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)$1 × 10 × 10^{-19}$

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