1. 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

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. 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)

4. Prove that for a many particle system1 3 2

5. Total linear momentum of a many-particle system

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

7. 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

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. Angular momentum of a many-particle system 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.