Dynamics: Force and Motion

A Rigorous Examination of Newton's Laws and Their Applications

1. The Axioms of Motion: Newton's Laws

Newton's First Law (Inertia)

An object's state of motion ($\vec{v}$) remains constant if and only if the net force on it is zero ($\sum \vec{F} = 0$). This defines the concept of an inertial reference frame.

Newton's Second Law (Dynamics)

The net force on an object is equal to the time rate of change of its linear momentum: $\sum \vec{F} = \frac{d\vec{p}}{dt}$. For constant mass, this becomes the second-order differential equation governing motion: $\sum \vec{F} = m\frac{d^2\vec{x}}{dt^2}$.

Newton's Third Law (Interaction)

Forces arise from interactions between bodies. If body A exerts force $\vec{F}_{AB}$ on body B, then B exerts an equal and opposite force $\vec{F}_{BA} = -\vec{F}_{AB}$ on A.

2. Work and Energy: An Integral Approach

Work-Kinetic Energy Theorem

The concept of work links force and energy. The work $W$ done by a net force $\sum \vec{F}$ in moving an object from position $\vec{r}_i$ to $\vec{r}_f$ is defined by the line integral:

$$ W_{net} = \int_{\vec{r}_i}^{\vec{r}_f} \sum \vec{F} \cdot d\vec{r} $$

We can derive its connection to motion by starting with Newton's second law:

$$ W_{net} = \int m\frac{d\vec{v}}{dt} \cdot d\vec{r} = \int m\frac{d\vec{v}}{dt} \cdot \vec{v} dt = \int m \vec{v} \cdot d\vec{v} $$

Since $\vec{v} \cdot d\vec{v} = \frac{1}{2} d(\vec{v} \cdot \vec{v}) = \frac{1}{2} d(v^2)$, the integral becomes:

$$ W_{net} = \int_{v_i}^{v_f} m d\left(\frac{1}{2}v^2\right) = \left[\frac{1}{2}mv^2\right]_{v_i}^{v_f} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 $$

This is the Work-Kinetic Energy Theorem: $W_{net} = \Delta K$. The net work done on an object equals the change in its kinetic energy.

3. Conservative Forces and Potential Energy

A force is conservative if the work it does on an object moving between two points is independent of the path taken. This is equivalent to stating that the work done over any closed path is zero: $\oint \vec{F}_{cons} \cdot d\vec{r} = 0$.

By Stokes' theorem, this path independence condition is satisfied if the curl of the force field is zero:

$$ \nabla \times \vec{F}_{cons} = 0 $$

For any force field that satisfies this condition, we can define a scalar potential energy function $U(\vec{r})$ such that the force is the negative gradient of this potential:

$$ \vec{F}_{cons} = -\nabla U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right) $$

For example, the gravitational potential energy is $U(r) = -G M m / r$. Taking the negative gradient in spherical coordinates correctly yields the force $\vec{F}_g = -G M m / r^2 \hat{r}$.

4. Non-Inertial Frames and Fictitious Forces

Newton's laws hold only in inertial frames. When we describe motion in an accelerating or rotating frame, we must introduce fictitious forces to make the second law valid in that frame.

If an object has acceleration $\vec{a}_{in}$ in an inertial frame and $\vec{a}_{rot}$ in a frame rotating with angular velocity $\vec{\omega}$, the true force $\vec{F}_{true} = m\vec{a}_{in}$. The effective force in the rotating frame is $\vec{F}_{eff} = m\vec{a}_{rot}$, and the two are related by:

$$ \vec{F}_{eff} = \vec{F}_{true} - \underbrace{m\vec{a}_0}_{\text{Translational}} - \underbrace{m\dot{\vec{\omega}} \times \vec{r}'}_{\text{Euler}} - \underbrace{m\vec{\omega} \times (\vec{\omega} \times \vec{r}')}_{\text{Centrifugal}} - \underbrace{2m(\vec{\omega} \times \vec{v}')}_{\text{Coriolis}} $$

The two most common fictitious forces are:

  • Centrifugal Force: Acts outwards from the axis of rotation. It is what pushes you to the side in a turning car.
  • Coriolis Force: Acts on moving objects in a rotating frame. It is perpendicular to both the velocity of the object and the axis of rotation. It is responsible for the large-scale rotation of hurricanes and ocean currents.

5. Application: Damped Oscillations

Consider a mass $m$ on a spring (constant $k$) subject to a linear drag force $\vec{F}_d = -b\vec{v}$. The net force is $\sum F_x = -kx - b v_x$. Applying Newton's second law:

$$ -kx - b\frac{dx}{dt} = m\frac{d^2x}{dt^2} $$

This gives the standard second-order homogeneous linear differential equation for a damped harmonic oscillator:

$$ m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = 0 $$

The solution to this equation describes how the system's oscillations decay over time due to the dissipative drag force.

6. Problem Set

Problem 1 (Intermediate):

A 5.0 kg block is pulled along a horizontal frictionless surface by a rope that makes an angle of 30° with the horizontal. The tension in the rope is 20 N. What is the acceleration of the block?

Solution: The horizontal component of the tension provides the acceleration. $\sum F_x = T\cos\theta = ma_x$. $a_x = \frac{T\cos\theta}{m} = \frac{20 \text{ N} \cdot \cos(30^\circ)}{5.0 \text{ kg}} = \frac{20 \cdot \sqrt{3}/2}{5} = 2\sqrt{3} \approx 3.46 \, \text{m/s}^2$.

Problem 2 (Intermediate):

Two masses, $m_1=2$ kg and $m_2=3$ kg, are connected by a light string that passes over a frictionless pulley (Atwood's machine). Find the acceleration of the system and the tension in the string. Use $g=9.8 \text{ m/s}^2$.

Solution: The net force is the difference in weights: $F_{net} = m_2g - m_1g$. The total mass is $M = m_1+m_2$. $a = \frac{F_{net}}{M} = \frac{(m_2-m_1)g}{m_1+m_2} = \frac{(3-2) \cdot 9.8}{3+2} = \frac{9.8}{5} = 1.96 \, \text{m/s}^2$. For $m_1$: $T - m_1g = m_1a \implies T = m_1(g+a) = 2(9.8+1.96) = 23.52 \, \text{N}$.

Problem 3 (Intermediate):

A 10 kg block rests on a horizontal plane. The coefficient of static friction is $\mu_s=0.5$ and kinetic friction is $\mu_k=0.3$. What is the frictional force if a horizontal force of 40 N is applied? What if 60 N is applied? ($g=9.8 \text{ m/s}^2$)

Solution: First, find the maximum static friction: $f_{s,max} = \mu_s N = \mu_s mg = 0.5 \cdot 10 \cdot 9.8 = 49 \, \text{N}$.
Case 1 (40 N applied): Since $40 \text{ N} < 49 \text{ N}$, the block does not move. The static friction matches the applied force: $f_s = 40 \, \text{N}$.
Case 2 (60 N applied): Since $60 \text{ N} > 49 \text{ N}$, the block moves. The friction is now kinetic: $f_k = \mu_k N = \mu_k mg = 0.3 \cdot 10 \cdot 9.8 = 29.4 \, \text{N}$.

Problem 4 (Intermediate):

A car of mass 1200 kg travels at a constant speed of 15 m/s around a flat circular track of radius 50 m. What is the minimum coefficient of static friction required between the tires and the road?

Solution: The centripetal force is provided by static friction. $f_s = mv^2/r$. The maximum static friction available is $f_{s,max} = \mu_s N = \mu_s mg$. To avoid slipping, we need $f_{s,max} \ge mv^2/r$. $\mu_s mg \ge mv^2/r \implies \mu_s \ge \frac{v^2}{gr} = \frac{(15)^2}{9.8 \cdot 50} = \frac{225}{490} \approx 0.46$.

Problem 5 (Intermediate):

A 2 kg block starts from rest on a rough inclined plane that makes an angle of 37° with the horizontal. The coefficient of kinetic friction is 0.25. Use the Work-Energy theorem to find the block's speed after it has slid 4 m down the plane.

Solution: $W_{net} = \Delta K = \frac{1}{2}mv_f^2$. The net work is done by gravity and friction. $W_{net} = W_g + W_f = (mg\sin\theta)d - f_k d$. $f_k = \mu_k N = \mu_k mg \cos\theta$. $W_{net} = (mg\sin\theta - \mu_k mg\cos\theta)d = mgd(\sin\theta - \mu_k\cos\theta)$. $\frac{1}{2}mv_f^2 = mgd(\sin\theta - \mu_k\cos\theta) \implies v_f = \sqrt{2gd(\sin\theta - \mu_k\cos\theta)}$. $v_f = \sqrt{2 \cdot 9.8 \cdot 4 (\sin37^\circ - 0.25\cos37^\circ)} \approx \sqrt{78.4(0.602 - 0.25 \cdot 0.799)} \approx 5.61 \, \text{m/s}$.

Problem 6 (Advanced):

A small object of mass $m$ is subject to a resistive force proportional to its velocity, $F_r = -bv$, in addition to a constant force $F_0$. It starts from rest. Find its velocity $v(t)$ as a function of time.

Solution: This is a differential equation problem. $\sum F = F_0 - bv = ma = m\frac{dv}{dt}$. Rearrange to separate variables: $m\frac{dv}{dt} = F_0 - bv \implies \frac{dv}{F_0 - bv} = \frac{1}{m}dt$. Integrate from $v=0$ at $t=0$: $\int_0^v \frac{dv'}{F_0 - bv'} = \int_0^t \frac{1}{m}dt'$. Let $u = F_0 - bv'$, so $du = -b dv'$. $\int_{F_0}^{F_0-bv} \frac{-1/b \, du}{u} = \frac{t}{m} \implies -\frac{1}{b}[\ln|u|]_{F_0}^{F_0-bv} = \frac{t}{m}$. $\ln(F_0-bv) - \ln(F_0) = -bt/m \implies \ln\left(\frac{F_0-bv}{F_0}\right) = -bt/m$. $1 - \frac{bv}{F_0} = e^{-bt/m} \implies v(t) = \frac{F_0}{b}(1 - e^{-bt/m})$. The velocity approaches a terminal velocity of $F_0/b$.

Problem 7 (Advanced):

A particle of mass $m$ moves in a potential $U(x) = U_0((\frac{a}{x})^2 - \frac{a}{x})$. Find the force $F(x)$ acting on the particle. Find the equilibrium position $x_0$ and determine if it is stable.

Solution: The force is the negative derivative of the potential: $F(x) = -\frac{dU}{dx}$. $F(x) = -U_0 \frac{d}{dx}(a^2x^{-2} - ax^{-1}) = -U_0(-2a^2x^{-3} - (-a)x^{-2}) = U_0(2a^2x^{-3} - ax^{-2})$. $F(x) = \frac{U_0 a}{x^3}(2a-x)$. Equilibrium is where $F(x_0)=0$. This occurs at $2a-x_0 = 0 \implies x_0 = 2a$. To check stability, we examine the second derivative of $U(x)$ at $x_0$. $\frac{d^2U}{dx^2} = -\frac{dF}{dx} = -U_0(-6a^2x^{-4} + 2ax^{-3})$. At $x_0=2a$: $\frac{d^2U}{dx^2}|_{2a} = -U_0(\frac{-6a^2}{16a^4} + \frac{2a}{8a^3}) = -U_0(\frac{-3}{8a^2} + \frac{1}{4a^2}) = \frac{U_0}{8a^2}$. Since $\frac{d^2U}{dx^2} > 0$, the potential has a local minimum, so the equilibrium is stable.

Problem 8 (Advanced):

A projectile is fired with initial speed $v_0$ at an angle $\theta$. Air resistance provides a drag force $\vec{F}_D = -c\vec{v}^2 \hat{v}$. Write down the coupled differential equations for the x and y components of the velocity, $v_x$ and $v_y$. (Do not solve).

Solution: The net force is $\vec{F}_{net} = \vec{F}_g + \vec{F}_D = -mg\hat{j} - c v^2 \hat{v}$. The velocity vector is $\vec{v} = v_x \hat{i} + v_y \hat{j}$. The speed is $v = \sqrt{v_x^2 + v_y^2}$. The unit vector is $\hat{v} = \frac{\vec{v}}{v} = \frac{v_x\hat{i} + v_y\hat{j}}{\sqrt{v_x^2 + v_y^2}}$. So, $\vec{F}_D = -c\sqrt{v_x^2 + v_y^2}(v_x\hat{i} + v_y\hat{j})$. Now apply $\vec{F}_{net} = m\vec{a} = m\frac{dv_x}{dt}\hat{i} + m\frac{dv_y}{dt}\hat{j}$. X-component: $m\frac{dv_x}{dt} = -c v_x \sqrt{v_x^2 + v_y^2}$. Y-component: $m\frac{dv_y}{dt} = -mg - c v_y \sqrt{v_x^2 + v_y^2}$.

Problem 9 (Irodov-like):

A particle of mass $m$ is located in a unidimensional potential field where the potential energy of the particle depends on the coordinate $x$ as $U(x) = A x^2 - B x^3$, where $A$ and $B$ are positive constants. Find the position of equilibrium and investigate its stability.

Solution: Equilibrium occurs where the force is zero, $F(x) = -dU/dx = 0$. $F(x) = -\frac{d}{dx}(Ax^2 - Bx^3) = -(2Ax - 3Bx^2) = 3Bx^2 - 2Ax = x(3Bx - 2A)$. Equilibrium positions are $x_1 = 0$ and $x_2 = 2A/(3B)$. To investigate stability, we check the sign of the second derivative of the potential, $U''(x)$. $U''(x) = \frac{d^2U}{dx^2} = 2A - 6Bx$. For $x_1 = 0$: $U''(0) = 2A$. Since $A>0$, $U''(0)>0$, so this is a position of stable equilibrium (a potential minimum). For $x_2 = 2A/(3B)$: $U''(2A/3B) = 2A - 6B(2A/3B) = 2A - 4A = -2A$. Since $A>0$, $U''(x_2)<0$, so this is a position of unstable equilibrium (a potential maximum).

Problem 10 (Irodov-like):

A uniform rod of mass $m$ and length $l$ rotates with a constant angular velocity $\omega$ in a horizontal plane about a vertical axis passing through one of its ends. Find the tension force $T(x)$ in the rod as a function of the distance $x$ from the axis of rotation.

Solution: Consider a small element of the rod of length $dx$ at a distance $x$ from the axis. Its mass is $dm = (\frac{m}{l})dx$. This element moves in a circle of radius $x$, so it has a centripetal acceleration $a_c = \omega^2 x$. The net force on this element is $dF = (dm)a_c = (\frac{m}{l}dx)\omega^2x$. The tension $T(x)$ at position $x$ must provide the centripetal force for the entire outer part of the rod (from $x$ to $l$). The tension is the integral of the required centripetal force for all elements from $x$ to $l$: $T(x) = \int_x^l dF = \int_x^l (\frac{m}{l}\omega^2 r) dr$, where $r$ is the integration variable for position. $T(x) = \frac{m\omega^2}{l} \int_x^l r dr = \frac{m\omega^2}{l} \left[\frac{r^2}{2}\right]_x^l = \frac{m\omega^2}{2l}(l^2 - x^2)$. The tension is maximum at the axis ($x=0$), $T(0) = m\omega^2l/2$, and zero at the free end ($x=l$), $T(l)=0$.