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 x O y m1 m2 h = 2 m l = 1 m

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

m g Example: Acceleration of C.G. of a multi-particle system.  C.G. m g 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)

Prove that for a many particle system 1 3 2

Total linear momentum of a many-particle system

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

Example: A 20-kg girl on a 80-kg boat is 10 m from the shore. The C. G 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: Shore B 10m x 2 - x before after The distance from the shore is: = 10 - (2 – 0.4) = 8.4 m

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

K.E. of a many-particle system:

Angular momentum of a many-particle system about P:

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

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