Physics Formulas and Concepts Summary

Posted on May 21, 2025 in Physics

Moments of Inertia

I = m * d². System for discrete masses: I = Σmᵢ * dᵢ². System for continuous bodies: I = ∫ (d² * dm). Steiner’s Theorem (Parallel Axis Theorem): Ia = Ib + m * d² (where Ib is the moment of inertia about an axis passing through the center of mass G, and the axis ‘a’ is parallel to ‘b’ at a distance ‘d’).

Kinematics of a Point

Position

Defined by a path-time law.

Path (Trajectory)

The set of positions of point P. Can be the intersection of two surfaces. Eliminate time ‘t’ from equations to get the path equation.

Time Function

Obtained by integrating v = ds/dt. Position vector: r(t) = x(t)i + y(t)j + z(t)k (where x(t), y(t), z(t) are functions of time).

Velocity

Vector: v(t) = dr(t)/dt. Magnitude (Speed): |v| = ds/dt (always positive). Direction: Tangent to the path.

Acceleration

Vector: a(t) = dv(t)/dt.

  • Tangential component magnitude: |at| = d|v|/dt
  • Normal component magnitude: |an| = v²/R (where R is the radius of curvature)

Uniform Rectilinear Motion (URM)

  • Acceleration a = 0
  • Velocity v = constant
  • Position s(t) = v₀t + s₀

Uniformly Accelerated Rectilinear Motion (UARM)

  • Acceleration a = constant
  • Velocity v(t) = v₀ + at
  • Position s(t) = v₀t + ½at² + s₀

Circular Motion

  • Arc length s = θ * R
  • Tangential velocity v = ω * R
  • Angular velocity ω = dθ/dt
  • Angular acceleration α = dω/dt
  • Normal acceleration an = v²/R = ω² * R

Kinematics of a Rigid Body in Plane Motion

Velocity Analysis

Velocity of point B relative to point A: vB = vA + ω × rAB (Translation + Rotation).

Velocity relative to the Instantaneous Center of Rotation (ICR): vP = ω × rIC P (where rIC P is the vector from the ICR to point P).

Degrees of Freedom and Constraints

  • Point in a plane: 2 DOF
  • Rigid body in a plane: 3 DOF
  • System of 2 rigid bodies in a plane: 6 DOF (before constraints)

Constraints (reducing DOF):

  • Simple Support: -1 DOF
  • Cable: -1 DOF
  • Pin Joint: -2 DOF
  • Fixed Support (Abutment): -3 DOF

Velocity Determination Methods

Method 1: Using Relative Velocity (vB = vA + ω × rAB)

Method 2: Using the Instantaneous Center of Rotation (ICR)

Locate the ICR (e.g., intersection of perpendiculars to known velocities). Determine the angular velocity ω using a point with known velocity: ω = vA / rIC A. Note the sign convention for ω. Calculate other velocities using vP = ω * rIC P (magnitude) and direction perpendicular to rIC P.

Special Cases

Body rolling in contact with the ground: If rolling without slipping, the ICR is at the contact point. If rolling and slipping, the velocity at the contact point is non-zero.

Acceleration Analysis

Acceleration of point B relative to point A: aB = aA + α × rAB – ω² * rAB.

If velocity depends on a parameter (e.g., angle θ), acceleration can be found using the chain rule: aB = dvB/dt = (dvB/dθ) * (dθ/dt) = (dvB(θ)/dθ) * ω.

Systems with Multiple Bodies

Analyze each body separately. Start with a body connected to a fixed point (its ICR is the fixed point). Find its angular velocity. Use velocity compatibility at connection points (velocity of the shared point is the same for both bodies).

Statics

Method for Solving Statics Problems

  1. Draw a Free-Body Diagram (FBD). Identify all external forces (weight, applied forces, spring forces) and reactions (supports, connections).
  2. Write the Equilibrium Equations (ΣF = 0, ΣM = 0).
  3. Check: Number of equations vs. number of unknowns. If the number of equations equals the number of unknowns, the system is statically determinate. If there are more unknowns than equations, it’s indeterminate. If there are more equations than unknowns, check the problem statement or FBD. Remember friction: Fr ≤ μ * N (for impending motion, Fr = μ * N).
  4. Solve the system of equations for the unknowns.

Analysis of Composite Structures

Analyze each component separately using FBDs. Internal reactions at connections are equal and opposite between connected components.

For complex systems, global equilibrium equations can be used to find some unknowns.

System Classification

  • Statically Determinate (Isostatic): 0 degrees of freedom (after constraints). Number of unknowns equals number of equilibrium equations.
  • Mechanism: Degrees of freedom > 0. System is free to move.
  • Statically Indeterminate: Degrees of freedom < 0. Number of unknowns exceeds number of equilibrium equations.

Principle of Virtual Work (PVW)

  1. Establish a coordinate system, often with a fixed point.
  2. Define a generalized coordinate (parameter), typically an angle (θ).
  3. Express the position vectors of the points where forces are applied (e.g., rP). If there’s a spring, calculate its length and force (F = k(x – L₀)).
  4. Express the force vectors (e.g., F).
  5. Calculate the virtual displacement vector for each point of application: δrP = (drP/dθ) δθ.
  6. Apply the Principle of Virtual Work: Σ (Fi ⋅ δri) = 0. This expands to Σ (Fi ⋅ (dri/dθ)) δθ = 0. Since δθ is arbitrary (and non-zero), the sum of the dot products must be zero: Σ (Fi ⋅ (dri/dθ)) = 0.
  7. Solve the resulting equation for the unknown force or parameter.

Fluid Mechanics

Fluid Properties

  • Density (ρ): ρ = dm/dV. For a homogeneous substance, ρ = m/V (Units: kg/m³). Examples: ρwater ≈ 1000 kg/m³, ρair ≈ 1.2 kg/m³ (at STP), ρmercury ≈ 13600 kg/m³.
  • Relative Density (Specific Gravity): RD = ρsubstance / ρreference (dimensionless). Reference is usually water for liquids/solids, air for gases.
  • Specific Weight (γ): γ = Weight/Volume = ρ * g (Units: N/m³).

Pressure

Pressure (p) = Fperpendicular / Area (Units: N/m² = Pascal, Pa).

Other Units:

  • 1 atm ≈ 1.01325 × 10⁵ Pa
  • 1 atm = 760 mmHg
  • 1 atm ≈ 1013.25 millibar
  • 1 atm ≈ 1.033 kgf/cm²

Pascal’s Principle

Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. Example: Hydraulic Press F₁/A₁ = F₂/A₂.

Hydrostatic Pressure

Pressure at a depth h in a fluid: p = ρgh (relative to the surface pressure).

Pressure difference between two points in a vertical column of fluid: p₂ – p₁ = -ρg(h₂ – h₁) (where h is height, positive upwards).

Absolute Pressure: pabs = pgauge + patmospheric.

Method for Hydrostatics Problems

  1. Identify fluids and their densities.
  2. Locate interfaces between different fluids or between fluid and atmosphere/solid.
  3. Choose reference points and apply pressure formulas (p = ρgh, pabs = pgauge + patm).
  4. Relate pressures at different points, often using the fact that pressure is constant at the same horizontal level within a continuous fluid.