Introductory Physics Study Notes - Work, Energy, and Rotational Dynamics

Fundamental Physical Constants and Expressions

In the study of introductory physics, several fundamental constants and mathematical expressions serve as the basis for all calculations. The acceleration due to gravity on Earth is taken as $g = 9.80\,\text{m/s}^2$ (or $9.81\,\text{m/s}^2$ in specific contexts). The universal gravitational constant is $G = 6.673 \times 10^{-11}\,\text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2}$. For planetary calculations, the mass of the Earth ($M_e$) is $5.98 \times 10^{24}\,\text{kg}$, the radius of the Earth ($R_e$) is $6.37 \times 10^6\,\text{m}$, and the mass of the Sun is $1.99 \times 10^{30}\,\text{kg}$.

Essential formulas for mechanics include the work done by a constant force, defined as $W = (F \cos(\theta)) \Delta x$, and for a varying force, $W = \int F_x \, dx$. Power is defined as the rate at which work is done, expressed as $P = F v$. For rotational dynamics, the moment of inertia for a solid cylinder or disk is $I = \frac{1}{2} M R^2$, and for a solid sphere, it is $I = \frac{2}{5} M R^2$. Angular momentum is given by $L = I \omega$. Kinetic energy of a translating object is $KE = \frac{1}{2} m v^2$, while rotational kinetic energy is $KE_{rot} = \frac{1}{2} I \omega^2$.

Work, Energy, and Power

Work is defined as the product of the displacement and the component of a force in the direction of that displacement ($W = F \cdot s \cos(\theta)$). It is measured in Joules ($\text{J}$). When calculating the work done on an object by multiple forces, one must consider the applied force, friction, gravity, and the normal force. For instance, a fisherman pulling a $50.0\,\text{kg}$ cooler box with a force of $120\,\text{N}$ at an angle of $30.0^\circ$ over a distance of $5.0\,\text{m}$ performs work equal to $W = (120\,\text{N})(\cos(30.0^\circ))(5.0\,\text{m}) = 520\,\text{J}$.

When friction is present, the work done by the frictional force is negative because it opposes motion. If the coefficient of kinetic friction ($\mu_k$) is $0.2$, the normal force on a level surface is $n = mg = (50.0\,\text{kg})(9.80\,\text{m/s}^2) = 490\,\text{N}$. The work done by friction as the box moves $5.0\,\text{m}$ is $W_{fric} = -\mu_k n \Delta x = -(0.2)(490\,\text{N})(5.0\,\text{m}) = -490\,\text{J}$. The net work done on the system is the algebraic sum of the work from individual forces: $W_{net} = W_{applied} + W_{fric} = 600\,\text{J} - 490\,\text{J} = 110\,\text{J}$.

Mechanical energy is the sum of kinetic energy ($KE$) and potential energy ($PE$). The law of conservation of mechanical energy states that in an isolated system (where friction is negligible), the initial total mechanical energy equals the final total mechanical energy ($PE_i + KE_i = PE_f + KE_f$). For a roller coaster car of mass $400\,\text{kg}$ moving at $25\,\text{m/s}$ at the top of a $50\,\text{m}$ hill, the total mechanical energy is $\frac{1}{2}(400\,\text{kg})(25\,\text{m/s})^2 + (400\,\text{kg})(9.8\,\text{m/s}^2)(50\,\text{m}) = 321,000\,\text{J}$. If this energy is conserved, the car would reach a height $h = \frac{321,000\,\text{J}}{(400\,\text{kg})(9.8\,\text{m/s}^2)} = 81.9\,\text{m}$ on a subsequent hill where it comes to rest.

The Work-Energy Theorem states that the net work done on an object equals the change in its kinetic energy ($W_{net} = \Delta KE$). If non-conservative forces like friction act on the system, the work done by these forces equals the change in the total mechanical energy ($W_{nc} = \Delta KE + \Delta PE$).

Power is the rate at which work is performed or energy is transformed. The unit of power is the Watt ($\text{W}$), and common conversions include $1\,\text{horsepower} (hp) = 746\,\text{W}$. To lift an elevator of mass $1.00 \times 10^3\,\text{kg}$ carrying an $8.00 \times 10^2\,\text{kg}$ load at a constant speed of $3.00\,\text{m/s}$ against a frictional force of $4.00 \times 10^3\,\text{N}$, the motor must deliver a specific minimum power. The tension ($T$) required is the sum of the total weight ($Mg$) and friction ($f$): $T = (1800\,\text{kg})(9.8\,\text{m/s}^2) + 4000\,\text{N} = 21,640\,\text{N}$. The power required is $P = T v = (21,640\,\text{N})(3.00\,\text{m/s}) = 64,920\,\text{W}$, or approximately $64.8\,\text{kW}$.

Momentum and Collisions

Linear momentum is defined as the product of an object's mass and its velocity ($p = mv$). The Law of Conservation of Momentum states that if no net external force acts on an isolated system, the total linear momentum remains constant. Impulse ($J$) is defined as a large force acting over a very short time interval, calculated as the change in momentum ($J = \Delta p = m v_f - m v_i$).

Collisions are categorized into elastic and inelastic types. In a perfectly elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved, but kinetic energy is not; the kinetic energy is lost to internal energy, heat, or deformation. A perfectly inelastic collision occurs when the colliding objects stick together after impact and move with a common final velocity ($v_f$).

In a crash test, if a car of mass $2000\,\text{kg}$ traveling west at $-20\,\text{m/s}$ rebounds east at $5.4\,\text{m/s}$, the impulse delivered is $J = 2000\,\text{kg}(5.4\,\text{m/s} - (-20\,\text{m/s})) = 50,800\,\text{kg} \cdot \text{m/s}$. If the collision lasts $0.15\,\text{s}$, the average force exerted is $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{50,800\,\text{kg} \cdot \text{m/s}}{0.150\,\text{s}} = 1.76 \times 10^5\,\text{N}$ directed eastward.

For two-dimensional collisions, momentum must be conserved independently in the x and y directions. Given an object $m_1 = 20.00\,\text{kg}$ at $25.00\,\text{m/s}$ at an angle of $150^\circ$ and $m_2 = 30.00\,\text{kg}$ at $20.00\,\text{m/s}$ at an angle of $20^\circ$, the initial components of velocity are found as: $u_{1x} = 25 \cos(150^\circ) = -21.65\,\text{m/s}$$u_{1y} = 25 \sin(150^\circ) = 12.5\,\text{m/s}$$u_{2x} = 20 \cos(20^\circ) = 18.794\,\text{m/s}$$u_{2y} = 20 \sin(20^\circ) = 6.840\,\text{m/s}$

Circular Motion and Gravitation

Uniform circular motion involves an object traveling in a circle at a constant speed. The time for one complete revolution is the period ($T$). The tangential speed is $v = \frac{2\pi r}{T}$. The centripetal acceleration ($a_c$) is siempre directed toward the center of the circle and is calculated as $a_c = \frac{v^2}{r}$. For example, a plane performing a "standard turn" of $2.00\,\text{minutes}$ at $250\,\text{m/s}$ has a radius $r = \frac{v T}{2\pi} = \frac{(250\,\text{m/s})(120\,\text{s})}{2\pi} = 4774.65\,\text{m}$ and a centripetal acceleration of $13\,\text{m/s}^2$.

In non-uniform circular motion, objects experience both centripetal acceleration and tangential acceleration ($a_t$). Total acceleration is the vector sum: $a = \sqrt{a_t^2 + a_c^2}$. For a boy of mass $35\,\text{kg}$ swinging on a rope of length $6.4\,\text{m}$ at a speed of $3.5\,\text{m/s}$, the tension at the lowest point accounts for both gravity and centripetal force: $T = mg + \frac{mv^2}{r} = (35)(9.8) + \frac{35(3.5^2)}{6.4} = 410\,\text{N}$.

Newton's Law of Universal Gravitation states that every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers ($F = \frac{G M m}{r^2}$). The term "Universal" implies this constant $G$ applies to any two bodies anywhere in the universe at any time.

Kepler's Laws of Planetary Motion describe the movement of celestial bodies:

1. Planets move in elliptical orbits with the Sun at one focus.
2. The radius vector from the Sun to a planet sweeps out equal areas in equal time intervals.
3. The square of the orbital period ($T^2$) is proportional to the cube of the semi-major axis ($r^3$) of the orbit ($T^2 = \frac{4\pi^2 r^3}{G M_s}$).

A geosynchronous satellite orbits the Earth with a period of $24\,\text{hours}$, keeping it fixed relative to a point on the surface. To achieve this, it must be placed at an orbital radius calculated by rearranging Kepler's 3rd law: $r = \sqrt[3]{\frac{G M_e T^2}{4\pi^2}}$. For Earth, this results in an orbital radius of approximately $4.21 \times 10^7\,\text{m}$, which corresponds to a height of roughly $36,000\,\text{km}$ above the surface.

Rotational Dynamics

Angular displacement ($\theta$) is measured in radians, where $1\,\text{revolution} = 2\pi\,\text{radians}$. Angular velocity ($\omega$) is the rate of change of angular displacement, and angular acceleration ($\alpha$) is the rate of change of angular velocity. The kinematic equations for constant angular acceleration mirror those for linear motion: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$. To find the time for a wheel bike component transitioning with $\omega_0 = 4\,\text{rad/s}$ and $\alpha = 1\,\text{rad/s}^2$ to complete $10\,\text{revolutions}$ ($20\pi\,\text{rad}$), we solve $20\pi = 4t + 0.5t^2$, yielding $t \approx 7.9\,\text{seconds}$.

Moment of inertia ($I$) is a measure of an object's rotational inertia. For a system of discrete particles, it is the sum of the products of each mass and the square of its distance from the axis ($I = \sum m r^2$). A system with $m_1 = 0.5\,\text{kg}$ at $(0,1)$, $m_2 = 1.0\,\text{kg}$ at $(2,0)$, and $m_3 = 1.5\,\text{kg}$ at $(2,2)$ has a moment of inertia about the x-axis of $I_x = (0.5)(1^2) + (1.0)(0^2) + (1.5)(2^2) = 6.5\,\text{kg} \cdot \text{m}^2$ and about the y-axis of $I_y = (0.5)(0^2) + (1.0)(2^2) + (1.5)(2^2) = 10\,\text{kg} \cdot \text{m}^2$.

In systems involving a torque ($\tau = I Α = R T$), such as a wheel of radius $0.4\,\text{m}$ and $I = 0.48\,\text{kg} \cdot \text{m}^2$ supporting a $3\,\text{kg}$ mass on a rope, the bucket's linear acceleration ($a$) and the wheel's angular acceleration ($\alpha$) are connected by $a = R Α$. Applying Newton's second law to the mass ($mg - T = ma$) and the torque equation to the wheel ($TR = I Α$), the resulting acceleration is found to be $a = 4.9\,\text{m/s}^2$, with an angular acceleration of $12.25\,\text{rad/s}^2$ and tension in the cord of $14.7\,\text{N}$.

Static Equilibrium

An object is in static equilibrium if it is at rest and remains at rest. This requires two necessary conditions:

1. The vector sum of all external forces acting on the object must be zero ($\sum F = 0$).
2. The vector sum of all external torques about any point must be zero ($\sum \tau = 0$).

These principles are used to calculate unknown tensions in cables or forces on hinges. For a $70\,\text{kg}$ box suspended by two cables at angles of $40^\circ$ and $30^\circ$ to a ceiling, the horizontal component condition ($T_2 \cos(30^°) - T_1 \cos(40^°) = 0$) and vertical component condition ($T_1 \sin(40^°) + T_2 \sin(30^°) = mg$) define a system of equations. Solving these gives $T_1 = 635\,\text{N}$ and $T_2 = 561\,\text{N}$.

In ladder problems, where a uniform ladder of mass $22\,\text{kg}$ and length $10\,\text{m}$ leans against a smooth wall, it will begin to slip when the friction between the ground and ladder is overwhelmed. If a $70\,\text{kg}$ person climbs $90\%$ of the way up, torque must be balanced at the base. By setting the torque about the base to zero ($\sum \tau_{base} = 0$), the normal force from the wall ($F_{wall}$) is calculated. The coefficient of static friction ($\mu_s$) is then found as the ratio of this wall force to the total normal force from the ground ($F_{wall} / F_g$—where $F_g$ is the total weight). For this specific configuration ($2.8\,\text{m}$ from the wall), the coefficient is calculated as $0.23$.

Questions & Discussion

Calculations for a Spinning Sphere: Question: A $0.5\,\text{kg}$ uniform sphere of $5.0\,\text{cm}$ radius spins at $1800\,\text{rev/min}$ on an axis through its center. Find its rotational kinetic energy and angular momentum. Answer: First, convert speed to frequency ($n = 30\,\text{rev/s}$). Rotational kinetic energy $E_k = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{5} m R^2)(2\pi n)^2 = \frac{4}{5} \pi^2 m R^2 n^2 = 8.9\,\text{J}$. Angular momentum $L = I \omega = (\frac{2}{5} m R^2)(2\pi n) = \frac{4}{5} \pi m R^2 n = 0.10\,\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1}$.

The Rubber Stopper Scenario: Question: A Physics lecturer swings a rubber stopper in a horizontal circle at the end of a string and tells Zuba he is going to let go when the stopper is directly in front of her face. Will the stopper hit Zuba's face or not? Discussion: No, the stopper will not hit Zuba's face. When the string is released, the velocity of the stopper is tangential to the circular path. At the point directly in front of Zuba, the tangent direction is parallel to her face, not toward it. Therefore, it will fly past her rather than hitting her directly.

Beads and Change in Momentum: Question: A stream of elastic glass beads ($0.50\,\text{g}$ each) fall $0.50\,\text{m}$ at a rate of $100\,\text{per second}$ to a balance pan and bounce back to their original height. What is the impact? Discussion: The velocity as they hit the pan is calculated via kinematics ($v = \sqrt{2gh} = 3.13\,\text{m/s}$). Since they bounce back to the original height, the final velocity is $-3.13\,\text{m/s}$. The change in momentum per bead is $\Delta p = m(v_f - v_i) = 0.0005\,\text{kg}(-3.13 - 3.13) = -0.00313\,\text{kg} \cdot \text{m/s}$. The force exerted is the rate of change of momentum: $F = 100 \times 0.00313 = 0.313\,\text{N}$.