1 Basic Newtonian Physics

1.1 Position and its derivatives

1.1.1 Position as function of time

x(t)  or  < x(t), y(t), z(t) >

1.1.2 Velocity as function of time is rate of change of Position with respect to time.

v(t) = dx(t)/dt

or < v_{x}(t), v_{y}(t), v_{z}(t) >

= d < x(t), y(t), z(t) > /dt

=< \frac{dx(t)}{dt}, \frac{dy(t)}{dt}, \frac{dz(t)}{dt} >

1.1.3 Acceleration as function of time is rate of change of Velocity with respect to time.

a(t) = dv(t)/dt = d^{2}x(t)/dt^{2}

or < a_{x}(t), a_{y}(t), a_{z}(t) >

=< \frac{d^{2}x(t)}{dt^{2}}, \frac{d^{2}y(t)}{dt^{2}}, \frac{d^{2}z(t)}{dt^{2}} >

1.1.4 Units m (meters) and s (seconds)

Position: m Velocity: m/s Acceleration: m/s/s = \frac{m}{s^{2}}

1.2 Force

Force: F = ma (Newton’s 2nd Law (with m = constant)) or ∑ F_{x} = ma_{x} ∑ F_{y} = ma_{y} ∑ F_{z} = ma_{z} Units: Newtons (N) = \frac{kgm}{s^{2}} Acceleration due to gravity, at sea level, on earth, is: 9.8m/s^{2} = g Force due to gravity (sea level, earth) is: m · g Newtons Frictional Forces: When stationary, the coefficient of static friction is active: μ_{s} When moving, the coefficient of kinetic (sliding) friction is active: μ_{k} Force due to friction acts opposite other forces, and is equal to F_{N}μ_{(s or k)} = NormalF orce · F rictionCoeff Example:

∑ F = ma = \{ \begin{matrix}F_{in} − m_{g} · μ_{s}, & if at rest and|F_{i}n| < m_{g}μ_{s} \\ F_{in} − m_{g} · μ_{k}, & ifF_{i}n > m_{g}μ_{k} \end{matrix}

1.3 Energy (work) and Energy Flow

Energy: W = F x (force through a distance, if F is constant) Units: Joules = Nm = kgm^{2}/s^{2} Power: = Energy flow, or the time rate of work = dW (t)/dt Units: Watts = Joules / s = kgm^{2}/s^{3} Potential Energy: Potential to do work because of position in a field

In a conservative field, potential (and work) only depends on initial and final positions, not on path or time taken to traverse it.

Example: Gravitational Potential Energy = W_{g} = m · g · h(h is height) Kinetic Energy: Potential to do work because of mass in motion W_{v} = 1/2mv^{2} Energy is Conserved, A Very Important Physical Concept!!! Example: Potential and Kinetic energy, dropping a ball InitialP otential + InitialKinetic

= F inalP otential + F inalKinetic m · g · h_{i} + 0 = 0 + 1/2mv_{f}^{2} Mass cancels!! (It took a long time to prove this) Mass is Conserved, Another Very Important Physical Concept!!! Example: If I pound water into a hose, it either comes out the other end

or the hose eventually blows up. Mass is conserved.

1.4 Rotational Frames of Reference

All of the above still works, just adjusted slightly Force → Torque τ = F_{t}r (tangential force times radius) Position → Angle θ Velocity → Angular velocity ω Mass → Moment of Inertia I =

mr^{2}  for a point mass at distance r
1/2mr^{2} for a disk about its center
2/5mr^{2} for a sphere about its central axis

Example: A fulcrum with a pivot at 1/3 its length will balance with twice as much mass on the short end as on the long end.

Assume it’s balanced and at rest (acceleration = 0) ∑ τ = I0 = 0 (∑ F = ma = 0) so m_{longend} · g · 2/3 = m_{shortend} · g · 1/3 thus 2m_{longend} = m_{shortend} Rotational Energy:

Kinetic Energy = 1/2Iω^{2}

1.5 The Basic 2nd Order Mechanical System

A spring with constant k F orce = −ky A Mass with mass m Some oil with constant damping R F orce = −Rv

y is the signed displacement from equilibrium (at rest with y = 0). Minus sign on spring term means force acts to restore mass to rest position.

Minus sign on damping term means force acts against motion, proportional to velocity.

∑ F = ma = −ky − Rv_{y} + mg = ma_{y} assume gravity negligible: ∑ F = ma = −ky − Rv_{y} = ma_{y} m\frac{d^{2}y}{dt^{2}} + R\frac{dy}{dt} + ky = 0 or \frac{d^{2}y}{dt^{2}} + \frac{R}{m} \frac{dy}{dt} + \frac{k}{m}y = 0 Solutions: Case 1: (R/m)^{2} − 4(k/m) < 0 (Under damped)

y = e^{−αt}(c_{1}sin(βt) + c_{2}cos(βt))

Case 2: (R/m)^{2} − 4(k/m) = 0 (Critically damped)

y = c_{1}e^{−αt} + c_{2}te^{−αt}

Case 3: (R/m)^{2} − 4(k/m) > 0 (Over damped) y = c_{1}e^{−α_{1}t} + c_{2}e^{−α_{2}t}

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