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Center of Mass Linear Momentum

Center of Mass Linear Momentum. Chapter 9. Center of Mass. the point in which the mass of an object can be considered to be concentrated. the objects motion acts as if it’s mass is concentrated at that point.

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Center of Mass Linear Momentum

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  1. Center of MassLinear Momentum Chapter 9

  2. Center of Mass the point in which the mass of an object can be considered to be concentrated the objects motion acts as if it’s mass is concentrated at that point the translational acceleration calculated by Newton’s second law Fnet = ma determines the motion of the object at it’s center of mass all gravitational forces exerted on the object is applied at that point

  3. Calculating Center of Mass The center of mass of an object can be found in all dimensions. (m1x1 + m2x2 + . . . + mnxn) → ^ ^ ^ ______________________ vectorcom = rcom = xcomi + ycomj + zcomk xcom = (m1 + m2 + . . . + mn) → → ___ 1 n Σ miri (m1y1 + m2y2 + . . . + mnyn) rcom = ______________________ M i = 1 ycom = (m1 + m2 + . . . + mn) (m1z1 + m2z2 + . . . + mnzn) ______________________ zcom = (m1 + m2 + . . . + mn)

  4. Newton’s Second Law for a System of Particles → → Fnet = Macom → Fnet : net force of all external forces (internal forces not included) M = total mass of the system (closed system) → acom = acceration of the center of mass of the system → → Fnet,x = Macom,x → → Fnet,y = Macom,y → → Fnet,z = Macom,z

  5. Conversion of Center of Mass Equation → rcom : gives position → → → → → dri → ___ Mrcom = m1r1 + m2r2 + . . . + mnrn vi= dt → → → → → [Mrcom]` = Mvcom = m1v1 + m2v2 + … + mnvn → dvi → ___ ai= → → → → → dt [Mvcom]` = Macom = m1a1 + m2a2 + … + mnan → → → → → → m1a1 + m2a2 + … + mnan = F1 + F2 + … + Fn → → Fi= miai

  6. Linear Momentum linear momentum of a particle = total mass of the system * velocity of its center of mass → → p = mv → the momentum of an particle acts in the same direction as the velocity of that particle because mass is a positive scaler quantity → → → P = p1 + p2 + … + pn → → → = m1v1 + m2v2 + … + mnvn → → P = Mvcom time rate of change of the momentum of a particle = net force acting on that particle and in the same direction → → dP ___ Fnet = dt → → → dp d → dvcom ___ ___ _____ → Fnet= = (Mvcom) = M = Macom dt dt dt therefore net external force changes the momentum of an object, thus momentum can only change due to an external force

  7. Impulse tf tf → ∫ → ∫ impulse is defined as: dp = F(t) dt ti ti → → → pf – pi = ∆p tf → → ∫ F(t)dt J = ti → → thus ∆p = J linear momentum – impulse theorem → → J = Favg∆t → → J = -n∆p n = # of projectiles series of collisions : → → → → → __ J -n __ -n __ ___ ∆m ∆p m∆v – ∆v Favg = = = = ∆t ∆t ∆t ∆t

  8. Conservation of Linear Momentum → no external forces : Fnet = 0 momentum stays constant in a closed system → → Pi = Pf law of conservation of linear momentum if a component of net external force in a closed system = 0 along an axis, then the component of the linear momentum of the system along that system can not change

  9. Collisions elastic collision: momentum AND kinetic energy conserved → → → → P1,i + P2,i = P1,f + P2,f conservation of momentum → → → → m1v1,i + m2v2,i = m1v1,f + m2v2,f K1,i + K2,i = K1,f + K2,f conservation of kinetic energy → → → → 1/2m1v1,i2 + 1/2m2v2,12 = 1/2m1v1,f2 + 1/2m2v2,f2 inelastic collision: momentum is conserved, kinetic energy IS NOT conserved → → → → P1,i + P2,i = P1,f + P2,f conservation of momentum → → → → m1v1,i + m2v2,i = m1v1,f + m2v2,f completely inelastic: bodies stick together

  10. Systems With Varying Mass: A Rocket → → velocity of products relative to frame velocity of rocket relative to frame velocity of rocket relative to products Pi = Pf + = → → → m vR-F = vR-P + vP-F v → vrel M + dM -dM v + dv U → → → → → → → → Mv = -dMU + (M +dM)(v + dv) (v + dv) = vrel + U → → → → → → U = v + dv - vrel P of gas P of rocket afterwards → → -dMvrel = Mdv → → → __ dv ___ → dM ____ -dM dv = – vrel Vrel = M dt M dt vf Mf ∫ → → ∫ → → ___ dM dv = – vrel Rvrel = Ma first rocket equation M vi Mi | | → → → Mi __ vf - vi = vrelln second rocket equation Mf

  11. What is Newton’s Second Law for a system of particles? → → Fnet = Macom

  12. How does Mrcom become Mvcom? → → → drcom → ____ Mvcom= M dt

  13. Why is the direction of a particle’s momentum the same as the direction of its velocity? mass is a scaler quantity in the equation: → → p = mv

  14. If there is no external force (Fnet = 0), P is constant in a _________ system. closed

  15. The equation for the linear momentum of a system is? → → p = mv

  16. Impulse is defined as? → tf → ∫ J = F(t)dt ti

  17. What is the impulse – momentum theorem? → → → → J = ∆P = Pf - Pi

  18. If there is no net external force momentum ( does, does not ) change.

  19. In an elastic collision both _______________ and _______________ are conserved. kinetic energy momentum

  20. In an inelastic collision _____________ is conserved and _____________ is not conserved. momentum kinetic energy

  21. In a completely inelastic collision what occurs? the bodies stick together

  22. What is the Law of Conservation of Momentum? Pi = Pf

  23. vR-F = _______ + ________where P = productsR = rocketF = frame of reference vR-P vP-F

  24. Vrel = ________ vR-G

  25. The vcom ( is, isn’t ) changed by a collision.

  26. In the absence of external forces a rocket accelerates at an instonaneous rate given by: (first rocket equation) Rvrel = Ma

  27. The point in which the mass of an object could be concentrated is? center of mass

  28. What is the equation for the center of mass of an object on the x axis? (m1x1 + m2x2 + . . . + mnxn) ____________________ xcom = (m1 + m2 + . . . + mn)

  29. The thrust of an object is given by: Rvrel

  30. All gravitation forces are exterted on what point? center of mass

  31. The path in which a baseball bat’s center of mass is thrown into the air is in the shape of a ______________. parabola

  32. Find the center of mass of these 3 objects: m 2m 2m d 3d d

  33. Find the impulse from 1 second to 5 seconds if Fnet(t) = t. 12

  34. In a completely elastic collision: v m 2m Find the final velocity of the second particle. → → V2 = 2v

  35. In a completely inelastic collision: v m 2m Find the final velocity. vf = 2/3v

  36. If a boat is moving at 10 mph against a river current of 2 mph, how fast is the boat going past the shoreline? 6 mph

  37. If a 2000 kg (including fuel) rocket burns 10 kg of fuel per second going at 10000 m/s past earth ejecting fuel at 500 m/s past earth. What is the rocket’s acceleration? 4.75 m/s 2

  38. A 2 kg mass hits a 4 kg mass at 5 m/s in an elastic collision. Find the final velocity of the 2 kg mass. – 1.67 m/s

  39. In a completely in elastic collision: 2v v 2m 4m What is the final velocity? 0 m/s

  40. Find the center of mass: m 2 2m 2 -2 4m 3m - 2 (0.1,- 0.8)

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