04_Harmonic

1. Equilibrium Overview

Conditions for Equilibrium

For a system to be in equilibrium, both its linear momentum (\(\vec{P}\)) and angular momentum (\(\vec{L}\)) must remain constant.

1. Translational Equilibrium: Net force must be zero.

   $$
   

   \sum \vec{F}_{net} = 0 \implies \vec{P} = \text{constant} $$

2. Rotational Equilibrium: Net torque must be zero, measured from any arbitrary point.

   \[ \sum \vec{\tau}_{net} = 0 \implies \vec{L} = \text{constant} \]

Static Equilibrium is a special case where the system is completely at rest:

\[ \vec{P} = 0 \quad \text{and} \quad \vec{L} = 0 \]

Proof: Independence of Torque Origin in Equilibrium

If a system is in translational equilibrium (\(\sum \vec{F}_i = 0\)) and rotational equilibrium about an origin \(O\) (\(\sum \vec{\tau}_{i, O} = 0\)), then the net torque about any other arbitrary point \(O'\) is also zero.

• schematic diagram from the hand-written notes of Mr. ztWang

Proof:

Let the position vector of point \(O'\) relative to \(O\) be \(\vec{r}'\). The position of the \(i\)-th force relative to \(O'\) is \(\vec{r}_{i, O'}\). Thus, the position relative to \(O\) is \(\vec{r}_{i, O} = \vec{r}' + \vec{r}_{i, O'}\).

\[ \sum_i \vec{\tau}_{i, O} = \sum_i \vec{r}_{i, O} \times \vec{F}_i \]

\[ = \sum_i (\vec{r}' + \vec{r}_{i, O'}) \times \vec{F}_i \]

\[ = \vec{r}' \times \left( \sum_i \vec{F}_i \right) + \sum_i \vec{r}_{i, O'} \times \vec{F}_i \]

Since \(\sum_i \vec{F}_i = 0\), the first term vanishes:

\[ 0 = 0 + \sum_i \vec{\tau}_{i, O'} \]

\[ \implies \sum_i \vec{\tau}_{i, O'} = 0 \]

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2. Center of Gravity (COG)

Definition: The point at which the total gravitational force \(\vec{F}_g\) effectively acts on a body.

Theorem: Under uniform gravity, the Center of Gravity (COG) coincides with the Center of Mass (COM).

Proof:

The actual net torque due to gravity on all individual mass elements is:

\[ \vec{\tau}_{net} = \sum_i \vec{r}_i \times m_i \vec{g}_i \]

The effective torque acting at the COG is:

\[ \vec{\tau}_{eff} = \vec{r}_{cog} \times \left( \sum_i m_i \vec{g}_i \right) \]

For uniform gravity (\(\vec{g}_i = \vec{g}\)):

\[ \vec{\tau}_{net} = \sum_i m_i \vec{r}_i \times \vec{g} \]

\[ \vec{\tau}_{eff} = \vec{r}_{cog} \times (M\vec{g}) \]

If we set \(\vec{r}_{cog} = \vec{r}_{com} = \frac{\sum m_i \vec{r}_i}{M}\):

\[ \vec{\tau}_{eff} = \frac{1}{M} \left( \sum_i m_i \vec{r}_i \right) \times (M\vec{g}) = \sum_i m_i \vec{r}_i \times \vec{g} = \vec{\tau}_{net} \]

Therefore, we can treat the entire gravitational force \(M\vec{g}\) as acting at a single point (the COG/COM) to calculate net torque.

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3. Static Equilibrium Examples

Example A: Standing on a Supported Beam

A uniform beam of length \(L\) and mass \(M\) is hinged to a wall and supported by a wire at an angle \(\theta\). A person of mass \(m\) stands at a distance \(d\) from the wall.

1. Torque Balance (around the hinge):

   $$
   

   mgd + Mg\frac{L}{2} = TL\sin\theta \implies T = \frac{mgd + Mg\frac{L}{2}}{L\sin\theta} $$

2. Force Balance:

   • Horizontal: \(N_x = T\cos\theta\)

   • Vertical: \(N_y = mg + Mg - T\sin\theta\)

Example B: A Leaning Ladder

A uniform ladder of length \(L\) and mass \(M\) leans against a perfectly smooth wall. What is the minimal angle \(\theta\) with the floor required to maintain static equilibrium without slipping?

• Method 1 (Standard Torque Balance):

  

  Take the torque around the top of the ladder (point A). The forces at the bottom are the normal force \(N_1\) and static friction \(\mu_s N_1\). The normal force from the wall is \(N_2\).

  • Vertical Forces: \(N_1 = Mg\)

  • Horizontal Forces: \(N_2 = \mu_s N_1 = \mu_s Mg\)

  • Torque around top: \(Mg \frac{L}{2} \cos\theta + \mu_s N_1 L \sin\theta = N_1 L \cos\theta\)

    Substitute \(N_1 = Mg\):

    \[ \mu_s Mg L \sin\theta = \left(Mg L - Mg \frac{L}{2}\right) \cos\theta \]
    \[ \mu_s \sin\theta = \frac{1}{2} \cos\theta \implies \tan\theta = \frac{1}{2\mu_s} \]

• Method 2 (Concurrent Forces):

  For a body in equilibrium under three non-parallel forces (Gravity, Wall Normal Force, and the Total Floor Reaction), their lines of action must intersect at a single point \(D\).

  

  Let \(\varphi\) be the angle of the floor reaction force.

  \[ \tan\varphi = \frac{\mu_s N_1}{N_1} = \mu_s \]

  From the geometry of the intersection point \(D\):

  \[ \tan\varphi = \frac{\frac{L}{2} \cos\theta}{L \sin\theta} = \frac{1}{2}\cot\theta \]

  Equating the two:

  \[ \mu_s = \frac{1}{2}\cot\theta \implies \tan\theta = \frac{1}{2\mu_s} \]

Example C: Wheel over a Step

A wheel of radius \(R\) and mass \(M\) is pulled by a horizontal force \(F\) at the top to overcome a step of height \(h\).

• Q1: Minimal \(F\)? Take the torque around the corner of the step.

  The lever arm for \(F\) is \((2R - h)\). The lever arm for gravity is determined by the Pythagorean theorem: \(\sqrt{R^2 - (R-h)^2}\).

  \[ F(2R - h) = Mg \sqrt{R^2 - (R-h)^2} \implies F = Mg\frac{\sqrt{R^2 - (R-h)^2}}{2R - h} \]

• Q2: Direction of normal force \(\vec{N}\) at the corner?

  For \(\vec{\tau}_{net, c} = 0\), the line of action of the normal force \(\vec{N}\) must pass through the intersection of the applied force \(F\) and gravity \(Mg\) (Concurrent forces theorem).

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4. Simple Harmonic Motion (SHM)

Stable Equilibrium and Taylor Expansion

A system is in stable equilibrium if:

1. \(F = -\frac{dU}{dx} = 0\) (Equilibrium state)

2. \(\frac{d^2U}{dx^2} > 0\) (Stable; corresponds to a minimum in the potential energy \(U(x)\))

For any potential energy function \(U(x)\) near a stable equilibrium point \(x_0\), we can apply a Taylor expansion:

\[ U(x) = U(x_0) + \left.\frac{dU}{dx}\right|_{x_0} (x-x_0) + \frac{1}{2} \left.\frac{d^2U}{dx^2}\right|_{x_0} (x-x_0)^2 + \dots \]

Since \(\frac{dU}{dx} = 0\) at equilibrium, for small displacements (\(x \sim x_0\)):

\[ U(x) \approx U(x_0) + \frac{1}{2}k(x-x_0)^2 \quad \text{where} \quad k = \left.\frac{d^2U}{dx^2}\right|_{x_0} \]

This represents a universal family of motion: Simple Harmonic Motion.

Equation of Motion

The restoring force is linear:

\[ F = -\frac{dU}{dx} = -k(x-x_0) \]

\[ a = \frac{F}{m} = -\frac{k}{m}(x-x_0) \implies \ddot{x} = -\frac{k}{m}(x-x_0) \]

Kinematic Solutions

The general solution to this differential equation is:

\[ x - x_0 = A \cos(\omega t + \phi) \]

• Angular Frequency: \(\omega \equiv \sqrt{\frac{k}{m}}\)

• Amplitude: \(A\)

• Period: \(T \equiv \frac{2\pi}{\omega}\)

• Frequency: \(f \equiv \frac{1}{T} = \frac{\omega}{2\pi}\) (Units: \(\text{s}^{-1}\) or Hertz, Hz)

• Phase Constant: \(\phi\)

By taking derivatives, we find velocity and acceleration:

• Velocity: \(v = -\omega A \sin(\omega t + \phi) = \omega A \cos(\omega t + \phi + \frac{\pi}{2})\)

• Acceleration: \(a = -\omega^2 A \cos(\omega t + \phi) = \omega^2 A \cos(\omega t + \phi + \pi)\)

Key Kinematic Notes:

1. \(a = -\omega^2(x-x_0)\) (Acceleration is strictly proportional to displacement, opposite in sign).

2. \(\{x, v, a\}\) share the same frequency \(\omega\) but have different phases (shifted by \(\pi/2\) increments).

Energy in SHM

• Kinetic Energy:

  \[ K = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t + \phi) = \frac{1}{2}kA^2 \sin^2(\omega t + \phi) \]

• Potential Energy:

  \[ U = \frac{1}{2}k(x-x_0)^2 = \frac{1}{2}kA^2 \cos^2(\omega t + \phi) \]

• Total Mechanical Energy:

  \[ E = K + U = \frac{1}{2}kA^2 \]

Energy Frequency (Double Angle Identity):

Using trigonometric identities, we can rewrite the energy terms:

\[ K = \frac{1}{2}kA^2 \left( \frac{1 - \cos(2\omega t + 2\phi)}{2} \right) \]

\[ U = \frac{1}{2}kA^2 \left( \frac{1 + \cos(2\omega t + 2\phi)}{2} \right) \]

Note: The kinetic and potential energies oscillate with a frequency of \(2\omega\) (twice the frequency of the position/velocity oscillation), transforming completely into one another twice every full period \(T\).