Center of Mass and Linear Momentum

1. Chapter 9 Center of Mass and Linear Momentum

2. 9.2 The Center of Mass The center of mass of a system of particles is the point that moves as though (1) all of the system’s mass were concentrated there and (2) all external forces were applied there. The center of mass (black dot) of a baseball bat flipped into the air follows a parabolic path, but all other points of the bat follow more complicated curved paths.

3. 9.2 The Center of Mass: A System of Particles Consider a situation in which n particles are strung out along the x axis. Let the mass of the particles are m1, m2, ….mn, and let them be located at x1, x2, …xn respectively. Then if the total mass is M = m1+ m2 + . . . + mn, then the location of the center of mass, xcom, is

4. 9.2 The Center of Mass: A System of Particles In 3-D, the locations of the center of mass are given by: The position of the center of mass can be expressed as:

5. 9.2 The Center of Mass: Solid Body In the case of a solid body, the “particles” become differential mass elements dm, the sums become integrals, and the coordinates of the center of mass are defined as where M is the mass of the object. If the object has uniform density, r, defined as: Then Where V is the volume of the object.

6. Calculations: First, put the stamped-out disk (call it disk S) back into place to form the original composite plate (call it plate C). Because of its circular symmetry, the center of mass comS for disk S is at the center of S, at x =-R. Similarly, the center of mass comC for composite plate C is at the center of C, at the origin. Assume that mass mS of disk S is concentrated in a particle at xS =-R, and mass mP is concentrated in a particle at xP. Next treat these two particles as a two particle system, and find their center of mass xS+P. Next note that the combination of disk S and plate P is composite plate C. Thus, the position xS+P of comS+P must coincide with the position xC of comC, which is at the origin; so xS+P =xC = 0. But, and xS=-R Sample problem, COM

7. The total mass M of the system is 7.1 kg. The coordinates of the center of mass are therefore: Note that the zcom = 0. Sample problem, COM of 3 particles We are given the following data:

8. 9.3: Newton’s 2nd Law for a System of Particles The vector equation that governs the motion of the center of mass of such a system of particles is: Note that: 1. Fnetis the net force of all external forces that act on the system. Forces on one part of the system from another part of the system (internal forces) are not included 2. M is the total mass of the system. M remains constant, and the system is said to be closed. 3. acomis the acceleration of the center of mass of the system.

9. 9.3: Newton’s 2nd Law for a System of Particles: Proof of final result • For a system of n particles, • where M is the total mass, and ri are the position vectors of the masses mi. • Differentiating, • where the v vectors are velocity vectors. • This leads to • Finally, • What remains on the right hand side is the vector sum of all the external forces that act on the system, while the internal forces cancel out by Newton’s 3rd Law.

10. Calculations: Applying Newton’s second law to the center of mass, Sample problem: motion of the com of 3 particles

11. Linear Momentum I) One particle system Definition of linear momentum: The linear momentum of an object is the product of its mass and velocity. Note that momentum is a vector—it has both a magnitude and a direction. SI unit of momentum: kg • m/s. This unit has no special name.

12. Linear momentum (Newton’s 2nd Law - acceleration) . The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force. (Newton’s 2nd Law - momentum)

13. Collision and Impulse The change in linear momentum is related to the force by Newton’s second law written in the form The let side of the equation is the change of the momentum of the particle: The right side of the equation is a measure of both the magnitude and the duration of the net force, and is called the impulse of the (net) force, J.

14. Impulse In this case, the impact is brief, and the ball experiences a force that is great enough to slow, stop, or even reverse its motion. The figure depicts the force of impact at one instant. The ball experiences a force F(t) that varies during the collision and changes the linear momentum of the ball.

15. Graphical Representation of Impulse Instead of the ball, one can focus on the bat. At any instant, Newton’s third law says that the force on the bat has the same magnitude but the opposite direction as the force on the ball. That means that the impulse on the bat has the same magnitude but the opposite direction as the impulse on the ball.

16. Impulse-Momentum Example: Bouncing vs. Stopping Actual contact times may be very short.

17. Example Tennis • A tennis ball weighing 50 g and moving horizontally with a velocity of 12 m/s, is hit by a racket such that its velocity becomes 20 m/s in exactly the opposite direction. a) What is the change in the momentum of the ball? b) What impulse is delivered by the racket to the ball? J = Δp = 1.6 Ns c) What force is exerted by the racket on the ball if the time of impact is 5 ms? F Δt = Δp therefore F x 0.005 s= =1.6 Ns, F = 320 N

18. Example Tennis pi • The tennis ball (50 g) moving with a velocity of 12 m/s, hits a wall at an angle of 30o with the normal and bounces with the velocity of 10 m/s at an angle of 45o with the normal. a) What is the change in the magnitude of the momentum of the ball? 30 45 pf b) What impulse is delivered by the wall to the ball? J = Δp = 0.875 Ns c) What force is exerted by the wall on the ball if the time of impact is 5 ms? F Δt = Δp therefore F x 0.005 s= 0.875 Ns, F = 174.9 N

19. Force & Impulse • Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each box for exactly 1 second. • Which box has the most momentum after the force acts? (a)heavier(b)lighter(c)same F F heavy light

20. We know so The boxes will have the same final momentum. Solution In this problem F and Dt are the same for both boxes! F F heavy light

21. II) Systems of Particles For a system of objects, the total momentum is the vector sum of each.

22. Suppose we have a system of two particles as shown. The particles interact with each other through equal and oppositeinternal forcesF, and in addition there are external forces pushing on particles 1 and 2. We will apply the impulse momentum theorem separately for each particle: System of Two Particles 1) (F1, EXT + F12)Δt = Δp1 2) (F2, EXT + F21)Δt = Δp2 F2,EXT Adding the two equations yields: (F1, EXT + F2, EXT+ F12 + F21)Δt = Δp1+ Δp2 But because of Newton’s 3rd Law: F12 = -F21, so F12 + F21 =0 F12 m1 F21 m2 (F1, EXT + F2, EXT)Δt = Δ(p1+ p2) F1,EXT If we denote: Ptot = p1+ p2 ΣFEXT= F 1, EXT + F 2, EXT and

23. System of Two Particles we get: (ΣFEXT) Δt = ΔPtot or (ΣFEXT) = ΔPtot/ Δt Impulse-Momentum Theorem for Systems of Particles All the “internal” forces cancel !! Only the “external” forces matter !! Note: Internal forces cannot change the total momentum of the system. If (ΣFEXT) =0 then ΔPtot=0, and Ptot=const.

24. Conservation of Linear Momentum-System If (ΣFEXT) =0 than ΔPtot=0, and Ptot=const. If there is no net external force acting on a system, its total momentum cannot change, and stays constant. This is the law of conservation of momentum. If there are internal forces, the momenta of individual parts of the system can change, but the overall momentum stays the same.

25. Example: A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s. What is the recoil speed of the rifle as the bullet leaves the barrel? As long as the rifle is horizontal, there will be no net external force acting on the rifle-bullet system and momentum will be conserved. 25

26. Examples: Conservation of Linear Momentum In this example, there is no external force, but the individual components of the system do change their momenta:

27. 9.6: Collision and Impulse: Series of Collisions Let n be the number of projectiles that collide in a time interval Dt. Each projectile has initial momentum mv and undergoes a change Dp in linear momentum because of the collision. The total change in linear momentum for n projectiles during interval Dt is nDp. The resulting impulse on the target during Dt is along the x axis and has the same magnitude of nDp but is in the opposite direction. In time interval Dt, an amount of mass Dm =nm collides with the target.

28. Sample problem: 2-D impulse, cont.

29. Sample problem: 1-D explosion

30. Sample problem: 2-D explosion

31. 9.8: Momentum and Kinetic Energy in Collisions In a closed and isolated system, if there are two colliding bodies, and the total kinetic energy is unchanged by the collision, then the kinetic energy of the system is conserved (it is the same before and after the collision). Such a collision is called an elastic collision. If during the collision, some energy is always transferred from kinetic energy to other forms of energy, such as thermal energy or energy of sound, then the kinetic energy of the system is not conserved. Such a collision is called an inelastic collision.

32. Sample problem: two pendulums

33. Collisions Collisions: an interaction that lasts a very short time

34. vi 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. Kbefore = Kafter • Carts colliding with a spring in between, billiard balls, etc. • A collision is said to be inelastic when kinetic energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter • Car crashes, collisions where objects stick together, etc.

35. Completely Inelastic Collisions Objects stick together after collision. m1+m2 m1 m2 v’ v2 v1 P before = P after p1 + p2 = P’ m1v1 + m2v2 = (m1 + m2) v’

36. Energy Total kinetic energy is not conserved in an inelastic collision.

37. 2D Inelastic Collisions Again, in a completely inelastic collision the objects stick together afterwards. vy’ vx’ p + P = P’ P =MV P’=(m+M)v’ p = mv

38. Inelastic collisions in 1-D: Velocity of Center of Mass Fig. 9-16 Some freeze frames of a two-body system, which undergoes a completely inelastic collision. The system’s center of mass is shown in each freeze-frame. The velocity vcom of the center of mass is unaffected by the collision. Because the bodies stick together after the collision, their common velocity V must be equal to vcom.

39. Before the explosion: M After the explosion: v1 v2 m1 m2 Explosion (inelastic un-collision)

40. v1 v2 m1 m2 Explosion... • No external forces, so P is conserved. • Initially: P = 0 • Finally: P = m1v1 + m2v2 = 0 m1v1 = - m2v2 M

41. v v m m v V v v m m m m Explosion 2 • A bomb explodes into 3 identical pieces. Which of the following configurations of velocities is possible? (a)1(b)2(c)both (1) (2)

42. v mv mv m mv v v m m Explosion 2 • No external forces, so P must be conserved. • Initially: P = 0 • In explosion (1) there is nothing to balance the upward momentum of the top piece so Pfinal 0. (1)

43. mv v mv mv m v v m m Explosion 2 • No external forces, so P must be conserved. • All the momenta cancel out. • Pfinal=0. (2)

44. Ballistic Pendulum L L V=0 L L m H v M + m V 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?

45. Ballistic Pendulum... • Two stage process: 1. mcollides 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 energyE. (no non-conservative forces acting after collision)

46. Ballistic Pendulum... • Stage 1: Momentum is conserved in x-direction: • Stage 2: K+U Energy is conserved Eliminating V’ gives:

47. The collision within the bullet– block system is so brief. Therefore: • During the collision, the gravitational force on the block and the force on the block from the cords are still balanced. Thus, during the collision, the net external impulse on the bullet–block system is zero. Therefore, the system is isolated and its total linear momentum is conserved. • (2) The collision is one-dimensional in the sense that the direction of the bullet and block just after the collision is in the bullet’s original direction of motion. • As the bullet and block now swing up • together, the mechanical energy of the bullet– block–Earth system is conserved: • Combining steps: Sample problem: conservation of momentum

48. Elastic Collisions For an elastic collision, both the kinetic energy and the momentum are conserved: Before After v1 v2 v1’ v2’ m2 m2 m1 m1

49. 2D Elastic Collisions (off Center) If m1 = m2 = m Pythagorean Theorem, angle 90o

50. Billiards. • The final directions are separated by 90o. pf pi vcm Pf F initial final