One-Dimensional KinematicsInstantaneous speed, velocity and accelerationv = dx/dt x = ∫v dt Graphical interpretationsa = dv/dt v = ∫a dtAverage velocity and accelerationv = ∆x/ ∆ta = ∆v/ ∆tThe Big Three for UAMFree Fall
3 D KinematicsPosition Vector = r = xî + yĵ + z kVelocity vector = v = vxî + vyĵ + vzkSpeed is magnitude of velocity = pythag of velocityEx 1: Sally is chasing Calvin.Calvin is moving along a path defined by rcalvin = (t-1) i+(11-t2) jand Sally is moving along the path rsally = (5-5t) i + (5+5t) ja) When does Sally catch Calvin?b) What will be the angle between their two velocities when this happens?Scalar Dot Product – a method of multiplying vectors that results in a scalar quantityA ∙ B = AB cos θA ∙ B = AxBx + AyBy + …
Ex 2: A particle is moving in three dimensions with a velocity given byv = 4t3 i – (cos t) j + 7 ka) What is its acceleration as a function of time?b) What is its speed at t = 2?c) What is its displacement between t = 0 and t = 3?Projectile MotionEx 1 An object is thrown off a cliff 50 m high with an initial velocity of 25 m/s and an angle of elevation of 30 degrees. How far from the base of the cliff will it land?Ex 2: Derive an equation for range as a function of angle and initial velocity for level ground.Ex 3: A ball is thrown up a hill with an initial speed of 20 m/s at an angle of elevation of 45 degrees. The hill is inclined at an angle of 30 degrees.
Uniform Circular MotionDirection of velocity and accelerationac = v2/rRelative velocityV object a relative to c = v object a relative to b + v object b relative to cEx. Boat crossing a river
Newton’s LawsNewton’s First LawEquilibrium – a = 0 and Net F = 0Dynamic vs Static EquilibriumNewton’s 2nd LawFnet = maNewton’s 3rd LawEqual and opposite reaction forcesMass vs weight
Common forces1) Gravitational force (weight) Magnitude = mg Direction = straight down2) Normal force Direction = perpendicular to surface Magnitude = weight (if there are no other forces acting on it!) 3) Friction Force Direction = parallel to surface, opposite to incipient motion Magnitude Fstatic friction <= µsFN Fkinetic friction <= µk FN 4) Tension Direction = in direction of rope (towards center of rope) Magnitude = determined by SIN2
Problem SolvingFBDApply SIN2Look for equilibrium in each direction (tilt axis if necessary)Call direction of acceleration positiveCircular motion = there is always an acceleration toward thecenter.ac = v2/r
1) Suppose you apply the minimum force required to move an objectinitially at rest on a horizontal surface. If you continue to apply thisforce, once the object moves what will be the acceleration of theobject in terms of m, g, µs and µk?2) A mass m is suspended by two ropes on either side of the mass.The angles the ropes make with the horizontal are θ and ø. Calculatethe tension in each rope in terms of m, g, θ, and ø3) An object is at rest on an inclined plane of angle θ. a) Draw a free body diagram of the object. b) Express the frictional force required to keep the object stationaryin terms of m, g, and θ. c) Calculate the normal force in terms of…. d) What is the minimal coefficient of static friction required to keepthe object from slipping?
4) The Atwood machine consists of two masses (m1>m2)attached to arope over a pulley. a) What is the acceleration of the system of masses (assuming therope and pulley are massless) b) What vertical force must the pin of the pulley exert on the wheelof the pulley to hold it up?5) A man of mass m is standing in an elevator that is acceleratingupward with an acceleration a. What is the reading on a scale the manis standing on in terms of m and g?6) A pendulum bob of mass m hanging in a truck is deflected throughan angle of θ as the truck accelerates. Find the tension and theacceleration of the truck in terms of m, g, and θ.
7) Two masses (m1 and m2) are connected by a rope hangingfrom a pulley on an incline (θ and µ). What is the accelerationof the masses?8) A car is driven around a curve of radius R that is banked atan angle of θ. If the coefficient of friction between the car’stires and the road is µ, what is the maximum speed the car cango without flying off the road?
Motion with air resistanceFBDSIN2Express variables in terms of velocity and time to obtaina differential equationSeparate and integrateAn object of mass m is released from rest at t=0 andexperiences a force of air resistance given by F = -cv, where c is a constant. Calculate the velocity as afunction of time. Sketch graphs of velocity andacceleration.
Work, Energy and PowerWork--Displacement must occur for work to be done--Path independent--Unit is joulesW = ∫F ds where s is position/displacement
Total work can be found by finding the work done by each forceand adding OR finding the net force and compute the work doneby it.If force is constant, W = F ∙ s ↖Dot productW = F s cos θ OR Fxsx + Fysy +…Example: A mass m is dragged a distance s along a rough surfaceby a force F applied at an angle of elevation θ. The coefficient offriction is µ. What is the work done by each force and the total work?
Work Energy Theorem When work is done on an object, energy is changed. W = ∆KE = ∆ ½ m v2Example: A force F = (4x)i acts on an object of mass 2 kg as itmoves from x = 0 to x = 5m. If the object started at rest, find thenet work and the final speed of the object.Conservation of EnergyConservative forces (no friction): kinetic potential kineticNonconservative forces: kinetic/potential heat, sound, etc
Gravitational PE Ug = mghSpring PE (derived from Hooke’s Law (F = -kx)) Us = ½ kx2F = - dU/dsEnergy Wells : Stable and unstable equilibrium
PowerRate at which work is done P = W/tOf course, any equation can be turned into a calculusversion:P = dW/dt = F v cos θ
MomentumVector in the direction of velocityp = mvFnet = ma = m (dv/dt) = dp/dtForce is slope of p vs t graphWhen the net force on an object or system is zero, thenet momentum is constant (conserved).
Impulse = ∫F dt (= F∆t if the force is constant)Impulse – area under F vs t graphExample 1: During a collision with a wall lasting from t =0 to t =2s, the force acting on a 2 kg object is given bythe equation F = 4t(2s-t) i. a) Calculate the impulse of the force on the object. 16/3 i b) What is the average force on the object? 8/3 i c) If the object starts from rest, what is its finalvelocity? 8/3 i
Elastic collision – kinetic energy is conserved. Energystored in deforming the object is completely recovered.Spring between, very special situationsInelastic collision – kinetic energy is not conserved(converted to heat, sound, deformation)Completely inelastic collision – objects stick togetherafter the collision, KE not conserved.Solving collision problems1) Decide if collision is elastic or inelastic.2) Momentum is a vector.3) Use separate cons of p equations for x and y directions.4) Change in KE in collision is final KE – initial KE
Example 1 A bullet of mass 0.005 kg moving at a speed of 100 m/s imbeds in a 1kg block of wood resting on a frictionless surface and attached to a horizontal spring of k = 50 N/m.a) What is the velocity of the block the instant after the bullet strikes it? v = 0.50 m/sb) What is the maximum compression of the spring? x = 0.07 mExample 2 A mass of 3 kg moving to the right at a speed of 5m/s collides elastically with a 10 kg mass moving at 1 m/s tothe left. What is the velocity of the 10 kg mass after thecollision? v = 1.769 m/s
Center of massThe center of mass is the weighted average of thelocation of masses in a system.rCM = ∑mr/ ∑ mrCM = ∫x dm/ ∫dm This is extremely rare. You knowwhat that means…A 1 m long linear segment of wire has a linear massdensity of ρ= a x2 where a is a constant. Find thelocation of the center of mass.xCM = ¾ m
Angular acceleration ( α ) α = d ω /dt = d2θ /dt2 ω = ∫ α dtUAMThe Big Three and the Forgotten Fourth (mutatedversion) ω = ωo + αt θ = θ o + ωo t + ½ αt2 ω2 = ωo2+ 2α θ θ = ((ω + ωo)/2) tRelationship between angular and linear quantities v=rω a=rα
Moment of Inertia (rotational inertia) I = ∫ r2dmKinetic Energy of a rotating object KE = ½ I ω2Torque-the ability of a force to cause an object to accelerate angularlyUnits – Newton metersτ = r x F = rF sin θWhere r is the vector from the axis of rotation to the point wherethe force is applied.
Vector cross productA method of multiplying vectors that results in a vectorA x B = A B sinθTorque can also be found if the moment arm (perpendiculardistance from the axis to the line of action of the force) is knownτ = (moment arm) FNewton’s Second Law analogτ=IαWork analog Power analogW=τθ P=τω
Rotational Motion 2Moment of InertiaI = ∫ r2dmReview the integral derivation of the MOI equation forthe rod.I = ∑mr2 for a mass moving around a pointParallel Axis TheoremIparallel axis = I CM + MD2Given that the I CM for a rod about its center is 1/12ml2, what is the MOI for a rod around its end?
Angular MomentumL = r x p = rp sin θL=IωConservation of Angular MomentumRolling without SlippingTorque Equilibrium problems