Module 5: Advanced Mechanics Projectile motion: - horizontal and vertical of motion are independent, they do not affect each other - horizontally no acceleration/deceleration ignoring air resistance, so it can be said that initial horizontal velocity ux is the same as any vx during the duration of the flight - vertical acceleration exists even without additional propulsion, vy=uy-9.8t (g=a=-9.8ms-2) - based on rectilinear motion, 2D motion includes (assume u is resultant velocity): - x-axis y-axis ux=ucosθ vx=ux (a=0) vx2=ux2 sx=uxt uy=usinθ vy=uy-9.8t (a=g=-9.8ms-2) vy2=uy2-19.6sy sy=uyt-4.9t2 (not in syllabus, but nice observation) assuming place of landing is level with place of launch (ie same acute angle), the launch angle is the same as the tan-1(4sy maximum/sx maximum), finding the average gradient between starting point and highest point, highest point having a gradient of 0, as there is 0 vertical speed instantaneously Circular motion: - accelerates towards centre, centripetal acceleration often manifesting as tension, (plus minus gravity in vertical circular however) friction etc, always changing direction, but not speed (ac=v2/r) - speed is revolutions per period, where period T is completion of one revolution (v=2𝝿r/T) - if centripetal force is suddenly cut off (eg tension in string lost as it snaps), spinning object follows a tangential path - -direction of acceleration and F derived as follows: 1. examine displacement vectors in circular motion 2. examine velocity factors 3. similar triangles - given change of velocity is acceleration, it can be viewed as velocity*frequency (how often the velocity changes) - just multiply by mass for centripetal force, Fnet=ΣF=ma, Fc=mv2/r Angular displacement and velocity: - angular change is just raw displacement over r, as displacement is a portion of 2𝝿r, if 2𝝿 is one revolution in radians (Δθ=s/r) - angular velocity should therefore be the angular displacement over a set time, hence ω=Δθ/Δt=s/(r*Δt)=v/r Horizontal circular bends: - on flat ground, perpendicular forces do not contribute to centripetal force - rather in a circular bend, friction provides Fc -ΣF=Friction=mv2/r, ∴we can see that it is dependent on velocity -if friction is less than centripetal force, object leaves circle Banked turns: - let mg=weight, N equals normal force to surface, centripetal force composed of friction on road (FNsinθ=Fc=mv2/r) to keep vehicle on the surface, reaction ‘force’ or normal to gravity (not surface), FNcosθ=weight=mg (balance out, net force up and down are equal) ∴ tanθ=v2/gr Horizontal circular motion on a string: -w=mg=Tsinθ=Fctanθ -Fc=Tcosθ Vertical circular motion on a string: - tension varies, with low tension at top of circle (weightlessness) and high tension on bottom (heaviness) however, centripetal acceleration and force is constant Work in circular motion: - work is equated to change in energy - when moving vertically, work=potential energy gained - no work is done in circular motion, as energy is constant and a revolution gets something to its original position, effectively as if no energy is put in centrifugal work is zero as force is drawn to centre and motion is perpendicular (W=Fs, s is displacement, cos90=0) sum of energy is KE and PE (gravitational potential energy), meaning for a constant internal energy, object is faster the lower it gets (potential energy is spent) Torque: - essentially the rotational equivalent of work - τ (tau, not T, do not confuse with tension) = r⊥F = r F sinθ (perpendicular to lever) - therefore, to generate the most torque, one must grip closer to the end away from the centre, or push directly at 90° (sinθmax=1, and only acute to perpendicular angles are considered, as supplementary angles yield same results) Motion in Gravitational Fields: - mass attracts mass - g field is a region of space around a body on another - F=GMm/r2, where F is gravitational force in newtons, G is universal gravitational constant G=6.67*10-11Nkg-2m2 - m is mass experiencing force (smaller mass), M exerting force (technically interchangeable, but on Earth, Earth >>>> smaller mass) - r is for radius, distance from centre to centre of mass, or Earth to height of orbit where distance to centre in smaller object is negligible - Given the format F=ma, g of earth (9.8ms-2) is equal to GM/r2 Geostationary, geosynchronous and near-earth satellites: - geostationary: always aligned with same point over the equator - geosynchronous: matches orbit on axis, or ~24hours for full rotation on axis - - near-earth: often used for nearby surveillance, low orbit of ~2000km above the surface, decays in energy due to friction so high initial velocity is needed for a long lasting orbit Hohmann transfer in orbit → path used to manoeuvre satellite one orbit to another → initially in LEO, then boosted up → specific orbital energy lies between lower and higher path → slowest at apogee (highest point, see Kepler’s #3 law) Here, the chaser is at the perigee, then approaches height at apogee to reach the target altitude Kepler’s Laws: - #1: planets are in elliptical orbit around a star (ie the Sun, if within Solar System), at one of two foci - #2: equal areas are swept in a path within the same time, regardless of arc length and distance/eccentricity from star - → A smaller angle/angle is needed to cover an area if far away, but is proportionally slower (taking the same time) due to high GPE and low kinetic energy, if energy is conserved as ΔU+ΔK=0 #3: average radius cubed over average period squared is constant for all distances from a star (R1 3/T12=R23/T22) →given v=√(GM/r) and v=2πr/T, this means R3 /T2=GM/4π2 Potential energy in a gravitational field: - W=Fs (vector work, not just scalar), where F=mg=GMm/r2, and s=radius - M exerts a force on m, limr^2→∞ Fg=0, ie gravitational force becomes negligible at enormous distances Total orbital energy: - defining W when U=0=mgh, →W=Fg*r=K=(GMm/r2)r=GMm/r - As work is change in energy, by conservation (assuming no additional input), E=K+U=0, where K=-U - therefore, U=-GMm/r, negative as energy ‘held from’ equilibrium of 0, the closer you get, the more GPE you lose Escape velocity: - W=GMEarthm/rEarth is work needed to escape earth →if Fg is equated to mv2/r or 2Kr, v2=2GM/r and escape velocity is √(2GM/r)
