Momentum Circular Motion and Gravitation Rotational Motion Fluid Mechanics

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Momentum Circular Motion and Gravitation Rotational Motion Fluid Mechanics

Momentum

Momentum Collisions between objects can be evaluated using the laws of conservation of energy and of momentum. Momentum is defined as p = mv (units: kg m/s) Think of momentum as a measure of a moving objects resistance to change in motion.

Momentum Changing the momentum of an object requires a force The change in momentum is called the impulse FΔt = Δp Area under force-time graph is impulse Conservation of Momentum the total momentum of a system is conserved p i = p f Momentum is always conserved in collisions Momentum is always conserved in propulsion (like rocketry or particle decay or nuclear fission)

Momentum Elastic Collisions Kinetic energy is conserved KE i = KE f Occur in objects hard enough for no thermal energy to be produced At the instant of collision, all energy is elastic potential Elastic collision occur on the atomic scale, but are an idealization for larger objects To solve, set up a conservation of momentum equation and a conservation of kinetic energy equation, then solve system of equations

Inelastic Collisions Momentum Objects collide and then move as separate object, most of the time moving in the same general direction Objects lose some kinetic energy during collision Becomes thermal or sound or maybe light Perfectly Inelastic Collisions Objects do not separate after collision Treat them as one object in the final terms of your equation

Glancing collisions Momentum Sometimes collisions will be in one dimension, but most of the time you will be asked to solve twodimensional collision problems Momentum is a vector Break it down into x- and y-components Set up conservation of momentum equation for each direction, solve Use vector addition (Pythagorean Theorem, inverse tangent) when needed

Momentum Multiple Choice Questions TIMED! Quietly answer the questions on your own.

Multiple Choice Answers 1. B 2. C 3. D 4. B 5. E 6. B 7. C 8. D 9. B 10. A

Circular Motion and Gravitation

Circular Motion and Uniform Circular Motion Movement of an object on a circular path with a constant speed Velocity vector is tangent to the circle Centripetal Acceleration Gravitation Velocity is always changing object always changes direction Acceleration points towards the center of the circle a c = v 2 /r Acceleration is constant for uniform circular motion

Circular Motion and Centripetal Force Gravitation The net force on an object moving in uniform circular motion is the centripetal force F = F c = ma c Points towards the center of the circle (same direction as the acceleration) Other forces can take the role of the centripetal force If force is removed, object will continue in the path of the velocity vector, tangent to the circle

Circular Motion and Non-Uniform Circular Motion When speed around the circle is not constant, there is a changing acceleration The net force will not point directly to the center Break the net force down to components Radial component Gravitation Tangential component Similarly, break the acceleration vector down to components

Circular Motion and Gravitation Newton s Law of Universal Gravitation For both masses, the magnitude of the attractive force between them is F G = G(m 1 m 2 /r 2 ) G is the universal constant 6.67x10-11 N m 2 /kg 2 This equation works for any two objects anywhere in the universe, even on the surface of Earth

Circular Motion and Gravitation Kepler s First Law of Planetary Motion Planetary pathways are elliptical, orbiting around the Sun, which is at one focus of the ellipse

Circular Motion and Gravitation Kepler s Second Law of Planetary Motion An imaginary line from the Sun to the planet sweeps out equal areas in equal period of time Planets move faster when they re closer to the Sun and slower when they re farther away

Circular Motion and Gravitation Kepler s Third Law of Planetary Motion The ratio of the cube of a planet s mean distance from the Sun to the square of its period is constant for all planets (R 1 ) 3 /(T 1 ) 2 = (R 2 ) 3 /(T 2 ) 2 Period time (in seconds) for one revolution Frequency number of revolutions per second Inversely related f = 1/T and T = 1/f

Circular Motion and Gravitation Multiple Choice Questions TIMED! Quietly answer the questions on your own.

Multiple Choice Answers 1. D 2. A 3. E 4. B 5. C 6. D 7. D 8. A 9. C 10. B

Rotational Motion

Rotational Motion Linear Displacement (m) Δx = x f x i Velocity (m/s) v = Δx/Δt Acceleration (m/s 2 ) a = Δv/Δt Rotational Angular Displacement (rad) Δθ = θ f - θ i Angular Velocity (rad/s) ω = Δθ/Δt Angular Acceleration (rad/s 2 ) α = Δω/Δt

Rotational Motion Relationships Between Linear and Rotational x = rθ v = rω a tan = rα a rad = ω 2 r Rotational Kinematic Equations ω = ω 0 + αt θ = ω 0 t + ½ αt 2 ω 2 = (ω 0 ) 2 + 2α(Δθ)

Rotational Motion Torque tendency of a force to cause rotation The perpendicular component of a force applied to an object causes rotational motion τ = F r sinθ Counterclockwise torque is, by convention, a positive quantity; clockwise is negative Use right-hand rule to determine direction (curl fingers in direction of rotation, thumb points in direction of torque)

Rotational Motion Rotational Analog to Newton s 2 nd Law F = ma τ = mr 2 α Moment of Inertia objects resistance to change in rotation I = mr 2 Depends on size, shape and density of the object In general, more mass further from the rotational axis means more rotational inertia

Rotational Motion

Rotational Motion Rotational kinetic energy KE rot = ½ Iω 2 Total kinetic energy of an object undergoing both linear and rotational motion is KE tot = ½ m(v cm ) 2 + ½ I cm ω 2 cm = center of mass Rotational analog of work-energy theorem W = ΔKE W = τδθ

Rotational Motion Angular Momentum L = Iω (units: kg m 2 /s) Rotational analog of Newton s 2 nd Law in terms of angular momentum τ = Iα = ΔL/Δt (equates torque with rate of change of angular momentum) Conservation of Angular Momentum total angular momentum is conserved as long as there s no net torque acting on the rotating object

Rotational Motion Rotational Equilibrium 1 st Condition net force is zero 2 nd Condition net torque is zero Solving rotational equilibrium problems Set up net force equation, set to zero Set up net torque equation, set to zero Solve system of equations

Rotational Motion Multiple Choice Questions TIMED! Quietly answer the questions on your own.

Multiple Choice Answers 1. D 2. D 3. B 4. C 5. E 6. D 7. E 8. E 9. B 10. A

Fluid Mechanics

Fluid Mechanics States of matter: solids, liquids, gases Solids: unchanging shape and volume, incompressible Liquids: assumes form of container, generally incompressible, fixed volume Gases: assumes shape and volume of container Fluids: liquids and gases

Fluid Mechanics Density ratio of mass to volume In a pure substance, density is uniform Temp. and pressure can affect density Specific Gravity ratio of substance s density to water s density (at 4.0 C)

Fluid Mechanics Pressure ratio of force to area, P = F/A Units: pascals, Pa 1Pa = 1N/m 2 Pressure is proportional to the depth at measurement P = ρgh Atmospheric pressure 1atm = 1.013x10 5 N/m 2 Pressure at sea level

Fluid Mechanics Pascal s Principle pressure is distributed equally throughout a fluid (at same height) Hydraulic lifts operate under this principle F 1 /A 1 = F 2 /A 2

Fluid Mechanics Buoyant force upward force on a object in a fluid Archimedes Principle the buoyant force is equal to the weight of the fluid displaced by the object F B = ρ f gv Object will float in a fluid of higher density, sink in a fluid of lower density

Fluid Mechanics Two types of flow Laminar flow smooth motion of adjacent layers of fluid - low viscosity Turbulent flow creation of small circular currents (eddies) high viscosity Equation of continuity relationship between speed of flow to cross-sectional area of pipe or tube A 1 v 1 = A 2 v 2

Fluid Mechanics Bernoulli s Principle velocity and pressure are inversely related When pressure is high, fluid flow is low When pressure is low, fluid flow is high Bernoulli s Equation (laminar flow) P 1 + ½ ρ(v 1 ) 2 + ρgy 1 = P 1 + ½ ρ(v 2 ) 2 + ρgy 2

Fluid Mechanics Multiple Choice Questions TIMED! Quietly answer the questions on your own.

Multiple Choice Answers 1. C 2. E 3. C 4. E 5. D 6. A 7. C 8. E 9. A 10. D