Comprehensive Lecture Notes on Mechanics: Work, Energy, Momentum, Rotational Dynamics, and Statics

Work, Kinetic Energy, and Power

• Definition of Work Done by a Constant Force:

  • Work ($W$) performed on an object by a constant force ($F$) acting parallel to the object's displacement ($d$) is defined as the product of the magnitude of the force and the magnitude of the displacement:         $W = F d$
  • When the applied force acts at an angle ($\theta$) relative to the direction of displacement, only the component of the force directed along the line of motion performs work:         $W = (F \cos(\theta)) d = F d \cos(\theta)$
  • Equivalently, work can be interpreted as the full magnitude of the force multiplied by the component of the displacement parallel to the force ($d_{\parallel} = d \cos(\theta)$).
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  • Vector Dot Product Formulation:
    • Work is a scalar quantity represented by the vector dot product of the force vector ($\mathbf{F}$) and the displacement vector ($\mathbf{d}$):             $W = \mathbf{F} \cdot \mathbf{d} = F_x d_x + F_y d_y = F d \cos(\theta)$
    • The angle ($\theta$) between two vectors (e.g., $\mathbf{A}$ and $\mathbf{B}$) can be determined directly from the dot product definition:             $\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{A B}$
    • Worked Vector Example: For vectors $\mathbf{A} = (2, -3)$ and $\mathbf{B} = (5, 1)$, the dot product is:             $\mathbf{A} \cdot \mathbf{B} = (2)(5) + (-3)(1) = 10 - 3 = 7$
    • The magnitudes are $A = \sqrt{2^2 + (-3)^2} = \sqrt{13}$ and $B = \sqrt{5^2 + 1^2} = \sqrt{26}$. The angle between them is:             $\cos(\theta) = \frac{7}{\sqrt{13}\sqrt{26}} = \frac{7}{\sqrt{338}} \approx 0.3807 \implies \theta = 67.2^\circ$
  • Sign Conventions of Work:
    • Positive Work ($W > 0$): Occurs when $-90^\circ < \theta < 90^\circ$ (the force has a component in the direction of motion). This increases the object's speed.
    • Zero Work ($W = 0$): Occurs when $\theta = \pm 90^\circ$ (force is strictly perpendicular to displacement) or when displacement is zero ($d = 0$). Perpendicular forces produce no change in speed.
    • Negative Work ($W < 0$): Occurs when $90^\circ < \theta < 270^\circ$ (force opposes motion). This decreases the object's speed.
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  • Units of Work and Energy:
    • The SI unit of work is the Newton-meter ($\text{N}\cdot\text{m} = \text{kg}\cdot\text{m}^2/\text{s}^2$), designated as the Joule ($\text{J}$).
    • Imperial and heat units: calorie ($1\,\text{cal} = 4.184\,\text{J}$). Food calories are kilocalories ($1\,\text{Cal} = 1\,\text{kcal} = 4184\,\text{J}$).

• Total Work Done by Multiple Forces:

  • When multiple forces act simultaneously on a body, the total work ($W_{\text{total}}$) can be calculated in two identical ways:
    1. Summing the individual work done by each respective force: $W_{\text{total}} = \sum W_i = W_1 + W_2 + W_3 + \dots$
    2. Taking the dot product of the net force ($\mathbf{F}_{\text{net}}$) with displacement:             $W_{\text{total}} = \mathbf{F}_{\text{net}} \cdot \mathbf{d} = (F_{\text{net}} \cos(\theta)) d$
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  • Car Rolling Down Incline Example: A car descending a hill of angle $\phi$ under gravity ($\mathbf{W} = m\mathbf{g}$), normal force ($\mathbf{N}$), and air drag ($\mathbf{F}_{\text{air}}$) experiences total work:         $W_{\text{total}} = \mathbf{N} \cdot \mathbf{d} + \mathbf{F}_{\text{air}} \cdot \mathbf{d} + m\mathbf{g} \cdot \mathbf{d}$         Because $\mathbf{N} \perp \mathbf{d}$, $W_N = 0\,\text{J}$. Air resistance acts antiparallel ($\theta = 180^\circ$), yielding $W_{\text{air}} = -F_{\text{air}} d$. Gravity acts at an angle $(90^\circ - \phi)$, yielding $W_g = mg d \sin(\phi)$.

• Kinetic Energy and the Work-Energy Theorem:

  • Kinetic Energy ($K$): The scalar energy associated with a particle's motion:         $K = \frac{1}{2}m v^2$         The SI unit of kinetic energy is the Joule ($\text{J} = \text{kg}\cdot\text{m}^2/\text{s}^2$).
  • The Work-Energy Theorem: The total work performed on an object by the net force equals the change in the object's kinetic energy:         $W_{\text{total}} = \Delta K = K_f - K_i = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2$
  • Physical Significance: Evaluating the total work done does not require detailed step-by-step knowledge of individual forces or exact trajectories; it depends purely on the initial and final states of speed.
  • Consequence for Constant Speed: If an object moves at a constant speed ($v_f = v_i$), $\Delta K = 0$, meaning the total net work done must be zero ($W_{\text{total}} = 0$).

• Work Done by Variable Forces:

  • On a force-versus-position plot, the work done by a force along displacement from $x_1$ to $x_2$ corresponds geometrically to the area under the curve.
  • For an arbitrary varying force, the displacement is partitioned into small segments ($\Delta x$) where the force is approximately constant:         $W = \sum (F_i \cos(\theta)) \Delta x$
  • As $\Delta x \to 0$, the discrete sum converges to the integral of the force over position:         $W = \int_{x_1}^{x_2} F(x)\,dx = F_{\text{av}} d$
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  • Application to an Ideal Spring (Hooke's Law):
    • The restoring force of an ideal spring stretched by displacement $x$ from equilibrium is $F_s = -kx$, where $k$ is the spring constant ($\text{N/m}$).
    • The external applied force required to stretch or compress the spring at each point is $F_{\text{app}} = kx$.
    • Because the force-displacement relationship is linear (starting at $0$ and reaching $kx$), the average force is:             $F_{\text{av}} = \frac{0 + kx}{2} = \frac{1}{2}kx$
    • The work done in stretching the spring from $x = 0$ to position $x$ is the triangular area under the curve:             $W = F_{\text{av}} x = \left(\frac{1}{2}kx\right) x = \frac{1}{2}kx^2$

• Power:

  • Power ($P$) is the time rate at which work is performed:         $P = \frac{W}{\Delta t}$
  • The SI unit of power is the Watt ($\text{W}$), defined as one Joule per second ($1\,\text{W} = 1\,\text{J/s} = 1\,\text{N}\cdot\text{m/s} = 1\,\text{kg}\cdot\text{m}^2/\text{s}^3$).
  • Horsepower ($\text{hp}$): Established historically by James Watt as the rate at which an average dray horse could lift a $550\,\text{lb}$ weight by a height of $1\,\text{ft}$ in $1\,\text{s}$:         $1\,\text{hp} = 550\,\text{ft}\cdot\text{lb/s} = 745.7\,\text{W} \approx 746\,\text{W}$
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  • Power at Constant Velocity: When an object moves at a constant speed ($v$) overcoming opposing resistance (drag, friction), the driving power exerted is:         $P = \frac{F d}{t} = F \left(\frac{d}{t}\right) = F v$
  • Automobile Engine Acceleration Example: An automobile has mass $m = 1500\,\text{kg}$ and engine power output $P = 70\,\text{kW} = 70,000\,\text{W}$. Ignoring friction:
    • At $v = 60\,\text{km/h} = \frac{60,000\,\text{m}}{3600\,\text{s}} = 16.7\,\text{m/s}$:             $F = \frac{P}{v} = \frac{70,000\,\text{W}}{16.7\,\text{m/s}} = 4.20 \times 10^3\,\text{N} = 4.20\,\text{kN}$$a = \frac{F}{m} = \frac{4,200\,\text{N}}{1,500\,\text{kg}} = 2.8\,\text{m/s}^2$
    • At $v = 100\,\text{km/h} = \frac{100,000\,\text{m}}{3600\,\text{s}} = 27.8\,\text{m/s}$:             $F = \frac{P}{v} = \frac{70,000\,\text{W}}{27.8\,\text{m/s}} = 2.52 \times 10^3\,\text{N} = 2.52\,\text{kN}$$a = \frac{F}{m} = \frac{2,520\,\text{N}}{1,500\,\text{kg}} = 1.68\,\text{m/s}^2$
    • Physical Insight: Because kinetic energy is proportional to the square of velocity ($K \propto v^2$), raising vehicle speed from $100\,\text{km/h}$ to $110\,\text{km/h}$ demands substantially more work than increasing speed from $60\,\text{km/h}$ to $70\,\text{km/h}$. At fixed power output, this greater work requires more elapsed time, diminishing the maximum achievable acceleration at higher speeds.
  • Electrical Energy Consumption Units: Utility billing measured in kilowatt-hours ($\text{kWh}$) represents energy, not power ($1\,\text{kWh} = (1000\,\text{W})(3600\,\text{s}) = 3.60 \times 10^6\,\text{J} = 3.60\,\text{MJ}$). Whether a high-power device (e.g., $1500\,\text{W}$ hair dryer) costs more to operate than a moderate-power device (e.g., $600\,\text{W}$ microwave) depends strictly on the total duration each appliance operates.

• Detailed Worked Applications and Conceptual Scenarios:

  • Sliding Giant Gumdrop: A gumdrop of mass $m = 200\,\text{kg}$ slides across a flat horizontal surface with $\mu_k = 0.20$, initial velocity $v_i = 10\,\text{m/s}$, and $g = 10\,\text{m/s}^2$. Determine stopping distance ($d$):
    • Via Kinematics: The normal force is $N = mg$. Friction force is $f_k = -\mu_k mg$. Acceleration is $a_x = -\mu_k g$.             $v_f^2 = v_i^2 + 2a_x d \implies 0 = v_i^2 - 2\mu_k g d \implies d = \frac{v_i^2}{2\mu_k g} = \frac{10^2}{2(0.20)(10)} = 25\,\text{m}$
    • Via Work-Energy Theorem: Friction is the sole horizontal force performing work:             $W_{\text{total}} = -f_k d = -\mu_k mg d$$W_{\text{total}} = K_f - K_i = 0 - \frac{1}{2}m v_i^2 \implies -\mu_k mg d = -\frac{1}{2}m v_i^2 \implies d = \frac{v_i^2}{2\mu_k g} = 25\,\text{m}$
  • Braking Truck: A truck with mass $m = 2100\,\text{kg}$ slows on a level road from $v_i = 22.0\,\text{m/s}$ to $v_f = 12.0\,\text{m/s}$ over $d = 50.0\,\text{m}$. Work done is:         $W = \Delta K = \frac{1}{2}(2100\,\text{kg})\left((12.0\,\text{m/s})^2 - (22.0\,\text{m/s})^2\right) = \frac{1}{2}(2100)(144 - 484) = -3.57 \times 10^5\,\text{J}$         The required braking force is:         $F = \frac{W}{d} = \frac{-3.57 \times 10^5\,\text{J}}{50.0\,\text{m}} = -7.14 \times 10^3\,\text{N} = -7.14\,\text{kN}$
  • Vehicle Braking Distance Scaling: If a car traveling at $60\,\text{km/h}$ stops over distance $d = 20\,\text{m}$ under constant braking force, doubling its initial speed to $120\,\text{km/h}$ quadruples the initial kinetic energy ($K \propto v^2$). Because work $W = F d = \Delta K$, the stopping distance quadruples to $d = 4 \times 20\,\text{m} = 80\,\text{m}$.
  • Block on an Incline: Pushing an object up a ramp of horizontal base $7\,\text{m}$ and vertical rise $2.5\,\text{m}$ with a strictly horizontal force of $F = 100\,\text{N}$: only the horizontal displacement component parallel to the force performs work:         $W = F d_x = (100\,\text{N})(7\,\text{m}) = 700\,\text{J}$

Potential Energy and Energy Conservation

• Conservative and Nonconservative Forces:

  • Conservative Forces: Store energy internally in reversible forms without thermal dissipation. Work done around any closed path is identically zero ($\oint \mathbf{F}_c \cdot d\mathbf{r} = 0$), and work between two endpoints is path-independent. Examples: Gravity ($mg$), ideal spring force ($kx$).
  • Nonconservative Forces: Dissipate mechanical energy into thermal, acoustic, or internal chemical energy. Work done depends directly on the path taken, and work around a closed path is non-zero ($\oint \mathbf{F}_{\text{nc}} \cdot d\mathbf{r} \ne 0$). Examples: Kinetic friction ($f_k$), tension ($T$), motor-applied forces, muscle contraction forces.

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• Path Comparison: Across a rectangular loop of width $d$ and height $h$:

  • Gravity yields $W_{\text{bottom}} = 0$, $W_{\text{right}} = -mgh$, $W_{\text{top}} = 0$, $W_{\text{left}} = +mgh$. Total round-trip work: $W_{\text{total}} = 0\,\text{J}$.
  • Friction acts continuously antiparallel to motion on each side: $W_{\text{total}} = -\mu_k mg d - \mu_k mg h - \mu_k mg d - \mu_k mg h = -2\mu_k mg(d + h) \ne 0$.

• Definition of Potential Energy ($U$):

  • Potential energy represents stored capacity to do work, defined exclusively for conservative forces. The work done by a conservative force equals the negative change in potential energy:         $W_c = -\Delta U = -(U_f - U_i) = U_i - U_f$
  • If a conservative force does positive work ($W_c > 0$), potential energy decreases ($\Delta U < 0$). If it does negative work ($W_c < 0$), potential energy increases ($\Delta U > 0$).
  • Because only the difference ($\Delta U$) is physically defined, the absolute zero reference level ($U(y_0) = 0$) can be positioned arbitrarily without affecting physical results.
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• Forms of Potential Energy:

  • Gravitational Potential Energy ($U_g$):$U_g(y) = mgy$
    • Work done by gravity during vertical displacement $\Delta y = y_f - y_i$:             $W_g = -mg\Delta y = mg(y_i - y_f)$
    • Path independence: Moving between two elevations via a steep direct path or a winding indirect path imparts the identical change in potential energy ($\Delta U = mgh$).
  • Elastic (Spring) Potential Energy ($U_s$):$U_s(x) = \frac{1}{2}kx^2$
    • Displacement $x$ is measured relative to the spring's uncompressed/unstretched equilibrium ($x = 0$).
    • Because $x$ is squared, $U_s \ge 0$ universally, whether extended ($x > 0$) or compressed ($x < 0$).
    • Work done by a spring moving from $x_i$ to $x_f$:             $W_s = -\Delta U_s = \frac{1}{2}k(x_i^2 - x_f^2)$

• Conservation of Mechanical Energy:

  • Total mechanical energy ($E$) is the sum of all kinetic ($K$) and potential ($U$) energies present in the system:         $E = K + U = \frac{1}{2}m v^2 + U_g + U_s$
  • When only conservative forces act on an isolated system, the total mechanical energy is conserved:         $E_i = E_f \iff K_i + U_i = K_f + U_f$

• Comprehensive Mechanical Energy Problems:

  • Vertical and Angled Rock Trajectories: A rock of mass $m = 0.26\,\text{kg}$ is thrown upward from the edge of a $32\,\text{m}$ cliff and hits the ground at $v_f = 29\,\text{m/s}$. Taking $y = 0$ at the base and $g = 9.81\,\text{m/s}^2$:
    • Initial Velocity: Compare cliff top ($y = h = 32\,\text{m}$) to base ($y = 0$):             $E_i = E_f \implies mgh + \frac{1}{2}m v_i^2 = 0 + \frac{1}{2}m v_f^2$$v_i = \sqrt{v_f^2 - 2gh} = \sqrt{(29\,\text{m/s})^2 - 2(9.81\,\text{m/s}^2)(32\,\text{m})} = \sqrt{841 - 627.84} = 14.6\,\text{m/s}$
    • Maximum Height: Compare peak ($y = y_{\text{max}}$, where $v_{\text{top}} = 0$) to base:             $mg y_{\text{max}} = \frac{1}{2}m v_f^2 \implies y_{\text{max}} = \frac{v_f^2}{2g} = \frac{(29\,\text{m/s})^2}{2(9.81\,\text{m/s}^2)} = 42.9\,\text{m}$
    • Directional Invariance: If thrown straight downward with identical launch speed ($14.6\,\text{m/s}$) or at any arbitrary angle $\theta$, the mechanical energy at the cliff ($mgh + \frac{1}{2}m v_i^2$) is identical, resulting in the exact same ground impact speed of $29\,\text{m/s}$.
  • Loop-the-Loop Release Height: A block of mass $m$ slides from rest at height $h$ down a frictionless track into a circular loop of radius $r$:
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    • Speed at Loop Apex (Point B, $y = 2r$):$mgh = mg(2r) + \frac{1}{2}m v_B^2 \implies v_B^2 = 2g(h - 2r)$
    • Dynamic Contact Requirement: At the apex, gravity and the downward normal force ($N$) provide the required centripetal acceleration:             $\sum F_y = -N - mg = -m\frac{v_B^2}{r} \implies N = m\left(\frac{v_B^2}{r} - g\right)$
    • To avoid falling off the track, the block must maintain contact ($N \ge 0$). The threshold boundary occurs at $N = 0$:             $v_{B,\text{min}}^2 = gr$
    • Equating the kinetic expressions:             $2g(h_{\text{min}} - 2r) = gr \implies 2h_{\text{min}} - 4r = r \implies h_{\text{min}} = \frac{5}{2}r = 2.5r$
    • Mass Cancels: Both potential and kinetic energy terms scale directly with mass ($m$); thus, the minimum release height is entirely independent of $m$.
  • Spring Launcher Vertical Launch: A $1.00\,\text{kg}$ projectile is loaded into a vertical spring cannon ($k = 50.0\,\text{N/m}$) compressed by $\Delta s = 2.00\,\text{m}$. Taking the spring equilibrium position as $y = 0$ ($g = 10\,\text{m/s}^2$):
    • Initial state ($y = -\Delta s = -2.00\,\text{m}$, at rest):             $E_1 = U_{g,1} + U_{s,1} + K_1 = mg(-\Delta s) + \frac{1}{2}k\Delta s^2 + 0$$E_1 = (1.00)(10)(-2.00) + \frac{1}{2}(50.0)(2.00)^2 = -20\,\text{J} + 100\,\text{J} = 80\,\text{J}$
    • State at barrel exit ($y = 0$, spring uncompressed):             $E_2 = U_{g,2} + U_{s,2} + K_2 = 0 + 0 + \frac{1}{2}m v^2$$80\,\text{J} = \frac{1}{2}(1.00\,\text{kg})v^2 \implies v^2 = 160\,\text{m}^2/\text{s}^2 \implies v = 12.6\,\text{m/s}$
    • Maximum elevation reached ($y_{\text{max}}$, where $v = 0$):             $E_3 = mgy_{\text{max}} = 80\,\text{J} \implies y_{\text{max}} = \frac{80\,\text{J}}{(1.00\,\text{kg})(10\,\text{m/s}^2)} = 8.00\,\text{m}$
  • Slide Geometry and Travel Time: Two frictionless water slides connect the same top platform of height $h$ to ground level. One slide drops steeply immediately before leveling out, while the other descends on a gradual incline. Objects sliding down both achieve the exact same exit speed ($v = \sqrt{2gh}$). However, the object on the steep drop accelerates rapidly at the beginning, maintaining a higher speed over most of the path, and finishes the slide earlier.

• Potential Energy Curves, Equipotentials, and Topography:

  • A graph plotting potential energy ($U(x)$) against position ($x$) describes the dynamical bounds of a system.
  • For an isolated system with fixed mechanical energy ($E_0$), the kinetic energy at any point is the vertical separation: $K(x) = E_0 - U(x)$.
  • Turning Points: The spatial coordinates where $U(x) = E_0$, forcing $K = 0$. The object reverses its direction of motion at these locations.
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  • Force Relationship: The conservative force corresponds to the negative slope of the potential curve:         $F(x) = -\frac{dU}{dx}$
  • Equipotential Lines and Contour Maps: On topographic maps, contour lines indicate paths of constant gravitational potential energy. Closely spaced contour lines denote a steep incline (large spatial gradient and large component of gravitational force along the surface).

• Work Done by Nonconservative Forces ($W_{\text{nc}}$):

  • When nonconservative forces act within an open mechanical system, mechanical energy is not conserved. Applying the Work-Energy Theorem:         $W_{\text{total}} = W_c + W_{\text{nc}} = \Delta K$         Because conservative work is $W_c = -\Delta U$:         $-\Delta U + W_{\text{nc}} = \Delta K \implies W_{\text{nc}} = \Delta U + \Delta K = \Delta E = E_f - E_i$
  • Connected Blocks on a Rough Table: A table block of mass $m_1 = 2.40\,\text{kg}$ ($\mu_k = 0.450$) is tied to a hanging mass $m_2 = 1.80\,\text{kg}$. The system starts from rest and moves through $d = 0.500\,\text{m}$ before $m_2$ strikes the floor ($g = 9.81\,\text{m/s}^2$):
    • Initial state: $E_i = m_1 gh + m_2 gd + 0$
    • Final state: $E_f = m_1 gh + 0 + \frac{1}{2}(m_1 + m_2)v_f^2$
    • Change in mechanical energy: $\Delta E = E_f - E_i = \frac{1}{2}(m_1 + m_2)v_f^2 - m_2 gd$
    • Nonconservative work performed by kinetic friction on $m_1$:             $W_{\text{nc}} = -f_k d = -\mu_k N_1 d = -\mu_k m_1 gd$
    • Equating expressions:             $\frac{1}{2}(m_1 + m_2)v_f^2 - m_2 gd = -\mu_k m_1 gd$$\frac{1}{2}(m_1 + m_2)v_f^2 = (m_2 - \mu_k m_1)gd$$v_f = \sqrt{\frac{2(m_2 - \mu_k m_1)gd}{m_1 + m_2}} = \sqrt{\frac{2[1.80 - (0.450)(2.40)](9.81)(0.500)}{2.40 + 1.80}} = 1.30\,\text{m/s}$
  • Skater in a Halfpipe with Friction: A skater of mass $m = 60\,\text{kg}$ drops from rest at the lip of a circular halfpipe ($R = 10\,\text{m}$). An average friction force of $f_s = 50\,\text{N}$ opposes travel ($g = 10\,\text{m/s}^2$):
    • Arc length of a quarter pipe: $d = \frac{2\pi R}{4} = \frac{\pi R}{2}$
    • Nonconservative work: $W_{\text{nc}} = -f_s d = -f_s \left(\frac{\pi R}{2}\right) = -\frac{1}{2}(50\,\text{N})\pi(10\,\text{m}) = -250\pi\,\text{J} \approx -785.4\,\text{J}$
    • Mechanical energy relation:             $W_{\text{nc}} = E_B - E_A = \left(\frac{1}{2}m v_B^2 + 0\right) - (0 + mgR)$$\frac{1}{2}m v_B^2 = mgR - \frac{1}{2}f_s \pi R \implies v_B = \sqrt{2gR - \frac{f_s \pi R}{m}}$$v_B = \sqrt{2(10)(10) - \frac{50\pi(10)}{60}} = \sqrt{200 - 26.18} = 13.2\,\text{m/s}$(Without friction, speed would be $v = \sqrt{2gR} = 14.1\,\text{m/s}$.)
    • Snowboarder Crossing to Opposite Lip: To travel from lip A across the bottom to opposite lip C at the same elevation ($\Delta U = 0$), the half-pipe arc is $d = \pi R$. Frictional dissipation is $W_{\text{nc}} = -f_k \pi R$. Initial kinetic energy must cover this loss:             $0 - \frac{1}{2}m v_A^2 = -f_k \pi R \implies v_A = \sqrt{\frac{2f_k \pi R}{m}} = \sqrt{\frac{2(50)\pi(10)}{60}} = 7.2\,\text{m/s}$
  • Air Resistance on Up-and-Down Projectiles: A ball thrown upward subject to aerodynamic drag loses mechanical energy along its entire flight ($W_{\text{nc}} < 0$). Because energy is continuously removed, its speed at any given elevation during the downward trip is lower than at the identical elevation during the ascent. With a lower average speed on the descent, the return trip takes longer than the upward trip.

Linear Momentum, Impulse, and Collisions

• Linear Momentum ($\mathbf{p}$):

  • Linear momentum is a vector quantity defined as the product of an object's mass ($m$) and its velocity ($\mathbf{v}$):         $\mathbf{p} = m\mathbf{v}$
  • The direction of $\mathbf{p}$ matches the direction of $\mathbf{v}$. The SI unit is kilogram-meter per second ($\text{kg}\cdot\text{m/s} = \text{kg}\cdot\text{m}\cdot\text{s}^{-1}$).
  • Change in Momentum ($\Delta \mathbf{p}$): Defined as $\Delta \mathbf{p} = \mathbf{p}_f - \mathbf{p}_i$.
    • Soft Clay / Teddy Bear Dropped on Floor: Approaches floor with $-v$, hits and comes to rest ($v_f = 0$):             $\Delta p = 0 - m(-v) = mv$
    • Elastic Rubber Ball Dropped on Floor: Approaches floor with $-v$, rebounds elastically upward with $+v$:             $\Delta p = mv - m(-v) = 2mv$
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• Newton's Second Law and Variable Mass Systems:

  • The fundamental formulation of Newton's second law defines net force as the time rate of change of linear momentum:         $\sum \mathbf{F} = \frac{\Delta \mathbf{p}}{\Delta t} = \frac{\mathbf{p}_f - \mathbf{p}_i}{\Delta t}$
  • When system mass ($m$) remains strictly constant:         $\sum \mathbf{F} = \frac{m\mathbf{v}_f - m\mathbf{v}_i}{\Delta t} = m\frac{\Delta \mathbf{v}}{\Delta t} = m\mathbf{a}$
  • If mass changes dynamically (e.g., fuel ejection in rocketry, rain accumulating in an open railroad car), the momentum derivative $\frac{\Delta \mathbf{p}}{\Delta t}$ must be applied.

• Impulse ($\mathbf{I}$) and the Impulse-Momentum Theorem:

  • Impulse is a vector defined as the product of the average force ($\mathbf{F}_{\text{av}}$) and the time duration ($\Delta t$) over which it acts:         $\mathbf{I} = \mathbf{F}_{\text{av}}\Delta t$
  • The SI unit of impulse is $\text{kg}\cdot\text{m/s}$, identical to momentum.
  • Impulse-Momentum Theorem:$\mathbf{I} = \mathbf{F}_{\text{av}}\Delta t = \Delta \mathbf{p} = \mathbf{p}_f - \mathbf{p}_i$
  • Significance: For rapidly varying impact forces (such as a bat striking a baseball), measuring instantaneous forces is difficult. The impulse equals the integral (area) of the force-time curve, which is mathematically equivalent to the rectangular area formed by the average force acting over duration $\Delta t$.
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  • Work vs. Impulse Analogy:
    • Work ($W = F_{\text{net}} d$) acts over spatial distance to change kinetic energy ($\Delta K$).
    • Impulse ($I = F_{\text{av}} \Delta t$) acts over temporal duration to change momentum ($\Delta p$).
  • Basketball Bounce Pass Problem: A basketball ($m = 0.60\,\text{kg}$) strikes the floor at $v_i = 5.4\,\text{m/s}$ at an angle of $65^\circ$ to the normal. It rebounds at the identical speed and angle. Define the normal to the floor as $+y$ and the horizontal motion as $+x$:
    • Horizontal velocity components: $v_{i,x} = v_0 \sin(65^\circ)$ and $v_{f,x} = v_0 \sin(65^\circ)$.             $\Delta p_x = m(v_{f,x} - v_{i,x}) = 0\,\text{kg}\cdot\text{m/s} \implies I_x = 0$
    • Vertical velocity components: $v_{i,y} = -v_0 \cos(65^\circ)$ (downward) and $v_{f,y} = +v_0 \cos(65^\circ)$ (upward).             $I_y = \Delta p_y = m[v_0 \cos(65^\circ) - (-v_0 \cos(65^\circ))] = 2m v_0 \cos(65^\circ)$$I_y = 2(0.60\,\text{kg})(5.4\,\text{m/s})\cos(65^\circ) = 2.7\,\text{kg}\cdot\text{m/s}$
  • Hail Versus Rain on an Umbrella: Liquid raindrops strike an umbrella and splatter or come to rest ($\Delta p \approx mv$), whereas hailstones of equivalent mass and speed bounce elastically ($\Delta p \approx 2mv$). Because the change in momentum of bouncing hail is nearly twice as large, the impulse and holding force required to support the umbrella in hail are significantly greater than in rain.

• Conservation of Linear Momentum:

  • Internal versus External Forces:
    • Internal forces act exclusively between particles inside the defined system boundary. According to Newton's third law, these forces occur in equal and opposite action-reaction pairs and sum identically to zero:             $\sum \mathbf{F}_{\text{int}} = 0$
    • The net force acting on a system consists solely of external forces:             $\mathbf{F}_{\text{net}} = \sum \mathbf{F}_{\text{ext}}$
  • The Principle of Conservation of Linear Momentum: If the net external force acting on a system is zero ($\sum \mathbf{F}_{\text{ext}} = 0$), the total linear momentum is conserved:         $\Delta \mathbf{p}_{\text{total}} = 0 \implies \mathbf{p}_{f,\text{total}} = \mathbf{p}_{i,\text{total}}$$\mathbf{p}_{1,f} + \mathbf{p}_{2,f} + \dots = \mathbf{p}_{1,i} + \mathbf{p}_{2,i} + \dots$
  • Pitcher and Catcher on Frictionless Ice Example:
    • A pitcher ($m_p = 80\,\text{kg}$) at rest on frictionless ice throws a baseball ($m_b = 0.2\,\text{kg}$) by exerting an average horizontal force $F_{\text{av}} = 50\,\text{N}$ for $\Delta t = 0.2\,\text{s}$.
    • Ball Impulse and Speed:             $p_b = I = F_{\text{av}}\Delta t = (50\,\text{N})(0.2\,\text{s}) = 10\,\text{kg}\cdot\text{m/s}$$v_b = \frac{p_b}{m_b} = \frac{10\,\text{kg}\cdot\text{m/s}}{0.2\,\text{kg}} = 50\,\text{m/s}$
    • Pitcher Recoil Speed: System starts at rest ($p_{\text{total}} = 0$):             $p_p + p_b = 0 \implies m_p v_p = -p_b = -10\,\text{kg}\cdot\text{m/s}$$v_p = \frac{-10\,\text{kg}\cdot\text{m/s}}{80\,\text{kg}} = -0.125\,\text{m/s}$
    • Catcher Recoil Speed: A catcher ($m_c = 80\,\text{kg}$) catches the incoming ball. Momentum is conserved during the catch:             $p_{\text{total},i} = p_b = 10\,\text{kg}\cdot\text{m/s}$$(m_c + m_b)v_{c+b} = p_b \implies v_{c+b} = \frac{10\,\text{kg}\cdot\text{m/s}}{80\,\text{kg} + 0.2\,\text{kg}} = \frac{10}{80.2} = 0.1247\,\text{m/s}$
    • Kinetic Energy Loss in Inelastic Catch:             $K_i = \frac{1}{2}m_b v_b^2 = \frac{1}{2}(0.2\,\text{kg})(50\,\text{m/s})^2 = 250\,\text{J}$$K_f = \frac{1}{2}(m_c + m_b)v_{c+b}^2 = \frac{1}{2}(80.2\,\text{kg})(0.1247\,\text{m/s})^2 = 0.62\,\text{J}$             Energy is dissipated as thermal and structural deformation ($\Delta K = -249.38\,\text{J}$).
  • Cardiovascular Recoil of the Human Body: At rest, a human heart pumps roughly $5\,\text{L/min}$ of blood vertically toward the head through the ascending aorta at an average speed of $v = 66\,\text{cm/s} = 0.66\,\text{m/s}$. Assuming blood density $\rho \approx 1.0\,\text{g/cm}^3$ ($1\,\text{kg/L}$):
    • Mass flow rate: $\frac{\Delta m}{\Delta t} = \frac{5\,\text{kg}}{60\,\text{s}} \approx 0.083\,\text{kg/s}$
    • Upward momentum per second: $p_{\text{blood}} = (0.083\,\text{kg})(0.66\,\text{m/s}) = 0.055\,\text{kg}\cdot\text{m/s}$
    • For a person of mass $M = 60\,\text{kg}$ on frictionless ice, momentum conservation requires:             $v_{\text{body}} = -\frac{p_{\text{blood}}}{M} = -\frac{0.055\,\text{kg}\cdot\text{m/s}}{60\,\text{kg}} = -0.9\,\text{mm/s}$
    • This recoil produces measurable periodic movement in rocking chairs and underlies sensitive mechanical motion detectors.

• Collisions in One and Two Dimensions:

  • Inelastic Collisions: Total momentum is conserved ($\mathbf{p}_f = \mathbf{p}_i$), but total kinetic energy is not conserved ($K_f \ne K_i$; typically $K_f < K_i$).
  • Completely Inelastic Collisions: Colliding bodies stick together, sharing a common final velocity ($\mathbf{v}_f$):         $\mathbf{p}_i = m_1\mathbf{v}_{1,i} + m_2\mathbf{v}_{2,i} = (m_1 + m_2)\mathbf{v}_f \implies \mathbf{v}_f = \frac{m_1\mathbf{v}_{1,i} + m_2\mathbf{v}_{2,i}}{m_1 + m_2}$
    • 
  • Ballistic Pendulum: A fast bullet of mass $m$ and speed $v_i$ embeds into a suspended pendulum bob of mass $M$:
    • Phase 1 (Inelastic Impact): Momentum is conserved during the brief impact:             $m v_i = (M + m)v_f \implies v_f = \left(\frac{m}{M + m}\right)v_i$             Kinetic energy immediately following collision:             $K_f = \frac{1}{2}(M + m)v_f^2 = \frac{1}{2}m v_i^2 \left(\frac{m}{M + m}\right)$
    • Phase 2 (Conservative Swing): Mechanical energy is conserved as the combined bob and bullet swing upward to height $h$:             $K_f = (M + m)gh \implies \frac{1}{2}(M + m)\left(\frac{m}{M + m}\right)^2 v_i^2 = (M + m)gh$$h = \left(\frac{m}{M + m}\right)^2 \frac{v_i^2}{2g}$
    • Numerical Scale: For bullet $m = 7\,\text{g} = 0.007\,\text{kg}$, bob $M = 0.95\,\text{kg}$, and initial speed $v_i = 280\,\text{m/s}$ ($g = 9.81\,\text{m/s}^2$):             $h = \left(\frac{0.007}{0.957}\right)^2 \frac{280^2}{2(9.81)} = (5.35 \times 10^{-5})(3996) \approx 0.21\,\text{m} = 21\,\text{cm}$
  • Elastic Collisions in One Dimension: Both linear momentum and kinetic energy are conserved ($\mathbf{p}_f = \mathbf{p}_i$ and $K_f = K_i$). No mechanical energy is lost to heat or sound.
    • When target mass $m_2$ is initially at rest ($v_{2,i} = 0$):             $v_{1,f} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)v_{1,i}$$v_{2,f} = \left(\frac{2m_1}{m_1 + m_2}\right)v_{1,i}$
    • Limiting Cases:
      1. Equal Masses ($m_1 = m_2$): $v_{1,f} = 0$ and $v_{2,f} = v_{1,i}$. The incoming mass stops dead; the target moves away with the incident speed.
      2. Target Extremely Massive ($m_2 \gg m_1$): $v_{1,f} \approx -v_{1,i}$ and $v_{2,f} \approx 0$. Light projectile rebounds with unchanged speed.
      3. Projectile Extremely Massive ($m_1 \gg m_2$): $v_{1,f} \approx v_{1,i}$ and $v_{2,f} \approx 2v_{1,i}$. Light target is propelled forward at double the projectile's speed.
    • Relative Speed Invariant: In every 1D elastic collision, the relative speed of separation after collision equals the relative speed of approach prior to collision:             $v_{2,f} - v_{1,f} = -(v_{2,i} - v_{1,i})$
  • Two-Dimensional Collisions: Evaluated by resolving momentum conservation into independent orthogonal axes:         $p_{x,i} = p_{x,f} \implies m_1 v_{1,i,x} + m_2 v_{2,i,x} = m_1 v_{1,f,x} + m_2 v_{2,f,x}$$p_{y,i} = p_{y,f} \implies m_1 v_{1,i,y} + m_2 v_{2,i,y} = m_1 v_{1,f,y} + m_2 v_{2,f,y}$
    • 
    • 2D Space Collision Example: In zero-gravity space, particle 1 ($m_1 = 0.130\,\text{kg}$) moving at $v_{1,i} = 1.11\,\text{m/s}$ collides elastically with particle 2 ($m_2 = 0.160\,\text{kg}$) moving at $v_{2,i} = 1.21\,\text{m/s}$.
    • Initial kinetic energy:             $K_i = \frac{1}{2}(0.130)(1.11)^2 + \frac{1}{2}(0.160)(1.21)^2 = 0.0800\,\text{J} + 0.1171\,\text{J} = 0.197\,\text{J}$
    • Particle 2 deflects at angle $42.0^\circ$ with final speed $v_{2,f} = 1.16\,\text{m/s}$. Its final kinetic energy is:             $K_{2,f} = \frac{1}{2}(0.160)(1.16)^2 = 0.108\,\text{J}$
    • Conservation of kinetic energy ($K_f = K_i$) yields final speed of particle 1:             $\frac{1}{2}m_1 v_{1,f}^2 + 0.108\,\text{J} = 0.197\,\text{J} \implies v_{1,f} = \sqrt{\frac{2(0.197 - 0.108)}{0.130}} = 1.17\,\text{m/s}$
    • Vertical momentum conservation ($p_{y,i} = 0$):             $0 = m_1 v_{1,f} \sin(\theta) - m_2 v_{2,f} \sin(42.0^\circ)$$\sin(\theta) = \frac{m_2 v_{2,f} \sin(42.0^\circ)}{m_1 v_{1,f}} = \frac{(0.160)(1.16)\sin(42.0^\circ)}{(0.130)(1.17)} = \frac{0.1242}{0.1521} = 0.817$$\theta = \sin^{-1}(0.817) = 54.8^\circ$

• Center of Mass (CM):

  • The center of mass is the unique point at which an entire system's mass can be treated as concentrated for translational dynamics.
  • For a system of discrete particles of masses $m_i$ at coordinates $(x_i, y_i)$ with total mass $M = \sum m_i$:         $X_{\text{cm}} = \frac{\sum m_i x_i}{M} = \frac{m_1 x_1 + m_2 x_2 + \dots}{m_1 + m_2 + \dots}$$Y_{\text{cm}} = \frac{\sum m_i y_i}{M} = \frac{m_1 y_1 + m_2 y_2 + \dots}{m_1 + m_2 + \dots}$
  • 
  • Velocity and Acceleration of the Center of Mass:$\mathbf{V}_{\text{cm}} = \frac{\sum m_i \mathbf{v}_i}{M} \implies M\mathbf{V}_{\text{cm}} = \sum \mathbf{p}_i = \mathbf{P}_{\text{total}}$$\mathbf{A}_{\text{cm}} = \frac{\sum m_i \mathbf{a}_i}{M} \implies M\mathbf{A}_{\text{cm}} = \sum \mathbf{F}_{\text{ext}} = \mathbf{F}_{\text{net,ext}}$
  • Newton's Second Law for a System: The center of mass accelerates exactly as a hypothetical point particle of mass $M$ subjected to the net external force. Internal forces cannot alter the motion of the center of mass.
  • 

• Rocket Propulsion (Systems with Changing Mass):

  • If a rocket expels fuel mass $\Delta m$ at a constant exhaust velocity $v$ relative to the rocket, the momentum transferred to the expelled fuel produces forward thrust force ($F$):         $F = \frac{\Delta p}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) v$
  • 
  • Apollo 16 Lunar Module Ascent Stage: The engine generates thrust $F = 15,600\,\text{N}$ while burning propellant at mass rate $\frac{\Delta m}{\Delta t} = 6.40\,\text{kg/s}$. The exhaust speed of the propellant gas is:         $v = \frac{F}{\Delta m / \Delta t} = \frac{15,600\,\text{N}}{6.40\,\text{kg/s}} = 2440\,\text{m/s}$
  • Biological Hydrodynamic Drag Reduction (Emperor Penguins): Adult Emperor penguins ($m \approx 38\,\text{kg}$) release micro air bubbles trapped within their plumage during ascent. The bubble cloud lowers water density and skin friction around their bodies, reducing drag and enabling them to reach exit speeds necessary to leap out of the water onto sea ice.

Rotational Kinematics and Energy

• Rotational Variables:

  • Angular Position ($\theta$): The angle measured relative to an arbitrary reference line. The SI unit is the radian ($\text{rad}$), a dimensionless ratio defined as:         $\theta = \frac{s}{r}$         where $s$ is arc length along a circular path of radius $r$.
  • 
  • Angular Conventions: Counterclockwise rotations are defined as positive ($\theta > 0$); clockwise rotations are negative ($\theta < 0$). Angular displacements are cumulative ($1\,\text{revolution} = 360^\circ = 2\pi\,\text{rad}$).
  • Angular Velocity ($\omega$):
    • Average angular velocity: $\omega_{\text{av}} = \frac{\Delta \theta}{\Delta t}$
    • Instantaneous angular velocity: $\omega = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t}$
    • SI Unit: $\text{rad/s} = \text{s}^{-1}$.
    • Period ($T$): The time required to complete one full revolution ($2\pi\,\text{rad}$):             $T = \frac{2\pi}{\omega} \iff \omega = \frac{2\pi}{T}$
    • Rotational frequency ($N$) in revolutions per second is $N = \frac{1}{T} = \frac{\omega}{2\pi}$.
  • Angular Acceleration ($\alpha$):
    • Average angular acceleration: $\alpha_{\text{av}} = \frac{\Delta \omega}{\Delta t}$
    • Instantaneous angular acceleration: $\alpha = \lim_{\Delta t \to 0} \frac{\Delta \omega}{\Delta t}$
    • SI Unit: $\text{rad/s}^2 = \text{s}^{-2}$.

• Rotational Kinematics Equations Under Constant Angular Acceleration ($\alpha = \text{constant}$):

  • Analogous to 1D linear kinematics:         $\omega = \omega_0 + \alpha t$$\theta = \theta_0 + \frac{1}{2}(\omega_0 + \omega)t$$\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$
  • Ceiling Fan Deceleration: A fan spinning at $\omega_0 = 0.96\,\text{rev/s}$ is switched off and slows uniformly to rest in $t = 2.4\,\text{min} = 144\,\text{s}$:
    • Total revolutions during deceleration:             $\Delta \theta = \frac{1}{2}(\omega_0 + \omega)t = \frac{1}{2}(0.96\,\text{rev/s} + 0)(144\,\text{s}) = 69\,\text{rev}$
    • Revolutions made while halving speed to $0.48\,\text{rev/s}$:             Because $\omega$ decreases linearly, halving speed requires half the time ($t = 1.2\,\text{min} = 72\,\text{s}$):             $\Delta \theta = \frac{1}{2}(0.96 + 0.48\,\text{rev/s})(72\,\text{s}) = \frac{1}{2}(1.44)(72) = 52\,\text{rev}$

• Connections Between Linear and Rotational Quantities:

  • Tangential Speed ($v_t$): Linear speed along the circular path:         $v_t = r\omega$
  • Centripetal Acceleration ($a_{\text{cp}}$): Direct result of continuous directional change toward the center of curvature:         $a_{\text{cp}} = \frac{v_t^2}{r} = \frac{(r\omega)^2}{r} = r\omega^2$
  • Tangential Acceleration ($a_t$): Linear acceleration tangent to the path due to changing angular speed:         $a_t = \frac{\Delta v_t}{\Delta t} = r\frac{\Delta \omega}{\Delta t} = r\alpha$
  • Total Linear Acceleration ($a_{\text{total}}$): Vector sum of perpendicular centripetal and tangential components:         $a_{\text{total}} = \sqrt{a_{\text{cp}}^2 + a_t^2} = \sqrt{(r\omega^2)^2 + (r\alpha)^2}$
  • 
  • Merry-Go-Round Comparison (Bonnie and Klyde): Bonnie sits at the outer edge ($r_B = 2.0\,\text{m}$) and Klyde sits midway ($r_K = 1.0\,\text{m}$). The platform rotates once every $T = 2\pi\,\text{s}$, so both share identical angular velocity:         $\omega = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1.0\,\text{rad/s}$
    • Tangential velocities: $v_{t,B} = r_B \omega = (2.0)(1.0) = 2.0\,\text{m/s}$; $v_{t,K} = r_K \omega = (1.0)(1.0) = 1.0\,\text{m/s}$.
    • Centripetal accelerations: $a_{\text{cp},B} = r_B \omega^2 = (2.0)(1.0)^2 = 2.0\,\text{m/s}^2$; $a_{\text{cp},K} = r_K \omega^2 = (1.0)(1.0)^2 = 1.0\,\text{m/s}^2$.
    • Under constant angular acceleration $\alpha = 0.20\,\text{rad/s}^2$ starting from rest over $t = 10\,\text{s}$:             $\Delta \theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(0.20)(100) = 10\,\text{rad}$$s_B = r_B \Delta \theta = (2.0)(10) = 20\,\text{m}; \quad s_K = r_K \Delta \theta = (1.0)(10) = 10\,\text{m}$$\omega(10\,\text{s}) = \alpha t = (0.20)(10) = 2.0\,\text{rad/s}$             Bonnie: $a_{\text{cp}} = (2.0)(2.0)^2 = 8.0\,\text{m/s}^2$; $a_t = (2.0)(0.20) = 0.40\,\text{m/s}^2 \implies a_{\text{total}} = \sqrt{8.0^2 + 0.40^2} = 8.01\,\text{m/s}^2$.             Klyde: $a_{\text{cp}} = (1.0)(2.0)^2 = 4.0\,\text{m/s}^2$; $a_t = (1.0)(0.20) = 0.20\,\text{m/s}^2 \implies a_{\text{total}} = \sqrt{4.0^2 + 0.20^2} = 4.005\,\text{m/s}^2$.
  • Earth Latitude Tangential Speeds: Earth's rotation period is $T = 24\,\text{h} = 86,400\,\text{s}$ and equatorial radius is $R_E = 6371\,\text{km}$. The perpendicular radius to the rotational axis at latitude $\phi$ is $r = R_E \cos(\phi)$:
    • Nairobi ($\phi = 0^\circ$): $v_t = \frac{2\pi(6371\,\text{km})}{24\,\text{h}} = 1670\,\text{km/h} = 460\,\text{m/s}$.
    • Mexico City ($\phi = 20^\circ\,\text{N}$): $v_t = \frac{2\pi(6371\cos(20^\circ))}{24} = 1570\,\text{km/h} = 435\,\text{m/s}$.
    • Edmonton ($\phi = 53.5^\circ\,\text{N}$): $v_t = \frac{2\pi(6371\cos(53.5^\circ))}{24} = 990\,\text{km/h} = 276\,\text{m/s}$.
  • Jeff of the Jungle Vine Swing: Jeff swings from a vine of length $r = 7.20\,\text{m}$. At the lowest point, his speed is $v = 8.50\,\text{m/s}$:         $\omega = \frac{v}{r} = \frac{8.50\,\text{m/s}}{7.20\,\text{m}} = 1.18\,\text{rad/s}$$a_{\text{cp}} = \frac{v^2}{r} = \frac{(8.50\,\text{m/s})^2}{7.20\,\text{m}} = 10.0\,\text{m/s}^2$Mechanical Implication: Tension in the vine must support both Jeff's weight and provide centripetal acceleration ($T - mg = m a_{\text{cp}} \implies T = m(g + a_{\text{cp}}) \approx 2mg$). A vine capable of supporting him statically can snap at the base of the swing.

• Rolling Motion Without Slipping:

  • Rolling combines translational and rotational motion. An object rolling without slipping moves its center of mass by $2\pi r$ in one revolution ($T$):         $v_{\text{cm}} = \frac{2\pi r}{T} = r\omega$
  • 
  • Velocity Profile of a Rolling Wheel:
    • Bottom (Contact Point): Pure translation ($+r\omega$) and pure rotation ($-r\omega$) cancel: $v = 0$.
    • Center of Mass: Unaffected by rotation: $v = v_{\text{cm}} = r\omega$.
    • Top of Wheel: Translation ($+r\omega$) and rotation ($+r\omega$) reinforce: $v = 2r\omega = 2v_{\text{cm}}$.
  • 
  • Contact Condition: Because the instantaneous velocity at the contact point with the pavement is zero, rolling without slipping engages static friction, rather than kinetic friction.
  • Cycloid Motion: A light source on the wheel's outer rim traces a cycloid curve through space.
  • 

• Moment of Inertia ($I$) and Rotational Kinetic Energy:

  • Moment of Inertia ($I$): Measures rotational inertia (resistance to angular acceleration) around a chosen axis:         $I = \sum m_i r_i^2$         SI unit: $\text{kg}\cdot\text{m}^2$.
  • Rotational Kinetic Energy ($K_{\text{rot}}$):$K_{\text{rot}} = \frac{1}{2}I\omega^2$
  • Moments of Inertia for Standard Uniform Rigid Bodies of Mass $M$ and Radius $R$:
    • Hoop or thin cylindrical shell: $I = M R^2$ ($\beta = 1$
    • Solid disk or cylinder (central axis): $I = \frac{1}{2}M R^2$ ($\beta = 1/2$
    • Solid disk or cylinder (rim axis): $I = \frac{3}{2}M R^2$
    • Thin spherical shell (hollow sphere): $I = \frac{2}{3}M R^2$ ($\beta = 2/3$
    • Solid uniform sphere (central axis): $I = \frac{2}{5}M R^2$ ($\beta = 2/5$
    • Solid sphere (rim axis): $I = \frac{7}{5}M R^2$
    • Long thin rod (axis through midpoint, length $L$): $I = \frac{1}{12}M L^2$
    • Long thin rod (axis through one end, length $L$): $I = \frac{1}{3}M L^2$
    • Solid rectangular plate (central axis in plane): $I = \frac{1}{12}M L^2$
    • Solid plate (perpendicular axis through center): $I = \frac{1}{12}M(L^2 + W^2)$
  • 
  • Total Kinetic Energy of Rolling Objects:$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}M v^2 + \frac{1}{2}I\omega^2$         Substituting non-slip rolling condition $\omega = v/R$ and $I = \beta M R^2$:         $K_{\text{total}} = \frac{1}{2}M v^2 + \frac{1}{2}(\beta M R^2)\left(\frac{v}{R}\right)^2 = \frac{1}{2}M v^2 (1 + \beta)$
  • The Incline Rolling Race: Releasing objects of equal mass and radius from rest at height $h$ down a ramp:         $Mgh = \frac{1}{2}M v^2 (1 + \beta) \implies v = \sqrt{\frac{2gh}{1 + \beta}}$
    • Objects with smaller $\beta$ convert a larger fraction of potential energy into translational motion rather than rotational storage.
    • Arrival order: Solid sphere ($\beta = 0.40$) arrives first, followed by solid disk ($\beta = 0.50$), hollow sphere ($\beta = 0.67$), and finally hoop/ring ($\beta = 1.00$).
  • Baseball Curve Ball Energy: Pitcher throws a $0.15\,\text{kg}$ solid sphere baseball ($R = 3.7\,\text{cm} = 0.037\,\text{m}$) at $v = 48\,\text{m/s}$ and $\omega = 42\,\text{rad/s}$:         $K_{\text{trans}} = \frac{1}{2}(0.15\,\text{kg})(48\,\text{m/s})^2 = 170\,\text{J}$$I = \frac{2}{5}M R^2 = \frac{2}{5}(0.15\,\text{kg})(0.037\,\text{m})^2 = 8.21 \times 10^{-5}\,\text{kg}\cdot\text{m}^2$$K_{\text{rot}} = \frac{1}{2}I\omega^2 = \frac{1}{5}(0.15)(0.037)^2(42)^2 = 0.072\,\text{J}$         Rotational energy constitutes only $0.04\%$ of total kinetic energy, yet the spin deflects the trajectory significantly through aerodynamic forces.
  • Falling Leg Exercise: A person lies flat and lets an upright leg fall freely to the floor. Modeling the leg as a uniform rod ($L = 0.95\,\text{m}$) pivoting about the hip ($I = \frac{1}{3}m L^2$):
    • The center of mass drops from $h_{\text{cm}} = \frac{L}{2} = \frac{0.95\,\text{m}}{2} = 0.475\,\text{m}$ to the floor ($y = 0$):             $E_i = E_f \implies mg h_{\text{cm}} = \frac{1}{2}I\omega^2$$mg \left(\frac{L}{2}\right) = \frac{1}{2}\left(\frac{1}{3}m L^2\right)\omega^2 \implies \omega = \frac{\sqrt{6gh_{\text{cm}}}}{L} = \frac{\sqrt{3gL}}{L} = \sqrt{\frac{3g}{L}}$$\omega = \frac{\sqrt{6(9.81\,\text{m/s}^2)(0.475\,\text{m})}}{0.95\,\text{m}} = 5.6\,\text{rad/s}$
    • Tangential foot speed ($r = L$) right before landing:             $v_t = L\omega = (0.95\,\text{m})(5.6\,\text{rad/s}) = 5.3\,\text{m/s}$
  • Bowling Ball Rolling Up a Ramp: A solid sphere ($I = \frac{2}{5}m r^2$) rolls without slipping up a ramp rising $h = 0.47\,\text{m}$ with initial speed $v_i = 2.62\,\text{m/s}$ ($g = 9.81\,\text{m/s}^2$):         $K_{i,\text{total}} = U_f + K_{f,\text{total}}$$\frac{1}{2}m v_i^2 + \frac{1}{2}\left(\frac{2}{5}m r^2\right)\left(\frac{v_i}{r}\right)^2 = mgh + \frac{1}{2}m v_f^2 + \frac{1}{2}\left(\frac{2}{5}m r^2\right)\left(\frac{v_f}{r}\right)^2$$\frac{7}{10}v_i^2 = gh + \frac{7}{10}v_f^2 \implies v_f = \sqrt{v_i^2 - \frac{10}{7}gh}$$v_f = \sqrt{(2.62\,\text{m/s})^2 - \frac{10}{7}(9.81\,\text{m/s}^2)(0.47\,\text{m})} = \sqrt{6.864 - 6.586} = 0.53\,\text{m/s}$