Newton s Second Law

8 Momentum Conservation: m ( v ) + m ( v ) = m ( v ) 1 fx 1 fx 1 ix 1 Energy Conservation: Elastic Collisions m1 ( v fx ) 1 + m ( v fx ) = m 1( v ) ix 1 m ( v ) = ( v )! ( v ) fx 1 ix 1 fx m1 m [( v )! m ( v ) ] + m ( v ) = m ( v ) ix 1 fx fx 1 ix 1 m1 [ m ( v )! m ( v ) ( v ) 1 1 ix 1 ix 1 fx m v fx m v fx m1 + ( ) + ( ) ] = m ( v ) 1 1 ix 1 m ( v ) [(1 + )( v )! ( v ) ] = 0 fx fx ix 1 m1 m ( v ) = ( v ) = ( v ) 1 fx ix 1 ix m / m1 m1 + m m m m m ( v ) ( v ) ( v ) ( v ) 1 1! fx 1 = ix 1! ix 1 = ix 1 m m m m1 + m

9 Three Elastic Collisions m ( v ) = ( v ) ; fx 1 ix 1 m1 + m m m ( v ) ( ) fx 1! 1 = vix 1 m1 + m m = m : ( v ) = ( v ) and ( v ) = 0 (knock-on) 1 fx ix 1 fx 1 m >> m : ( v ) = ( v ) and ( v ) = ( v ) (boost-ahead) 1 fx ix 1 fx 1 ix 1 m << m : ( v ) = 0 and ( v ) =!( v ) (bounce-off) 1 fx fx 1 ix 1

10 9.4 Two-Dimensional Collisions The momentum is conserved in all directions Use subscripts for identifying the object, indicating initial or final values, the velocity components If the collision is elastic, use conservation of KE as a second equation Particle 1 is moving at velocity v 1i and particle is at rest In the x-direction, the initial momentum is m 1 v 1i In the y-direction, the initial momentum is 0

11 9.4 Two-Dimensional Collisions After the collision, the momentum in the x-direction: m 1 v 1f cosθ + m v f cosφ After the collision, the momentum in the y-direction: m 1 v 1f sinθ + m v f sinφ

12 -Dimensional Collision Example Before the collision, the car has the total momentum in the x and the van has the total momentum in the y After the collision, both have x- and y-components

13 9.5 The Center of Mass There is a special point in a system or object, called the center of mass, that moves as if all mass of the system is concentrated at that point The coordinates of the center of mass are where M is the total mass of the system The center of mass can be located by its position vector, r CM r i is the position of the i th particle, defined by

14 Center of Mass, Example Both masses are on the x-axis Center of mass is on the x-axis Center of mass is closer to the particle with the larger mass

15 Center of Mass, Extended Object, Position The position of the center of mass can also be found by: The center of mass of any symmetrical object lies on an axis of symmetry and on any plane of symmetry

16 Center of Mass, Example An extended object can be considered a distribution of small mass elements, Δm The center of mass is located at position r CM 1 1 xcm = " xi! m = x dm M i M # 1 1 ycm = " yi! m = y dm M M # i As Fig shows, we can divide an extended object into many small cells or boxes, each with the very small mass Δm

17 The Center of Mass of a Rod Find the center of mass of a thin uniform rod of length L and mass M. Find the tangential acceleration of one tip of a 1.6 m rod that rotates about its center of mass with an angular speed of 6.0 rad/s. dm = dx so dm = M M L L 0 dx 1 1 M xcm = x dm x dx M $ = M $ L L 1 1 = = = L $! 1 L 1 xdx " x L # 0 L a = r! = L! t = 1 (1.6 m)(6.0 rad/s ) 1 = 4.8 m/s

18 9.6 Motion of a System of Particles Assume the total mass, M, of the system remains constant. We can describe the motion of the system in terms of the velocity and accel. of the C.O.M. of the system. We can also describe the momentum of the system and Newton s nd law for the system. The velocity of the center of mass of a system of particles is The momentum can be expressed as The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass

19 Acceleration of the Center of Mass The acceleration of the center of mass can be found by differentiating the velocity with respect to time Forces In a System of Particles The acceleration can be related to a force If we sum over all the internal forces, they cancel in pairs and the net force on the system is caused only by the external forces Momentum of System of Particles The total linear momentum of a system of particles is conserved if no net external force is acting on the system Mv CM = p tot = constant when ΣF ext = 0

20 Motion of the Center of Mass, Example A projectile is fired into the air and suddenly explodes With no explosion, the projectile would follow the dotted line After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion

21 9.7 Rocket Propulsion The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel The initial mass of the rocket plus all its fuel is M + Δm at time t i and velocity v The initial momentum of the system: p i = (M + Δm) v At some time t + Δt, the rocket s mass has been reduced to M and an amount of fuel, m has been ejected The rocket s speed has increased by Δv

22 Rocket Propulsion, Because the gases are given some momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction Therefore, the rocket is accelerated as a result of the push from the exhaust gases The basic equation for rocket propulsion: The increase in rocket speed is proportional to the speed of the escape gases (v e ) The increase in rocket speed is also proportional to the natural log of the ratio M i /M f So, the ratio should be as high as possible, meaning the mass of the rocket should be as small as possible and it should carry as much fuel as possible

23 Thrust The thrust on the rocket is the force exerted on it by the ejected gases Thrust: The thrust increases as the exhaust speed increases The thrust increases as the rate of change of mass increases The rate of change of the mass is called the burn rate

24 End of Lecture 11 Homework Assignment #5 will be due on 10/9 (Tuesday) Next Class: Thursday, 10/9