Comprehensive Lecture Notes on General Physics 1: Mechanics, Work-Energy, Gravitation, and Elasticity

General Course Information and Mechanics Curriculum

  • Course Title: General Physics 1 (3 Units)
  • Course Code: PHS 101 (PART A) / Physics
  • Instructor: Professor Akinlami Johnson O.
  • Platform: SONRAL LEARNING MANAGEMENT SYSTEM (LMS)
  • Core Discipline: Mechanics
  • Curriculum Module Outline:
    • Units and Dimension
    • Rectilinear Motion
    • Newton's Laws
    • Friction
    • Work, Energy and Conservation of Energy
    • Gravitation
    • Elasticity, Young Modulus and Bulk Modulus

Principles of Work

  • Fundamental Definition of Work:
    • Work is said to be done only when the point of application of the force moves through a distance.
    • Basic scalar formulation:     Work done=force×distance\text{Work done} = \text{force} \times \text{distance}
  • Work with Force at an Angle:
    • For a body acted upon by a force inclined at an angle θ\theta to the displacement, the work done is given by:     W=Fscos⁡(θ)W = F s \cos(\theta)
  • Vector Formulation of Work:
    • Force vector in three-dimensional Cartesian coordinates:     F=Fxi+Fyj+Fzk\mathbf{F} = F_x \mathbf{i} + F_y \mathbf{j} + F_z \mathbf{k}
    • Displacement vector in three-dimensional Cartesian coordinates:     s=Δxi+Δyj+Δzk\mathbf{s} = \Delta x \mathbf{i} + \Delta y \mathbf{j} + \Delta z \mathbf{k}
    • Total work done represented by the scalar (dot) product of force and displacement:     W=F⋅sW = \mathbf{F} \cdot \mathbf{s}W=FxΔx+FyΔy+FzΔzW = F_x \Delta x + F_y \Delta y + F_z \Delta z

Gravitational Work and Integral Formulations

  • Work Done by Gravity:
    • Gravitational work near the surface of the Earth:     W=mghW = mgh
    • Elevation change between two vertical coordinates h1h_1 and h2h_2:     W=mg(h2−h1)W = mg(h_2 - h_1)
    • Path Independence: The work done by the force of gravity depends only on the initial and final vertical coordinates, not on the path taken.
  • Integral Form of Work Done:
    • When a force FxF_x acts along a continuous path from an initial position xAx_A (point AA) to a final position xBx_B (point BB), the total work done is the line integral:     WA→B=∫xAxBFx dxW_{A \rightarrow B} = \int_{x_A}^{x_B} F_x\,dx

Kinetic Energy and the Work-Energy Relation

  • Derivation of the Kinetic Energy Expression:
    • Applying Newton's second law along the direction of displacement:     W=FΔxW = F \Delta xW=maΔxW = m a \Delta x
    • For a body undergoing uniform acceleration, the acceleration aa is constant, satisfying the kinematic equation:     v22=v12+2a(Δx)v_2^2 = v_1^2 + 2 a (\Delta x)
    • Rearranging the kinematic equation to solve for the product aΔxa \Delta x:     aΔx=12(v22−v12)a \Delta x = \frac{1}{2} (v_2^2 - v_1^2)
    • Substituting this relation directly into the work expression:     W=12m(v22−v12)W = \frac{1}{2} m (v_2^2 - v_1^2)W=12mv22−12mv12W = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2
  • Definition of Kinetic Energy:
    • The physical quantity 12mv2\frac{1}{2} m v^2 is called the Kinetic energy.

Potential Energy and Power

  • Potential Energy:
    • Gravitational potential energy relationship:     W=mgh=F⋅xW = mgh = \mathbf{F} \cdot \mathbf{x}
    • Concept: Potential energy is the energy associated with the relative positions of two or more interacting particles.
  • Power Definitions:
    • Power is defined as the rate at which work is done.
    • Average Power (PavP_{av}):     Pav=ΔWΔtP_{av} = \frac{\Delta W}{\Delta t}
    • Instantaneous Power (PinstP_{inst}):     Pinst=dWdtP_{inst} = \frac{dW}{dt}
    • Infinitesimal work element dWdW expressed in terms of force and displacement:     dW=F⋅dsdW = \mathbf{F} \cdot d\mathbf{s}
    • Instantaneous power expressed in terms of velocity:     P=F⋅dsdtP = \mathbf{F} \cdot \frac{d\mathbf{s}}{dt}
  • Alternative Definitions of Power:
    • Power can also be defined as the rate of energy transfer from one body to another.
    • Power can be defined as the rate at which energy is transferred from one form to another:     P=dEdtP = \frac{dE}{dt}

Water Pump Power Calculation

  • Example Problem Statement:
    • A pump raises water from a well of depth 20 m20\,\text{m} at a rate of 10 kg s−110\,\text{kg}\,\text{s}^{-1} and discharges it at 6 m s−16\,\text{m}\,\text{s}^{-1}. What is the power of the motor?
  • Formulation and Mathematical Solution:
    • Total work performed in raising and discharging the mass of water:     Total Work=mgh+12mv2\text{Total Work} = mgh + \frac{1}{2} m v^2
    • Determining the motor power by differentiating total work with respect to time:     P=ddt(mgh+12mv2)P = \frac{d}{dt}\left(mgh + \frac{1}{2} m v^2\right)P=(dmdt)(gh)+12(dmdt)v2P = \left(\frac{dm}{dt}\right)(gh) + \frac{1}{2}\left(\frac{dm}{dt}\right) v^2
    • Given parameters:
    • Mass flow rate: dmdt=10 kg s−1\frac{dm}{dt} = 10\,\text{kg}\,\text{s}^{-1}
    • Well depth: h=20 mh = 20\,\text{m}
    • Acceleration due to gravity: g=10 m s−2g = 10\,\text{m}\,\text{s}^{-2}
    • Discharge speed: v=6 m s−1v = 6\,\text{m}\,\text{s}^{-1}
    • Intermediate calculations:     gh=(10 m s−2)(20 m)=200 m2 s−2gh = (10\,\text{m}\,\text{s}^{-2})(20\,\text{m}) = 200\,\text{m}^2\,\text{s}^{-2}12v2=12(6 m s−1)2=12(36 m2 s−2)=18 m2 s−2\frac{1}{2} v^2 = \frac{1}{2}(6\,\text{m}\,\text{s}^{-1})^2 = \frac{1}{2}(36\,\text{m}^2\,\text{s}^{-2}) = 18\,\text{m}^2\,\text{s}^{-2}
    • Numerical substitution and final value:     P=10(200+18)P = 10(200 + 18)P=10(218)=2180 WP = 10(218) = 2180\,\text{W}

Gravitation and Gravitational Field Strength

  • Newton's Law of Gravitation:
    • Statement: The force of attraction between two masses m1m_1 and m2m_2 separated by a distance rr is proportional to the product of their masses and inversely proportional to the square of the distance of separation between them.
    • Mathematical expression:     F=−Gm1m2r2F = -\frac{G m_1 m_2}{r^2}
  • Gravitational Field Strength or Intensity (gg):
    • Definition: Gravitational field strength is the gravitational force per unit mass:     F=mgF = mgg=Fmg = \frac{F}{m}
    • Substituting Newton's gravitational force into the field equation:     g=−GMmmr2g = -\frac{G M m}{m r^2}g=−GMr2g = -\frac{G M}{r^2}
    • Surface of the Earth: The magnitude of the field strength at the surface of the earth is:     g(RE)=GMERE2g(R_E) = \frac{G M_E}{R_E^2}     where MEM_E is the mass of the Earth and RER_E is the radius of the Earth.

Kepler's Laws of Planetary Motion

  • First Law:
    • States that the planets move around the sun in elliptical orbits with the sun at one focus.
  • Second Law:
    • States that the line joining the sun to a planet sweeps out equal areas in equal times.
  • Third Law:
    • States that the square of the period of a planet is proportional to the cube of the semi-major axis of its orbit:     T2∝a3T^2 \propto a^3

Elasticity and Elastic Moduli

  • Mechanics of Elastic Deformation:
    • A force applied to an object can change its shape.
    • In general, the response of a material to a given type of deforming force is characterized by an elastic modulus.
    • General definition of an elastic modulus:     Elastic modulus=stressstrain\text{Elastic modulus} = \frac{\text{stress}}{\text{strain}}Elastic modulus=stresstensilestraintensile\text{Elastic modulus} = \frac{\text{stress}_{\text{tensile}}}{\text{strain}_{\text{tensile}}}
  • Fundamental Quantities:
    • Tensile Stress:
    • Defined as a force per unit area.
    • Measured in units of N m−2\text{N}\,\text{m}^{-2}.
    • Tensile Strain:
    • Defined as a fractional change in a dimension or volume.
    • Being a ratio of identical physical quantities, strain is dimensionless.
  • Three Primary Elastic Moduli:
    • Young's modulus (governs changes in length in solids).
    • Shear modulus (governs shear distortion and tangential force response in solids).
    • Bulk modulus (governs uniform volume changes in solids and fluids).

Young's Modulus and Longitudinal Strain

  • Physical Meaning:
    • Young's modulus is the measure of the resistance of a solid to a change in its length when a force is applied perpendicular to a face.
  • Theoretical Setup:
    • Consider a solid rod with an unstressed length l0l_0 and cross-sectional area AA.
    • The rod changes in length by Δl\Delta l when subjected to equal and opposite forces FF directed along its axis and perpendicular to the end faces.
    • These axial tensile forces tend to stretch the rod.
  • Mathematical Expressions:
    • Tensile stress on the rod:     Tensile stress=FA\text{Tensile stress} = \frac{F}{A}
    • Tensile strain on the rod:     Tensile strain=Δll0\text{Tensile strain} = \frac{\Delta l}{l_0}
    • Young's Modulus definition:     Y=Tensile stressTensile strainY = \frac{\text{Tensile stress}}{\text{Tensile strain}}Y=FAΔll0=Fl0AΔlY = \frac{\frac{F}{A}}{\frac{\Delta l}{l_0}} = \frac{F l_0}{A \Delta l}

Shear Modulus and Tangential Stress

  • Physical Meaning:
    • The shear modulus of a solid indicates its resistance to a shearing force, which is a force applied tangentially to a surface.
  • Shear Stress:
    • Defined as the ratio of tangential deforming force to the surface area over which it is applied:     Shear stress=Tangential Force FArea A=FA\text{Shear stress} = \frac{\text{Tangential Force } F}{\text{Area } A} = \frac{F}{A}
  • Shear Strain:
    • Defined as the lateral displacement divided by the transverse height:     Shear strain=Δxh\text{Shear strain} = \frac{\Delta x}{h}     where Δx\Delta x is the horizontal displacement of the sheared surface and hh is the vertical separation between the top and bottom surfaces.
  • Shear Modulus (SS):
    • Defined as the ratio of shear stress to shear strain:     S=shear stressshear strainS = \frac{\text{shear stress}}{\text{shear strain}}S=FAΔxh=FhAΔxS = \frac{\frac{F}{A}}{\frac{\Delta x}{h}} = \frac{F h}{A \Delta x}

Bulk Modulus and Compressibility

  • Physical Meaning:
    • The bulk modulus of a solid or a fluid indicates its resistance to a change in volume.
  • Model System:
    • Consider a cube of a material (solid or fluid) whose faces each experience the same uniform force FF applied normally across each face of area AA.
  • Pressure:
    • The pressure on the cube is the normal force per unit area:     P=FAP = \frac{F}{A}
  • Response to Pressure Changes:
    • When the pressure on a body is increased, its volume decreases.
    • Volume Stress: The change in pressure ΔP\Delta P applied to the body is called the volume stress.
    • Volume Strain: The resulting fractional change in volume ΔVV\frac{\Delta V}{V} is called the volume strain.
  • Bulk Modulus (BB) Expression:
    • The bulk modulus of the material is defined as:     Bulk Modulus=Volume StressVolume Strain\text{Bulk Modulus} = \frac{\text{Volume Stress}}{\text{Volume Strain}}B=−ΔPΔVVB = -\frac{\Delta P}{\frac{\Delta V}{V}}
  • Compressibility (KK):
    • Compressibility is the inverse of the bulk modulus:     K=1BK = \frac{1}{B}