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Example : The center of mass of a seesaw.

m 2. m 1. y. x. O. l = 1 m. x. y. h = 2 m. O. Chapter 5 Kinetics of many particles. Example : The center of mass of a seesaw. Find C.G. when: m 1 = m 2 m 1 = 70 kg, m 2 = 80 kg m 1 = 80 kg, m 2 = 70 kg. An object can be regarded as a collection of point masses.

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Example : The center of mass of a seesaw.

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  1. m2 m1 y x O l = 1 m x y h = 2 m O Chapter 5 Kinetics of many particles Example : The center of mass of a seesaw. • Find C.G. when: • m1 = m2 • m1 = 70 kg, m2 = 80 kg • m1 = 80 kg, m2 = 70 kg

  2. An object can be regarded as a collection of point masses. The center of gravity (C.G.) or center of mass (C.M.) is:

  3. C.G. m m  g  Example: Acceleration of C.G. of a multi-particle system. Total F = mg + mg = 2 mg Total mass = 2 m Acceleration of C.G. aG = g F = M aG(similar to F = ma for a point mass)

  4. 1 2 3 Prove that for a many particle system

  5. Total linear momentum of a many-particle system

  6. Impulse  change in linear momentum Fext max. tolerable force t Another form of Newton’s 2nd law: Example: Jumping down from an elevated position

  7. Shore B before 10m B after x 2 - x Example: A 20-kg girl on a 80-kg boat is 10 m from the shore. The C.G. of the boat and the girl approaches by 2 m. How far is the girl from the shore? Fext = 0. Take moment about CG: The distance from the shore is: = 10 - (2 – 0.4) = 8.4 m

  8. Linear momentum is conserved if the duration of interaction t2 - t1 is extremely short ----- collision Example: Find the initial velocity of the bullet u. Fast collision  After the collision, K.E. = energy dissipation due to friction

  9. K.E. of a many-particle system:

  10. F2-1 F1-2 Angular momentum of a many-particle system about P: O Moment about P

  11. If • The reference frame (P) is an inertial frame (aP = 0), or • P is the C.G., or • //

  12. Conservation of angular momentum occurs if: • Time of interaction is extremely short.

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