Theoretical Physics Document: Force and Dynamics

1. Newton's Laws and Momentum

In physics, a Force ($\vec{F}$) is defined as an interaction that causes an object with mass to accelerate. It is a fundamental vector quantity that changes the momentum of a system.

1.1. Newton's Second Law: The ODE of Motion

The Second Law is the core equation in classical dynamics. It states that the net external force ($\vec{F}_{net}$) equals the time rate of change of linear momentum ($\vec{p}$).

$$\vec{F}_{net} = \frac{d\vec{p}}{dt} \quad \text{where } \vec{p} = m\vec{v} \quad \text{(1.1)}$$

For objects with constant mass ($m$), this simplifies to the product of mass and acceleration ($\vec{a} = d\vec{v}/dt$), resulting in a second-order Ordinary Differential Equation (ODE):

$$\vec{F}_{net}(t, \vec{r}, \dot{\vec{r}}) = m \frac{d\vec{v}}{dt} = m \frac{d^2\vec{r}}{dt^2} \quad \text{(1.2)}$$

Discussion: The force $\vec{F}_{net}$ can depend on time ($t$), position ($\vec{r}$), and velocity ($\dot{\vec{r}}$). Solving this ODE, $\vec{r}(t)$, completely determines the trajectory of the particle, given the initial position $\vec{r}(t_0)$ and initial velocity $\vec{v}(t_0)$.

1.2. Derivation of the Impulse-Momentum Theorem

The Impulse-Momentum Theorem connects the total effect of a force over time (Impulse, $\vec{J}$) to the resulting change in momentum ($\Delta \vec{p}$).

$$\vec{J} = \int_{t_i}^{t_f} \vec{F}_{net}(t) dt = \vec{p}_f - \vec{p}_i = \Delta \vec{p} \quad \text{(1.3)}$$

Proof of (1.3):

We start with Newton's Second Law (1.1): $\vec{F}_{net} = d\vec{p}/dt$. Rearranging and integrating both sides over the time interval from $t_i$ to $t_f$:

$$\int_{t_i}^{t_f} \vec{F}_{net}(t) dt = \int_{t_i}^{t_f} \frac{d\vec{p}}{dt} dt$$

The integral on the right-hand side is simplified by the Fundamental Theorem of Calculus (since $d\vec{p}$ is an exact differential), resulting in the net change in momentum:

$$\int_{t_i}^{t_f} \vec{F}_{net}(t) dt = [\vec{p}(t)]_{t_i}^{t_f} = \vec{p}(t_f) - \vec{p}(t_i) = \vec{p}_f - \vec{p}_i$$

Thus, Impulse $\vec{J}$ is equal to the change in momentum $\Delta \vec{p}$.

2. Gravitational Analysis

2.1. Law of Universal Gravitation

The gravitational force is always attractive and acts along the line connecting the centers of two masses ($m_1$ and $m_2$) separated by distance $r$.

$$\vec{F}_{12} = - G \frac{m_1 m_2}{r^2} \hat{r} \quad \text{(2.1)}$$

Discussion: The force is an inverse-square law force ($\propto 1/r^2$). The negative sign signifies that the force is attractive, pointing opposite to the radial unit vector $\hat{r}$.

2.2. The Equivalence Principle and Gravitational Acceleration

Applying Newton's Second Law to the gravitational force shows that the mass $m$ of the test object cancels out, confirming the Weak Equivalence Principle.

$$\vec{g}(r) = - \frac{G M}{r^2} \hat{r} \quad \text{(2.2)}$$

3. Conservative Fields and Potential Energy

A force $\vec{F}(\vec{r})$ is conservative if the work done in a closed loop ($\oint \vec{F} \cdot d\vec{r}$) is zero. This means the force can be derived from a scalar Potential Energy function ($U(\vec{r})$) using the negative gradient:

$$\vec{F}(\vec{r}) = -\nabla U(\vec{r}) \quad \text{(3.1)}$$

3.2. Proof: The Curl Condition for Conservatism

A force field $\vec{F}$ is conservative if and only if its curl is zero ($\nabla \times \vec{F} = \vec{0}$), confirming it is an irrotational field.

$$\nabla \times \vec{F} = \vec{0} \quad \text{(3.2)}$$

Proof using the definition of potential (3.1):

We substitute $\vec{F} = -\nabla U$ into the curl operation. The $\hat{i}$ component of $\nabla \times \vec{F}$ is:

$$(\nabla \times \vec{F})_x = \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial}{\partial y} \left( -\frac{\partial U}{\partial z} \right) - \frac{\partial}{\partial z} \left( -\frac{\partial U}{\partial y} \right)$$ $$(\nabla \times \vec{F})_x = -\frac{\partial^2 U}{\partial y \partial z} + \frac{\partial^2 U}{\partial z \partial y}$$

By Clairaut's Theorem (equality of mixed partial derivatives), this component must be zero. The same holds for the other two components, proving $\nabla \times \vec{F} = \vec{0}$.

4. Solving Ordinary Differential Equations (ODEs) in Dynamics

4.1. First-Order ODE: Terminal Velocity in Viscous Drag

For a mass $m$ falling under gravity and linear drag ($\vec{F}_D = -b\vec{v}$), the equation of motion is a first-order separable ODE:

$$m \frac{dv}{dt} = mg - bv \quad \text{(4.1)}$$

Solution Process:

Separating variables and integrating from $v_0$ to $v$ and $0$ to $t$ yields:

$$\frac{dv}{mg - bv} = \frac{dt}{m} \implies \int_{v_0}^{v} \frac{dv'}{mg - bv'} = \frac{t}{m}$$

The resulting velocity function, where $\tau = m/b$ is the characteristic time constant, is:

$$v(t) = \frac{mg}{b} - \left(\frac{mg}{b} - v_0\right) e^{-t/\tau} \quad \text{(4.2)}$$

The terminal velocity is $v_{term} = mg/b$, which is reached as $t \to \infty$.

4.2. Second-Order ODE: Simple Harmonic Motion (SHO)

The SHO equation (mass $m$, spring constant $k$) is a second-order linear homogeneous ODE, with $\omega_0 = \sqrt{k/m}$.

$$m \frac{d^2x}{dt^2} + kx = 0 \implies \ddot{x} + \omega_0^2 x = 0 \quad \text{(4.3)}$$

Solution Process:

Assuming a solution $x(t) = e^{\lambda t}$ leads to the characteristic equation $\lambda^2 + \omega_0^2 = 0$, giving imaginary roots $\lambda = \pm i\omega_0$. The general real solution is a superposition of sine and cosine functions:

$$x(t) = A_{amp} \cos(\omega_0 t + \phi) \quad \text{(4.4)}$$

Where $A_{amp}$ is the amplitude and $\phi$ is the phase, both determined by the initial conditions.

5. Practice Problems

The following problems range from intermediate calculus applications to advanced theoretical challenges.

5.1. Intermediate Problems (5)

  1. A particle of mass $m$ moves in a force field given by $\vec{F}(x, y) = (4xy^2)\hat{i} + (4x^2y)\hat{j}$. Calculate the work done by the force when the particle moves from $P(1, 1)$ to $Q(2, 4)$ along the parabolic path $y=x^2$.
  2. A projectile of mass $m$ is launched with initial velocity $v_0$ at an angle $\theta$ in a medium that exerts a linear drag force $\vec{F}_D = -k\vec{v}$. Set up the coupled system of differential equations for the horizontal ($x$) and vertical ($y$) motion. Find the explicit expression for $v_x(t)$.
  3. A potential energy function is defined as $U(r) = - \frac{A}{r} + \frac{B}{r^2}$, where $A$ and $B$ are positive constants, and $r = \sqrt{x^2 + y^2 + z^2}$. Determine the magnitude of the force $|\vec{F}|$ acting on a particle at a distance $r$.
  4. A bead of mass $m$ is constrained to move along a smooth, helical wire defined by $\vec{r}(t) = (R\cos(\omega t), R\sin(\omega t), c\omega t)$. Assuming only the constraint force and gravity (in the $-\hat{k}$ direction) act, find the magnitude of the constraint force vector $|\vec{N}|$ at any time $t$.
  5. A small block of mass $m$ is pushed against a spring (spring constant $k$) and compresses it by distance $x$. The block is released on a rough horizontal surface with coefficient of kinetic friction $\mu_k$. Use the work-energy theorem to determine the speed of the block exactly when the spring reaches its natural length.

5.2. Advanced Problems (3)

  1. The force acting on a particle is $\vec{F} = K \vec{r}$, where $K$ is a positive constant and $\vec{r}$ is the position vector from the origin.
    a) Is this force conservative? Prove your answer using the curl operator (Eq. 3.2).
    b) Find the potential energy function $U(\vec{r})$.
  2. A point mass $m$ is subject to a central force $\vec{F}(\vec{r})$ proportional to the position vector and inversely proportional to its magnitude squared, $\vec{F} = C \frac{\vec{r}}{|\vec{r}|^2}$ where $C$ is a constant. Determine the angular momentum ($\vec{L}$) of the particle. Is angular momentum conserved? Prove your assertion using the torque $\vec{\tau}$.
  3. A particle of mass $m$ moves in one dimension under a time-dependent, non-conservative force $F(t) = F_0 \sin(\omega t)$. The particle starts from rest at $x=0$.
    a) Find the velocity $v(t)$.
    b) Calculate the instantaneous power exerted by the force, $P(t) = \vec{F} \cdot \vec{v}$.

5.3. Irodov-like Problems (2)

  1. A chain of length $L$ and mass $M$ lies on a smooth horizontal table. One end of the chain is pulled by a constant force $\vec{F}_0$. Determine the tension $\vec{T}(x)$ in the chain as a function of the distance $x$ from the end where the force is applied.
  2. A small ball of mass $m$ is attached to a light, rigid rod of length $L$. The rod is hinged at its other end $O$ to a vertical axis. The ball is made to rotate uniformly in a horizontal circle with an angular speed $\omega$. The only forces acting on the ball are gravity and the tension in the rod. Find the angle $\theta$ that the rod makes with the vertical axis under steady-state rotation.

6. Solutions to Practice Problems

Detailed solutions are provided below. Reviewing these derivations will help solidify the underlying principles.

6.1. Intermediate Solutions

Problem 1: Work Done along a Path

Path: $y=x^2 \implies dy = 2x dx$. The work is $W = \int_{P}^{Q} \vec{F} \cdot d\vec{r} = \int (F_x dx + F_y dy)$.

$$W = \int_1^2 \left( 4x(x^2)^2 dx + 4x^2(x^2)(2x dx) \right) = \int_1^2 (4x^5 + 8x^5) dx$$ $$W = \int_1^2 12x^5 dx = 2 [x^6]_1^2 = 2 (64 - 1) = \mathbf{126 \text{ Joules}}$$

Problem 2: Projectile with Linear Drag

The total force $\vec{F} = m\vec{g} + \vec{F}_D = -mg\hat{j} - k(v_x \hat{i} + v_y \hat{j})$. Applying $\sum \vec{F} = m\vec{a}$:

$$m \frac{dv_x}{dt} = -k v_x \quad \text{and} \quad m \frac{dv_y}{dt} = -mg - k v_y$$

Solving the separable ODE for $v_x(t)$, with $v_x(0) = v_0 \cos \theta$:

$$v_x(t) = \mathbf{(v_0 \cos \theta) e^{-kt/m}}$$

Problem 3: Force from a Potential Energy Function

The force magnitude is $|\vec{F}| = |-\partial U/\partial r|$.

$$\frac{\partial U}{\partial r} = \frac{\partial}{\partial r} \left( -Ar^{-1} + Br^{-2} \right) = \frac{A}{r^2} - \frac{2B}{r^3}$$ $$|\vec{F}| = \mathbf{\left| \frac{2B}{r^3} - \frac{A}{r^2} \right|}$$

Problem 4: Constraint Force on a Helix

The constraint force $\vec{N}$ is calculated from $\vec{N} = m\vec{a} - \vec{F}_g$. The acceleration is $\vec{a} = (-R\omega^2 \cos(\omega t), -R\omega^2 \sin(\omega t), 0)$.

$$|\vec{N}| = \sqrt{(mR\omega^2)^2 + (mg)^2} = \mathbf{m\sqrt{R^2\omega^4 + g^2}}$$

Problem 5: Work-Energy on a Rough Surface

Work-Energy Theorem: $W_{nc} = \Delta K + \Delta U$. $W_{nc} = W_{friction} = -\mu_k mg x$. $\Delta K = \frac{1}{2}mv_f^2$. $\Delta U = 0 - \frac{1}{2}kx^2$.

$$-\mu_k mg x = \frac{1}{2}mv_f^2 - \frac{1}{2}kx^2 \implies v_f^2 = \frac{k x^2 - 2\mu_k mg x}{m}$$ $$v_f = \mathbf{\sqrt{\frac{k x^2}{m} - 2\mu_k g x}}$$

6.2. Advanced Solutions

Problem 6: Curl and Potential for $\vec{F} = K \vec{r}$

a) Conservatism: $\vec{F} = Kx\hat{i} + Ky\hat{j} + Kz\hat{k}$. Since all mixed partial derivatives are zero, $\nabla \times \vec{F} = \vec{0}$. The force is conservative.

b) Potential Energy: $U(\vec{r}) = -\int F_x dx + C'$, etc. Integrating and combining yields:

$$U(\vec{r}) = \mathbf{-\frac{1}{2}K|\vec{r}|^2 + C}$$

Problem 7: Angular Momentum and Central Force

The torque is $\vec{\tau} = \vec{r} \times \vec{F}$. Since $\vec{F}$ is parallel to $\vec{r}$ (a central force), their cross product is zero:

$$\vec{\tau} = \vec{r} \times \left( C \frac{\vec{r}}{|\vec{r}|^2} \right) = \vec{0}$$

Since $\vec{\tau} = d\vec{L}/dt = \vec{0}$, the angular momentum $\vec{L}$ is conserved.

Problem 8: Non-Conservative Time-Dependent Force

a) Velocity $v(t)$: Integrating $m(dv/dt) = F_0 \sin(\omega t)$ with $v(0)=0$:

$$v(t) = \mathbf{\frac{F_0}{m\omega} (1 - \cos(\omega t))}$$

b) Instantaneous Power $P(t)$: $P(t) = \vec{F} \cdot \vec{v}$.

$$P(t) = \mathbf{\frac{F_0^2}{m\omega} \left( \sin(\omega t) - \frac{1}{2}\sin(2\omega t) \right)}$$

6.3. Irodov-like Solutions

Problem 9: Tension in a Chain with Constant Pull

Total acceleration is $a = F_0/M$. The tension $T(x)$ is the force required to accelerate the mass segment $m_{seg} = \frac{M}{L}(L-x)$ ahead of it.

$$T(x) = m_{seg} a = \frac{M}{L}(L-x) \frac{F_0}{M} = \mathbf{F_0 \left( 1 - \frac{x}{L} \right)}$$

Problem 10: Conical Pendulum

Vertical equilibrium: $T \cos\theta = mg$. Radial force: $T \sin\theta = m a_c = m L \omega^2 \sin\theta$.

Dividing the two equations (or solving the radial equation for $T$ and substituting into the vertical equation) gives:

$$\cos\theta = \frac{g}{L\omega^2} \implies \theta = \mathbf{\arccos\left(\frac{g}{L\omega^2}\right)}$$