Simplified Notes

Chapter 4: Solids, Forces, and Deformations

Ball-Spring Model of a Solid

• Balls - Atoms or molecules

• Springs - Interatomic/intermolecular forces

• When solids contact

  • Compressed Springs → Repulsive force

  • Stretched Springs → Attractive force

Applying the Momentum Principle (At Rest)

• At rest: \overrightarrow p = 0, so \frac {d\overrightarrow p}{dt} = 0

• Thus, net force is zero: \overrightarrow F_{net} = \frac {d\overrightarrow p}{dt} = 0

Estimating Interatomic Bond Length

• M = molar mass

• \rho = Density

• N_A = Avogadro’s Number

• Steps:

  • 1. Volume per atom:

    • V_{atom} = \frac M{N_A\rho}

  • 2. Radius

    • r = \sqrt[3]{\frac {3V_{atom}}{4\pi}}=\sqrt[3]{\frac {3M}{4\pi N_A \rho}}

  • 3. Bond length:

    • L = 2r =2\sqrt[3]{\frac {3M}{4\pi N_A \rho}}

Spring Stiffness Relationships

• Parallel springs:

  • k_{eq} = n \cdot k

• Series springs:

  • k_{chain} = \frac kn

Finding Interatomic Bond Stiffness from Wire Stretching

• When given

  • F

  • \Delta L

  • L

  • A

  • d (Bond length)

• Steps:

  • 1. Wire Stiffness

    • k_{wire} = \frac F{\Delta L}

  • 2. Series bonds:

    • n = \frac Ld

  • 3. Parallel chains

    • m = \frac A{d²}

  • 4. Atom stiffness

    • k_{atom} = \frac{F\cdot L}{\Delta L \cdot A}

Youngs Modulus (E)

• E = \frac \sigma\epsilon = \frac {F/A}{\Delta L/L_0} \Rightarrow \Delta L = \frac {FL_0}{AE}

• F = mg

• A = \frac {\pi d²}{4}

• Young’s modulus is a material property - it does not depend on size or shape

Static Friction

• f_s \le \mu_s N

• If F_{applied} < \mu_s N, then f_s = F_{applied}

• If F_{applied} \ge \mu_s N, object starts moving

  • * N = normal force

Kinetic Friction

• f_k = \mu_k N

• Always opposes motion

• Plug into:

  • F_{net} = ma = F_{applied} - f_k

Friction via Momentum Principle

f_k = F_{net} - ma

Periodic Motion from Displacement-Time Graph

• Amplitude (A) - Max displacement from equilibrium

• Period (T) - Time for one full cycle

Mass-Spring System

• Angular frequency:

  • \omega = \sqrt {\frac km}

• Period:

  • T = \frac {2\pi}\omega = 2\pi \sqrt{ \frac{m}{k}}

• Position function

  • x(t) = A\cos(\omega t + \phi)

Chapter 5: Determining Forces from Motion

Define the System

• Choose the object(s)

• Identify all external forces

• Draw a free-body diagram

Finding an Unknown Force (No Momentum Change)

• \frac {d\overrightarrow p}{dt} = \overrightarrow F_{net} = 0 \Rightarrow \sum \overrightarrow F = 0

• F_{unknown} = -\sum F_{known}

Rate of Change Momentum

• \frac {d\overrightarrow p}{dt} = \overrightarrow F_{net} = m\overrightarrow a

• Magnitude of net force = magnitude of momentum rate of change

• Direction = same as acceleration of \overrightarrow F_{net}

Infer Net Force from Momentum Change

• \overrightarrow F_{net} = \frac {d\overrightarrow p}{dt}

Unknown Force with Momentum Change (1D)

• F_{net} = \frac{dp}{dt} = m\cdot a = \frac {m\Delta v}{\Delta t}

• F_{net} = F_{known} + F_{unknown} \Rightarrow Solve for F_{unknown}

Force Components Relative to Momentum

• Given

  • \overrightarrow F and \overrightarrow p

• Step 1, Normalize momentum

  • \overline p = \frac {\overrightarrow p}{|\overline p|}

• Step 2, Parallel component

  • \overrightarrow F_{||} = (\overrightarrow F \cdot \overline p) \overline p

• Step 3, Perpendicular component

  • \overrightarrow F_{\pm} = \overrightarrow F - \overrightarrow F_{||}

Role of Force Components

• Parallel (\overrightarrow F_{||})

  • Changes Magnitude of momentum

  • Speeds up/slows down

• Perpendicular (\overrightarrow F_{\pm})

  • Changes direction of momentum

  • Causes curved motion

Dot product (Two Methods)

• Geometric Form:

  • \overrightarrow A \cdot \overrightarrow B = |\overrightarrow A||\overrightarrow B|\cos\theta

• Component Form:

  • \overrightarrow A \cdot \overrightarrow B = A_xB_x + A_y+B_y

Circular Motion & Momentum

• 1. Direction of the Net Force for an Object Moving in a Circle at Constant Speed

  • The net force is always directed toward the center of the circle — called the centripetal force

  • Even though the speed is constant, the velocity vector changes direction, so acceleration and net force points inward

• 2. Magnitude and Direction of Net Force on Circular Path at Constant Speed

  • Magnitude of centripetal force F_{net} = \frac {mv²}r

    • m= Mass

    • v=  Constant speed

    • r=Radius of circle

  • Direction: toward the center of the circular path (radially inward).

• 3. Situation Where Both Parallel and Perpendicular Components of Momentum change

  • When an object moves along a curved path with changing speed (e.g.,  a car accelerating while turning),

    • The parallel component changes changes because speed changes (momentum magnitude changes).

    • The perpendicular component changes because the direction of momentum changes (due to turning).

• 4. Rate of Change of Momentum at an Instant for Object on Circular Arc at Varying Speed

  • Use vector decomposition:

    • \frac {d\overrightarrow p}{dt} = \overrightarrow F_{||} + \overrightarrow F_{\pm}

  • Parallel component (tangential acceleration):

    • F_{||}=m\frac{dv}{dt}

  • Perpendicular component (centripetal force):

    • F_{\pm} = m\frac{v²}{r}

  • Total rate of change of momentum is the vector sum of these

• 5. Finding an Unknown Force at an Instant for Object Moving on Circular Arc at Varying Speed

  • Identify known forces and accelerations

  • Apply momentum principle:

    • \overrightarrow F_{net} = \frac {d\overrightarrow p}{dt} = m\overrightarrow a

  • Resolve net force into known and unknown components.

  • Solve for unknown components using vector sums.

• 6. Momentum Principle Explaining Sensations Moving in a Circle

  • You feel a force pushing you outward (centrifugal sensation) even though the real force (net force) points inward.

  • This is due to your inertia wanting to maintain the current velocity vector (tangential), but the net inward force changes direction

  • The inward net force causes centripetal acceleration; your body resists this change, creating the sensation of being pushed outward

Chapter 6: The Energy Principle

Total, Rest, and Kinetic Energy Near Speed of Light

• Total energy:

  • E = \gamma mc²

  • \gamma = \frac {1}{\sqrt{1-\frac{v²}{c}}}

• Rest Energy:

  • E_0 = mc²

• Kinetic Energy:

  • K = E - E_0 = (\gamma -1)mc²

Approximate Kinetic Energy at Low Speeds

• For v«c:

  • K =_{approx} \frac 12 mv²

Work Done by Constant Force

• W = F\cdot d

• F = Force magnitude

• d= Displacement along force direction

Determining Sign of Work

• Positive work: Force has a component in the direction of displacement

• Negative work: Force has a component opposite displacement

• Zero work: Force perpendicular to displacement

Energy Principle (Difference and Update Form)

• Difference form:

  • W_{net} = \Delta K = K_f - K_i

• Update form:

  • K_{new} = K_{old}+W_{net}

Apply Energy Principle for Change in Kinetic Energy

• Calculate net work done on the system to find change in kinetic energy

Processes Involving Change in Rest Energy

• Examples: Nuclear reactions, particle-antiparticle annihilation or creation, chemical reactions where mass-energy equivalence is significant.

Energy Principle for Systems Changing Identity

• When particles are created or destroyed, rest energy changes

• Total energy conservation still applies, accounting for rest mass changes

Energy Unit Conversion: Joules and Electron Volts (eV)

• 1 eV = 1.602 × 10^{-19} J

• To convert:

  • E(J) = E(eV)1 × 10 × 10^{-19}

  • E(eV) =\frac {E(J)}{1.602 × 10^{-19}}