1. Center of Gravity/Center of Mass
Terms are used interchangeably when in a uniform gravitational field
Center of Gravity is the point at which all of the weight of an object appears to be concentrated.
If an object rotates when thrown, the center of gravity is also the center of rotation. When an object is suspended so that it can move freely, its center of gravity is always directly below the point of suspension.

2. CPO website

3. Rotational Mechanics
Torque is the tendency of a force to produce rotation around an axis.
Torque = Force x distance arm(lever/moment arm)
τ = F x r
The unit for torque is the Newton-meter (N-m)
Distance is measured from the pivot point or fulcrum to the location of the force on the lever arm.
The longer the lever arm, the greater the torque.
There are two ways to solve equilibrium problems:
Translational Equilibrium
Rotational Equilibrium

4. Balancing Torques
When torques are balanced or in equilibrium, the sum of the torques = 0
ΣΤ = (m1g*r1) + (m2g*r2) + … = (0 Rotational Equil.)
Careful on Rotational Equilibrium to account for rotational direction assuming its rotational direction were to occur(“-”)
Rotational equilibrium occurs when the angular velocity is constant or at rest while all forces are balanced.

5. Balancing Torques
Transitional Equilibrium occurs when the velocity of the system is constatnt or at rest in which all linear aspects of the systems force are balanced
ΣF = (m1g) + (m2g) … = 0 (Transitional Equilibrium)
There is no acceleration in their respective frame of reference.

6. Torque Example
A pair of adult nitwits sit balanced on a teeter totter type device which is 5 m long and has no mass. One of them, who’s mass is 45.2 kg, is 1.30 m from the point of balance (assumed to be at the center of the teeter totter). The other chowder-head is 2.15 m from the point of balance. What is the mass of the second person? If it is translational equilibrium, what is the force exerted by the fulcrum?

7. Torque Example
A seesaw is 4.5 m across, and is pivoted in the middle. You have a mass of 50 kg and sit 1.0 m from the left end of the seesaw. Your buddy wants to play, and he sits 3.8 m from the left end. How big (in kg) is your buddy if the system is balanced?

8. Torque Example
A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 205 N and is 3.00 m long. What is the force on each rope when the 675 N worker stands 1.00 m from one end of the scaffold?

9. Rotational Inertia
Rotational Inertia or moment of inertia is the resistance of an object to changes in its rotational motion. (rotating objects keep rotating, non-rotating objects tend to stay still)
The further the mass is located from the axis of rotation, the greater the rotational inertia.
Equations can vary based on shape of the rotating object (ie solid cylinder I = ½ mr2)
Greater rotational inertia means more laziness per mass.
I = mr2
Unit = kg∙m2

10. Rotational Inertia
All objects of the same shape have the same laziness per mass.
You can change your rotational inertia when spinning by extending your arms or legs.
Angular momentum is the measure of how difficult it is to stop a rotating object.
Angular momentum = mass x velocity x radius
L = mvr (or later in life L = ωr)
where v = ωr and mr(ωr) is mr2(ω), and I = mr2
Unit = kg∙m2/s

11. Example
Consider a light wooden rod with two 1 kg iron masses. When the masses are 0.1 m from the center, what is the moment of inertia?
Now consider what the moment of inertia for the pair of masses when they exist 1 m from the axis(center)?
By what magnitude has the moment of inertia increased?

12. Example
Remember, that tangential velocity is 2Πr/T. Calculate the angular momentum of a 100 g mass revolving in a circle once per second at the end of a .75 m string.