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Physics 1D03 - Lecture 161 Serway and Jewett 9.6 - 9.7 Centre of Mass Definition Total momentum of a system of particles Motion of the centre of mass.

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Presentation on theme: "Physics 1D03 - Lecture 161 Serway and Jewett 9.6 - 9.7 Centre of Mass Definition Total momentum of a system of particles Motion of the centre of mass."— Presentation transcript:

1 Physics 1D03 - Lecture 161 Serway and Jewett 9.6 - 9.7 Centre of Mass Definition Total momentum of a system of particles Motion of the centre of mass

2 Physics 1D03 - Lecture 162 Review: Newton’s Second Law For a particle: (Net external force) = ma For a particle or a system of particles: (Net external force) = F = dp/dt (Net external impulse) = I =  p

3 Physics 1D03 - Lecture 163 Apply Newton’s Laws to objects that are not particles: F or F How will an extended body move (accelerate) when a force is applied at an arbitrary location? The motion of the centre of mass is simple; in addition, various parts of the object move around the centre of mass. e.g.,

4 Physics 1D03 - Lecture 164 (Recall the position vector r has components x, y, z.) x CM m1m1 m2m2 m3m3 Centre of Mass Recall: Definition: or, r CM For continuous objects,

5 Physics 1D03 - Lecture 165 Dynamics of a system of particles CM definition: Differentiate with respect to time: The total momentum of any collection of particles is equal to the total mass times the velocity of the CM point—amazing but true! So Newton’s second law gives (differentiate above): This is remarkable. The motion of the centre of mass is the same as if the object were a single particle at the CM, with all external forces applied directly to it.

6 Physics 1D03 - Lecture 166 Example A neutron (mass m) travels at speed v 0 towards a stationary deuteron (mass 2m). What is the initial velocity of the CM of the system (neutron plus deuteron)? v0v0 x CM Since:

7 Physics 1D03 - Lecture 167 v0v0 CM 1/3v0 1/3v0 1/3v0 1/3v0

8 Physics 1D03 - Lecture 168 Examples: 1)A springboard diver does a triple reverse dive with one and a half twists. Her CM follows a smooth parabola (external force is gravity). 2)A paddler in a stationary canoe (floating on the water, no friction) walks from the rear seat to the front seat. The CM of the canoe plus paddler moves relative to the canoe, but not relative to the land (the canoe moves backwards). Here there is no external force. 3)Pendulum cart (demonstration).

9 Physics 1D03 - Lecture 169 Quiz A space station consists of a 1000-kg sphere and a 4000-kg sphere joined by a light cylinder. A rocket is fired briefly to provide a 100-N force for 10 seconds. Compare the velocity (CM) change if the rocket motor is A) at the small sphere B) at the large sphere C) same for either F F In which case will the space station get to the moon faster? MOON 384,000 km

10 Physics 1D03 - Lecture 1610 Quiz A space station consists of a 1000-kg sphere and a 4000-kg sphere joined by a light cylinder. A rocket is fired briefly to provide a 100-N force for 10 seconds. In which case will the space station rotate faster? A) at the small sphere B) at the large sphere C) same for either F F MOON 384,000 km

11 Physics 1D03 - Lecture 1611 1) Kinetic Energy = ½ Mv 2 CM + (K.E. relative to CM) (e.g., rigid body: K = ½ Mv 2 CM + ½ I CM  2 ) 2) ( Torque about CM) = I CM , even for an accelerated body Still more amazing CM theorems: “alpha” (angular acceleration)

12 Physics 1D03 - Lecture 1612 For practice: Chapter 9 Problems 39, 41, 58 (6 th ed) Problems 41, 43, 57, 69 Summary p total = M v CM (net external force) = M a CM


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