Displacement (d)

d = Δx = final position - initial positon = net distance (plus direction)

Average velocity (v)

v = Δx/Δt = d/Δt

Average acceleration (a)

a = Δv/Δt

What assumption do we make with the BIG 5 equations? What are the 3 most frequently used equations?

That acceleration is constant (uniformly accelerated motion)

2. Missing d

3. Missing vf (v final, v)

5. Missing t

2. Missing d

3. Missing vf (v final, v)

5. Missing t

BIG 5:

1. Missing a

1. Missing a

d = ½(v₀ + v)t

BIG 5:

2. Missing d

2. Missing d

v = v₀ + at

BIG 5:

3. Missing v (v final)

3. Missing v (v final)

d = v₀²t + ½at²

BIG 5:

4. Missing v₀

4. Missing v₀

d = vt - ½at²

BIG 5:

5. Missing t

5. Missing t

v² = v₀² + 2ad

For projection motion, we assume that the downward direction is...

NEGATIVE direction.

DOWNWARD = NEGATIVE DIRECTION.

DOWNWARD = NEGATIVE DIRECTION.

PROJECTILE MOTION - Horizontal Motion

(in the x-direction)

Give the equation for:

1. Displacement

2. Velocity

3. Acceleration

(in the x-direction)

Give the equation for:

1. Displacement

2. Velocity

3. Acceleration

1. d = v₀t (in the x-direction)

2. v₀x = vx (velocity is CONSTANT in the x-direction!!)

v₀x = v₀cosθ₀

3. ax = 0 (acceleration in the x-direction is ZERO!)

2. v₀x = vx (velocity is CONSTANT in the x-direction!!)

v₀x = v₀cosθ₀

3. ax = 0 (acceleration in the x-direction is ZERO!)

PROJECTILE MOTION - Vertical Motion

Give the equation for:

1. Displacement

2. Velocity

3. Acceleration

Give the equation for:

1. Displacement

2. Velocity

3. Acceleration

1. y = v₀yt + ½(-g)t² (in the y-direction)

2. vy = v₀y + (-g)t (in the y-direction)

v₀y = v₀sinθ₀

3. ay = -g (accerleration in the y-direction)

2. vy = v₀y + (-g)t (in the y-direction)

v₀y = v₀sinθ₀

3. ay = -g (accerleration in the y-direction)

Newton's 1st Law

Fnet = 0

velocity (v) = constant!

∴a = 0 --> F = 0

velocity (v) = constant!

∴a = 0 --> F = 0

Newton's 2nd Law

Fnet = m a

Newton's 3rd Law

F₁₂ = F₂₁

Action-Reaction pair

Action-Reaction pair

Weight

w = mg

Force due to gravity (Fg) = weight = mg

Force due to gravity (Fg) = weight = mg

Gravitational force (Fg)

Fg = (GMm)/r²

since w = Fg,

mg = (GMm)r²

∴g = GM/r²

since w = Fg,

mg = (GMm)r²

∴g = GM/r²

Kinetic friction

Ff = μkFN

μk = coefficient of KINETIC friction

FN = normal force

μk = coefficient of KINETIC friction

FN = normal force

Static friction

Ff, max = μsFN

μs = coefficient of STATIC friction

FN = normal force

μs = coefficient of STATIC friction

FN = normal force

What is the relationship between μk and μs, max?

μs, max > μk

ALWAYS!

ALWAYS!

What is the force due to gravity acting PARALLEL to the inclined plane? (sliding down the plane)

F = mgsinθ

θ = angle between incline and horizontal

θ = angle between incline and horizontal

What is the force due to gravity acting PERPENDICULAR to the inclined plane?

F = mgcosθ

θ = angle between incline and horizontal

θ = angle between incline and horizontal

Center of mass xCM

xCM = (m₁x₁ + m₂x₂ + m₃x₃...)/(m₁ + m₂ + m₃...)

Center of gravity xCG

xCG = (w₁x₁ + w₂x₂ + w₃x₃...)/(w₁ + w₂ + w₃...)

In a uniform gravitational field (constant g)...

xCM = xCG

Centripetal acceleration (ac)

ac = v²/r

Centripetal force (Fc)

Fc = mac = mv²/r

Torque (τ)

τ = rFsinθ

r = distance from pivot point to F

F = force applied (Fapp)

θ = angle between r and F

r = distance from pivot point to F

F = force applied (Fapp)

θ = angle between r and F

When does max torque occur?

When θ =90°

τ= r F

τ= r F

What does it mean to be in ROTATIONAL EQUILIBRIUM?

τcw = τccw

∴τnet = 0 (net torque = 0)

clockwise torque = counterclockwise torque

∴τnet = 0 (net torque = 0)

clockwise torque = counterclockwise torque

τnet (net torque)

τnet = I α

I = inertia

α = rotational acceleration

*compare to Fnet = ma

I = inertia

α = rotational acceleration

*compare to Fnet = ma

Is the inertia greater for a solid sphere or a hollow sphere?

If the same net torque was applied to each sphere, which would spin faster?

If the same net torque was applied to each sphere, which would spin faster?

Hollow sphere has GREATER inertia than solid sphere.

Solid sphere would spin faster because it has less inertia!

Solid sphere would spin faster because it has less inertia!

What does inertia depend on?

It depends on how a mass is distributed in an object.

Work (W)

W = Fdcosθ

Unit = Joules (J)

F = force applied

d = displacement

θ = angle between F and d

Wmax occurs when θ = 0

Unit = Joules (J)

F = force applied

d = displacement

θ = angle between F and d

Wmax occurs when θ = 0

Power (P)

P = W/t

Unit: J/s = Watts (W)

W = Work

t = time

Unit: J/s = Watts (W)

W = Work

t = time

Kinetic Energy (KE)

KE = ½mv²

Unit: Joules (J)

KE always has a + value; object has to be in motion!

Unit: Joules (J)

KE always has a + value; object has to be in motion!

Work-energy theorem (Wtotal)

Wtotal = ΔKE = KEf - KE₀

Gravitational Potential Energy (PE)

PE = mgh

m = mass

g = acceleration due to gravity

h = height

m = mass

g = acceleration due to gravity

h = height

Total Mechanical Energy (E)

E = KE + PE

Conservation of Energy (ideal frictionless world)

E₀ = E

KE₀ + PE₀ = KE + PE

KE₀ + PE₀ = KE + PE

Work Energy Theorem (including PE)

W = ΔKE = -ΔPE

Conservation of Energy (when friction is taken into account)

KE₀ + PE₀ - Wf = KE + PE

Wf = work done by FRICTION!

Wf = work done by FRICTION!

Momentum (p)

p = mv

*p is always in the same direction as v

Unit = kg∗m/s

*p is always in the same direction as v

Unit = kg∗m/s

Conservation of total momentum

p₀ = p

m₁v₁ = m₁v₁'

total p₀ = total pf

m₁v₁ = m₁v₁'

total p₀ = total pf

Impulse-momentum theorem (J)

J = Δp = FΔt

change in momentum (Δp) = impulse (J) = FΔt

change in momentum (Δp) = impulse (J) = FΔt

During a collision (short duration of interaction between 2 objects)...

1. Total momentum (p) is ALWAYS conserved!

2. KE is NOT always conserved!

2. KE is NOT always conserved!

1. Elastic collision

(think of 2 pool balls)

1. p is conserved

2. KE is ALSO conserved!

∴p₀ = pf

∴KE₀ = KEf

1. p is conserved

2. KE is ALSO conserved!

∴p₀ = pf

∴KE₀ = KEf

2. Inelastic collision

(think of a car crash)

1. p is conserved

2. KE is NOT conserved: one object becomes deformed; loss in energy

m₁v₀₁ + m₂v₀₂ = m₁v₁' + m₂v₂'

1. p is conserved

2. KE is NOT conserved: one object becomes deformed; loss in energy

m₁v₀₁ + m₂v₀₂ = m₁v₁' + m₂v₂'

3. Perfectly inelastic collision

(think of 2 objects becoming stuck together)

1. p is conserved

2. objects get stuck together and move with the same v' (final velocity!)

m₁v₀₁ + m₂v₀₂ = (m₁ + m₂)v'

1. p is conserved

2. objects get stuck together and move with the same v' (final velocity!)

m₁v₀₁ + m₂v₀₂ = (m₁ + m₂)v'

During the collision, there is a short duration of interaction between the 2 objects so we can apply...

Newton's 3rd Law: action-reaction pair!

F₁₂ = F₂₁

F₁₂ = F₂₁

Angular momentum (L)

L = Iω

I = moment of inetia

ω = angular velocity

*compare to p = mv

I = moment of inetia

ω = angular velocity

*compare to p = mv

Conservation of total angular momentum

L₀ = L'

Density (ρ)

ρ = m/V

m = mass

V = volume

m = mass

V = volume

Specific gravity (sp gr)

sp gr = ρ/ρH₂O

ρ = density of object or fluid

ρH₂O = density of water

ρ = density of object or fluid

ρH₂O = density of water

What is the ρ of water?

ρH₂O = 1000 kg/m³ = 1 g/cm³ = 1 kg/L

Pressure (P)

F/A

F = perpendicularly applied force

A = area

Unit = N/m₂ = Pascal (Pa)

1atm = 100 kPa

F = perpendicularly applied force

A = area

Unit = N/m₂ = Pascal (Pa)

1atm = 100 kPa

Gauge Pressure (Pg)

Pressure felt at a particular depth = hydrostatic pressure

Pg = ρ g D

ρ = density of fluid

g = accl due to gravity

D = depth

Pg = ρ g D

ρ = density of fluid

g = accl due to gravity

D = depth

Total hydrostatic pressure (Ptotal)

Ptotal = Patm + Pg

Patm = pressure at the surface

Pg = gauge pressure

Patm = pressure at the surface

Pg = gauge pressure

Buoyant Force (Fb)

*Archimedes Principle

*Archimedes Principle

Fb = ρ V g

ρ = density of fluid

V = Vsub = volume of object submerged = volume of fluid displaced

g = accl due to gravity

ρ = density of fluid

V = Vsub = volume of object submerged = volume of fluid displaced

g = accl due to gravity

Archimedes Principle

There is an upward acting force on an object that is rising, sinking, floating, or completely submerged in a fluid. This force is called the buoyant force (Fb) and it equals the weight of the fluid displaced.

Case 1: Object is sinking

ρ object > ρ fluid

Case 2: Object is completely submerged and reaches the bottom of the tank

Fnet = Fn + Fb - Fg = 0

∴Fn = Fg - Fb

ρ object > ρ fluid

*Fn = normal force: weight of the object inside the fluid = apparent weight

∴Fn = Fg - Fb

ρ object > ρ fluid

*Fn = normal force: weight of the object inside the fluid = apparent weight

Case 3: Object is rising up in the fluid

Fnet = Fb - Fg = ma

ρ fluid > ρ object

ρ fluid > ρ object

Case 4: Object is floating on the surface

Fnet = Fb - Fg = 0

∴Fb = Fg (= weight of object)

ρ V g = m g

ρ V = m = ρ₀V₀ (object)

ρ fluid > ρ object

V/V₀ = ρ₀/ρ = specific gravity

∴Fb = Fg (= weight of object)

ρ V g = m g

ρ V = m = ρ₀V₀ (object)

ρ fluid > ρ object

V/V₀ = ρ₀/ρ = specific gravity

Remember, ice is always

90% submerged! Think of the Titanic...

What are the 4 conditions to have an IDEAL FLUID?

1. No or negligible viscosity.

2. Laminar flow (continuous, streamline flow).

3. Fluid must be incompressible.

4. Flow must be continuous.

2. Laminar flow (continuous, streamline flow).

3. Fluid must be incompressible.

4. Flow must be continuous.

Flow rate (f)

f = A v

A: cross-sectional area

v: velocity of fluid

Unit: m³/s

A: cross-sectional area

v: velocity of fluid

Unit: m³/s

Continuity equation (f₁ = f₂)

A₁v₁ = A₂v₂

Bernoulli's equation (ideal fluid)

describes ideal fluid flow through a pipe

*looks similar to conservation of energy equation

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

*pgh is NOT gauge pressure, but the PE of the fluid at a certain height, h.

*looks similar to conservation of energy equation

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

*pgh is NOT gauge pressure, but the PE of the fluid at a certain height, h.

Apply Bernoulli's equation to a HORIZONTAL PIPE.

Since h₁ = h₂, the equation looks like this:

P₁ + ½ρv₁² = P₂ + ½ρv₂²

P₁ + ½ρv₁² = P₂ + ½ρv₂²

For a horizontal pipe, describe the relationships between pressure and velocity.

1. If v₁ > v₂, then P₁ < P₂.

2. If v₂ > v₁, then P₂ < P₁.

2. If v₂ > v₁, then P₂ < P₁.

Objects will move from _______ pressure to _______ pressure. Think of an example.

High P --> Low P

Think of air planes and the lift force!

Think of air planes and the lift force!

Toricelli's result (velocity of fluid coming out of a hole in a tank)

*v efflux

*v efflux

v = √(2gD)

Stress

stress = F/A

Strain

strain = ΔL/L₀

ΔL = change in length

L₀ = initial length

ΔL = change in length

L₀ = initial length

Hooke's Law

stress = modulus ∗ strain

Tension or compression (ΔL)

ΔL = FL₀/EA

*FLEA

*FLEA

Shear (X)

X = FL₀/AG

*FLAG

*FLAG

What is the elementary charge of an e-?

1.6 ∗ 10⁻¹⁹ C

Coulomb's law (Electric force: force on a charged particle)

F = kQq/r²

k = Coulomb's constant = 9 ∗10⁹ Nm²/C²

Q = charge that applies force

q = test charged being acted on by Q

r = distance between 2 charges

Positive F = repelling force

Negative F = attractive force

k = Coulomb's constant = 9 ∗10⁹ Nm²/C²

Q = charge that applies force

q = test charged being acted on by Q

r = distance between 2 charges

Positive F = repelling force

Negative F = attractive force

1 microcoulomb equals how many Coulomb?

1μC = 10⁻⁶ C

1C = 10⁶μC

1C = 10⁶μC

State the principal of superposition.

The net electric force on a charge (q) due to a collection of other charges (Q's) is equal to the sum of the individual forces that each of the Q's alone exerts on q