Unit 2 (2.8-2.10)

2.8 Springs

F = -kx or F = kx

  • whenever the displacement is to the right the force is to the left

Example 1

Solution

Example 2

Solution

Second Order Differential functions have solutions that are cosine/sin

We found the second differential equation with example 2

Example 4

Solution

Springs in Parallel

X is constant, K is variable

Distances stretched/compressed are equal

  • X is constant in F= kx

Spring Constant, K, is total of two springs

  • Ktotal = K1 + K2

Example 1

Springs in Series

Distance (X) is totaled between the two springs

  • Xtotal = X1 + X2

The inverse of Keq is total of the inverse of the sum of the spring constants

  • 1Keq=1K1+1K2\frac{1}{K_{eq}}=\frac{1}{K_1}+\frac{1}{K_2}

  • The Keq will never be greater than the individual spring constants

  • Still have to solve the problem using Keq, so you have to take the inverse of 1keq\frac{1}{k_{eq}}

2.9 - Resistive Force

Resistive Force = velocity-dependent force that resists motion.∑\sum_{}^{}

  • Changes with speed

  • EX: air resistance, drag, etc

Fr = -kv

  • always opposite of motiong

  • k is a CONSTANT & dependent on MANY things

    • surface area

    • viscosity

    • density

    • etc

At t = 0, there is little FR since there is little velocity yet

At t > 0, there is some FR but not enough to balance gravity

After a long time, FR BALANCES the Fg

  • resistive force balances force of gravity

  • FR = Fg ———→ ∑F\sum_{}^{}F = Fg - FR

  • ∑F\sum_{}^{}F = 0, since Fg = Fr

  • ma = 0, so a = 0

Creating Differential Equations

1.) ∑F\sum_{}^{}F = ma & FR = -bv

2.) ∑F=mg−bv\sum_{}^{}F=mg-bv

3.) ma = mg - bv

4.) mdvdt\frac{dv}{dt}= mg -bv

1.) Sum of forces is equal to mass*acceleration & Resistive Force = -bv or -kv

2.) In free fall, the only forces are gravity and FR

  • Fr is always opposite of motion, so it is subtracted from mg (downwards motion)

3.) Sub in sum of forces with mass*acceleration

4.) Acceleration is EQUAL to the derivative of velocity

mdvdt\frac{dv}{dt}= mg -bv

Terminal Velocity

Terminal Velocity: Velocity reached when Resistive Forces balances Gravitational Force in Free Fall

  • Net force acting on object = 0

  • Reaches this state of equilibrium EVEN IF it initially was Larger OR Smaller than FG

Method 1:

Use Sum of Forces

  • ∑F=mg−bv\sum_{}^{}F=mg-bv

    • Where ∑F=0\sum_{}^{}F=0

      • v = </u></strong></span><spanstyle="color:rgb(251,130,218);"><strong><u>mgb</u></strong></span><span style="color: rgb(251, 130, 218);"><strong><u>\frac{mg}{b}

Graphing Velocity v Time graph under Resistive Force

mgk=vT\frac{mg}{k}=v_{T}

  • object never REALLY reaches Terminal Velocity

    • so VT is an asymptote

Graphing Acceleration v Time Graph under Resistive Force

Acceleration is initially gravity

  • As time passes, acceleration slowly decreases until it EQUALS 0

    • VT is reached

    • Resistive Force BALANCES gravity

Graphing Position v Time Graph under Resistive Force

X is positive because downward is t

Slope at t = 0 must be 0 AND x must be 0

  • although slope curves at start, slope EVENTUALLY BECOMES CONSTANT (linear)

    • Because acceleration becomes 0 after a long time, resistive forces balance out gravity’s acceleration

      • THE CONSTANT SLOPE = VT (mgk\frac{mg}{k} )

2.10 - Circular Motion

Centripetal Force, Centripetal Acceleration & Tangential Motion

Centripetal Force is the Force that points to the center of the Circular Motion

  • Centripetal Acceleration is in the same direction as the Centripetal Force

*Centripetal Force depends on MANY variables to keep the motion circular

  • Mass

  • speed

  • circle radius

UNDER UCM

  • velocity magnitude is constant

  • acceleration only redirects the velocity vector

    • does not change the magnitude

Tangential Velocity is the velocity at a certain frame

  • Magnitude is always the same

  • Direction changes (towards the center of the circle)

    • a→=vt→\overrightarrow{a}=\overrightarrow{\frac{v}{t}}

    • Acceleration and Velocity share the same direction,

    • Centripetal Acceleration means that the tangential velocity always changes direction towards the center

NOT UCM

   

When a tangential force changes the tangential velocity’s magnitude, (creates a tangential acceleration)

For instance, speeding up a ball from rest to start making it go into a circle

Centripetal Acceleration

velocity = distance/ time ———> distance = circumference

  • circumference =2πr\pi r

  • Ac = 2π(2πrt)t\frac{2\pi\left(\frac{2\pi r}{t}\right)}{t} = 4π2rt2\frac{\pi^2r}{t^2}

  • T = 2πrv\frac{2\pi r}{v} ———→ a=2πv2πrv=v2ra=\frac{2\pi v}{\frac{2\pi r}{v}}=\frac{v^2}{r}

Example 1

Solution

v=2πrtv=\frac{2\pi r}{t} v = 2⋅π⋅0.30.5\frac{2\cdot\pi\cdot0.3}{0.5} = 37.56 m/s

a = 2πvt\frac{2\pi v}{t} = 2⋅π⋅37.560.5\frac{2\cdot\pi\cdot37.56}{0.5} =

  • 4π2rt2\frac{\pi^2r}{t^2} = 47 m/s

Critical Speed

VCritical is the speed needed for the cart to make the full circular path

  • vcritical is tangential velocity

SO Velocity at the top CANNOT EQUAL 0

example 1

Solution

vcrit = gr\sqrt{gr}

Barely completes the loop = FN is 0

so centripetal force = mg

mv2r=mg\frac{mv^2}{r}=mg

Conical Pendulum

Example 1

Solution

Find the centripetal force using the X-Component of FT

Use the Y-Component of FT to solve for FT

Input that into original equation to find V

Finding Net Acceleration when there is atan

Full Body Diagram

FG is the TANGENTIAL ACCELERATION

  • Changes MAGNITUDE Of tangential velocity

Find acceleration of FN and FG

Find NET acceleration using Pythagorean Theorem

  • acceleration is a vector