Linear Motion
Kinetic Energy in J if m is in kg and V is in m/s KE = ½mV² Potential Energy in J PE = mgh g is the acceleration of gravity (9.8 m/s² on surface of earth) KE = P²/2m P = mV = momentum Impulse (N•s or kg•m/s) is change in momentum, dP Force = dP/dt Impulse dP = ∫ F dt Equations of motion (straight line, constant acc) d = ½at² + v₀t + d₀ d is displacenemt v₀ is initial velocity d₀ is initial position v = v₀ + at v² = v₀² + 2ad F = ma a = Δv/Δt body thrown upwards with initial velocity d = –½gt² + v₀t time to return to ground 0 = –½gt² + v₀t t = 2v₀/g g = 9.8 m/s² or 32 ft/s² body thrown upwards with initial v and d d = –½gt² + v₀t + d₀ maximum height reached by a vertically projected body with velocity v h = v²/2g v = √(2gh) v² = 2gh or 2ad g = 9.8 m/s² gravitational acceleration falling object starting from rest h is height in meters, t is time falling in seconds, g is acceleration of gravity, usually 9.8 m/s² v is velocity in m/s h = ½gt² t = √(2h/g) v = √(2gh) h = v²/2g v = gt Above modified for large distances v = √(2gh) becomes v = √[2Gm((1/r)–(1/(r+d))] distance d on planet with mass m and radius r G = 6.674e-11 m³/kgs² t = √(2h/g) becomes The time t taken for an object to fall from a height r to a height x, measured from the centers of the two bodies, is given by: t = (arccos√(x/r) + √((x/r)(1–(x/r))(R^3/2) / √(2µ) µ = G(m₁+m₂) m₁, m₂ masses of the two bodies falling body with initial velocity h = ½gt² + v₀t range of object projected at an angle θ R = (V²/g)sin(2θ) V is initial velocity in m/s g is the acceleration of gravity 9.8 m/s² Time of flight = R/(Vcosθ) Max height h = (V²/2g)sin²θ R/h = cotθ at 45º these reduce to R = (V²/g) t = (V/g)√2 h = (V²/4g) different heights: R = (v₀²/2g)(1 + √(1 + (2gy₀/v₀²sin²θ))) sin(2θ) d is the total horizontal distance travelled by the projectile. v is the velocity at which the projectile is launched g is the gravitational acceleration—usually taken to be 9.81 m/s² near the Earth's surface θ is the angle at which the projectile is launched y₀ is the initial height of the projectile Elastic collisions v is velocity after the collision, u before v₁ = (u₁(m₁–m₂) + 2m₂u₂) / (m₁ + m₂) v₂ = (u₂(m₂–m₁) + 2m₁u₁) / (m₁ + m₂) friction on a ramp the component of object's weight parallel to the ramp is mg•sinθ friction force is mg•µcosθ where µ is coef. of friction. when you combine these you get F = mg(sinθ – µcosθ) and acceleration is F/m = g(sinθ – µcosθ) terminal velocity Vt = √(2mg/ρACd) Vt = terminal velocity, m = mass of the falling object, g = gravitational acceleration, Cd = drag coefficient, ρ = density of the fluid/gas A = projected area of the object. force of drag for high speed turbulent flow Fd = ½ρv²ACd in Newtons ρ is the density of the fluid in kg/m³ (for air it is 1.293 kg/m³ at 0°C) v is the speed of the object relative to the fluid in m/s A is the reference area in m² (πr² for a sphere) Cd is the drag coefficient (0.47 for a sphere) For laminar: Under conditions of laminar flow, the force required to move a plate at constant speed against the resistance of a fluid is proportional to the area of the plate and to the velocity gradient perpendicular to the plate. The constant of proportionality is called the viscosity. power to push thru the drag Pd = ½ρv³ACd Low speed drag (laminar flow) Fd = -bv b is constant dependent on fluid and dimensions |
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