## I. 4.1: CENTER OF MASS

Special point where the object would behave like a single particle; sometimes not even located within the body of the object

### A. xCM = (m1x1 + m2x2….)/(m1 + m2…)

1. x = some point of reference from which to measure distance to masses = 0

a) If the stick in the example has a mass M = 4, it needs to be accounted for; distance from x would be its center of gravity, so if stick is 100, distance = 50 cm

b) xCM = (m1x1 + m2x2 + m3x3 + MxCM(M))/(m1 + m2 + m3 + M)

c) xCM = (50*2 +70*5 + 80*3 + 50*4)/(2+5+3+4) = (100 + 350 + 240 + 200)/14 = 790/14 = 56 cm

### B. How to calculate center of mass in non-linear problem: make it linear

### C. CENTER OF GRAVITY

Essentially same as center of mass (replace mass with weight)

1. xCG = (w1x1 + w2x2 …)/(w1 + w2 …) → (gm1x1 + gm2x2 …)/(gm1 + gm2 …) →

2. g(m1x1 + m2x2…)/

## II. 4.2: UNIFORM CIRCULAR MOTION (UCM)

### A. An object moving in a circular path is said to execute uniform circular motion if speed is constant

1. Velocity is not constant (it is a vector)

2. Velocity is always tangent to its path

3. Object moving in UCM is experiencing acceleration, which always points to the center of the circle

a) a = Δv/Δt; Δv = v2 – v1

b) Centripetal – used to describe the acceleration of an object undergoing UCM, denoted ac

(1) v is always perpendicular to a_{c}

(2) **a _{c} = v^{2}/r**→ this is the magnitude of a

_{c}

4. F_{c} (centripetal force) = m*a_{c} = mv^{2}/r → **F _{c} = mv^{2}/r**

5. Gravity provides a centripetal force

a) F_{grav} = mv^{2}/r → we also know that F_{grav} = G(mM)/r^{2}

b) mv^{2}/r = GmM/r^{2 } → v^{2 }= GM/r → **v = √(GM/r)**

(1) since m cancels out, you know that any object of any mass orbiting the earth will have the same speed

## III. 4.3 TORQUE

The measure of a forces’s effectiveness at making an object spin (or, more accurately, accelerate rotationally)

### A. Terms

1. Pivot point – center of rotation

### B. Wrench:

1. F = force applied, has torque (torque is not a force itself, but a property of a force)

2. r (radius vector) = point from center of rotation to point of application of force

3. θ = angle between vectors r and F

4. When F ⊥ r, sinθ = 1 and torque is greatest

### C. Torque (τ) = rFsinθ

1. Amount of torque depends on the magnitude of F, length of R, and angle θ

2. Torque units are N·m

### D. Torque (τ) = lF → lever arm method

1. Lever arm (l) is the shortest distance from the pivot point to the line of action of F

## IV. 4.4: EQUILIBRIUM – zero acceleration

### A. Translational equilibrium → F_{net} = 0

### B. Rotational equilibrium → τ_{net} = 0

1. A situation can have a non-zero torque but zero net force

### C. Static equilibrium vs dynamic equilibrium:

1. Static equilibrium means all forces = 0, and velocity = zero

2. Dynamic equilibrium means all forces = 0, but velocity is not zero

## V. 4.5: ROTATIONAL INERTIA – moment of inertia

### A. F_{net} = ma is analogous to τ_{net} = Iα → I = rotational inertia, α = rotational acceleration

### B. Rotational inertia depends on how an object’s mass is distributed around axis of rotation:

1. The farther away the mass is from the axis of rotation, the greater the rotational inertial will be

2. Rotational inertial will be the smallest when the rotation axis passes through the object’s center of mass

## CHAPTER 4 SUMMARY

- Center of mass →
*x*_{CM}= (m_{1}x_{1}+ m_{2}x_{2}….)/(m_{1}+ m_{2}…) - Center of gravity →
*x*_{CG}= (w_{1}x_{1}+ w_{2}x_{2}…)/(w_{1}+ w_{2}…) - Centripetal acceleration →
**a**_{c}= v^{2}/r - Centripetal force →
**F**_{c}= ma_{c}= mv^{2}/r - Torque
→ θ = angle between*τ = rFsinθ**r*and*F*→*τ = lF**l*= lever arm of force)

- Equilibrium
(translational equilibrium*F*_{net}= 0(rotational equilibrium*τ*_{net}= 0

- Rotational inertia (I)
- Mass closer to rotational axis gives smaller I
- Mass farther away from rotation axis gives larger I