1 / 24

Lecture 7 Chapter 9 Systems of Particles Wednesday 7-16-03

Lecture 7 Chapter 9 Systems of Particles Wednesday 7-16-03. Warm-up problem. Puzzle Question: Cite two possible reasons why it appears that some basket ball players and dancers have a greater hang time. Physlets. Topics. Center of mass Newton’s 2nd law for a system of particles

macaulay-peck
Download Presentation

Lecture 7 Chapter 9 Systems of Particles Wednesday 7-16-03

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 7 Chapter 9Systems of Particles Wednesday7-16-03 Warm-up problem Puzzle Question: Cite two possible reasons why it appears that some basket ball players and dancers have a greater hang time. Physlets

  2. Topics • Center of mass • Newton’s 2nd law for a system of particles • Linear momentum, 2nd law in terms of P • Conservation of P

  3. Center of mass The center of a body or a system of bodies is the point that moves as though all of the mass were concentrated there and all external forces were applied there.

  4. Center of Mass Why is it important? For any rigid body the motion of the body is given by the motion of the cm and the motion of the body around the cm. As an example find the center of mass of the following system d . x m M xcm m xcm = M (d-xcm) xcm = M/(m + M) d

  5. Problem 9.3 2 dimensions . Find xcm and ycm ycmM = m1 y1 + m2 y2 + m3 y3 xcmM = m1 x1 + m2 x2 + m3 x3 ycm 15 = 3*0 +4*1 + 8*2 xcm 15 = 3*0 + 4*2 + 8*1 ycm = 20/15 = 1.33 m xcm = 16/15 = 1.1 m

  6. If the 8 kg mass increases, how does the cm change? xcm =(m1 x1 + m2 x2 + m3 x3)/(m1+m2+m3) When m3 gets very large, suppose we neglect the other masses. xcm ~(m3 x3)/(m3) ~ x3 The center of mass moves towards the large object as it should

  7. Center of Mass for a system of particles xcm = (m1 x1 + m2 x2 + m3 x3)/M

  8. Newton’s Second Law for a System of particles: Fnet= Macm take d/dt on both sides take d/dt again Identify ma as the force on each particle

  9. Linear Momentum form of Newton’s 2nd Law Important because it is a vector quantity that is conserved in interactions. Now take derivative d/dt of

  10. Law of Conservation of Linear Momentum If Fnet = 0 on a closed system where no mass enters or leaves the system, then dP/dt = 0 or P = constant. Pi = Pf for a closed isolated system Also each component of the momentum Px,Py,Pz is also constant since Fx, Fy, and Fz all = 0 If one component of the net force is not 0, then that component of momentum is not a constant. For example, consider the motion of a horizontally fired projectile. The y component of P changes while the horizontal component is fixed after the bullet is fired.

  11. Air track examples Two carts connected by a spring. Set them into oscillation by pulling them apart and releasing them from rest. Note cm does not move. Now repeat with spring between two carts. Analogous to exploding mass. Again observers in two different inertial reference frames will measure different values of momentum, but both will agree that momentum is conserved. Note if the net force vanishes in one inertial frame it will vanish in all inertial frames

  12. Change in momentum of ball on left or right. F(t) -F(t)

  13. Andy Rodick has been clocked at serving a tennis ball up to 149 mph(70 m/s). The time that the ball is in contact with the racquet is about 4 ms. The mass of a tennis ball as is about 300 grams. What is the average force exerted by the racquet on the ball? Favg = J/Dt = Dp/ Dt Dp =70( 0.3) - 0 = 21 kg-m/s Favg = 21/0.004 = 5000 Newtons What is the acceleration of the ball? a = Favg /m = 5000N/0.3 kg = 16,667 m/s2 What distance the racquet go through while the ball is still in contact? V2f - V2i =2ax x = V2f /2a = (70)2/2(16667) = 0.15 m

  14. BOUNCING BALL hi -vf vf After bounce Before bounce Initial E = PE = mgh hf E = KE = 1/2mv2

  15. Measuring velocities and heights of balls bouncing from a infinitely massive hard floor Almost elastic collision C.O.R. = Vf/Vi R.E./C.E.= Hi/Hf Almost inelastic collision

  16. Conservation of Momentum In a closed isolated system containing a collision, the linear momentum of each colliding body may change but the total linear momentum P of the system can not change, whether the collision is elastic or inelastic. Fnet = 0, dP/dt = 0, Hence, P = constant . Each x,y, and z component of the momentum is a constant.

  17. One Dimension Collision m1v1i + m2v2i = m1v1f + m2v2f 1/2 m1v1i2 = 1/2 m1v1f2 + 1/2m2v2f2 Kinetic energy is conserved too. Total momentum before = Total momentum after

  18. m1 Balls bouncing off massive floors, we have m2 >>m1 m2

  19. Lecture 8 Chapter 10Collisions Thursday 7-17-03

  20. Colliding pool balls The excutive toy m2 = m1

  21. -v -v -v v v1f v Two moving colliding objects: Problem 45 m1v1i + m2v2i = m1v1f + m2v2f 1/2 m1v1i2 + 1/2 m2v2i2 = 1/2 m1v1f2 + 1/2m2v2f2 Just after the little ball bounced off the big ball Just after the big ball bounced Just before each hits the floor m1 m2

  22. -v v V1f=2V V2f =0 For m2 =3m1 , v1f = 8/4V =2V superball has twice as much speed. How high does it go? 4 times higher

  23. -v v V1f=3V V2f = -V V1f=3V For maximum height consider m2 >>m1 V2f = -V How high does it go? 9 times higher

More Related