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PHY 107 1st Edition Lecture 7 Outline of Last Lecture I. Free fall motion - “Round trip” free fallII. Graphical integration in motion analysis (non-constant acceleration)Outline of Current Lecture III. Motion in 2 and 3 dimensionsIV. Projectile motionV. Uniform circular motionCurrent LectureMotion in 2 and 3 Dimensions:- Two-dimensional – motion in a plane- Three-dimensional – motion through space- Relative motion – transformation of velocities between 2 reference systems- Position vector (´r) of a particle is a vector whose tail is at a reference point (usually the origin, O) and its tip is at the particle at point P´r = x^i + y^j + z^k- Displacement vector – for a particle that changes its position vector from ´r1 to ´r2, the displacement vector ∆´r is defined as: ∆´r = ´r2 - ´r1- Position vectors written in terms of components: ´r1 = x1^i + y1^j + z1^k´r2 = x2^i + y2^j + z^k ∆´r = (x2 - x1)^i + (y2 - y1)^j + (z2 - z1)^k = ∆x^i + ∆y^j + ∆z^k- Speed = |´v| = √vx2+vy2+vz2- Average and instantaneous velocity: Average velocity = displacementtime intervalo´vavg = ∆ ´r∆t = ∆ x^i+∆ y^j+∆ z^k∆ t = ∆ x^i∆t + ∆ y^j∆ t + ∆ z^k∆ t Instantaneous velocity as the limit: (∆t → 0)o´v = lim∆ ´r∆t = d ´rdt - As ∆t approaches zero;These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. Vector ´r2 moves towards vector ´r1 and ∆´r → 0 Direction of the ratio ∆ ´r∆t (and thus ´vavg) approaches the direction of the tangent to the path at position 1´vavg → ´v o´v = ddt(x^i + y^j + z^k) = dxdt^i + dydt^j + dzdt^k = vx^i + vy^j + vz^k- Average and instantaneous acceleration Average acceleration = c h ange∈velocitytime interval → ´aavg = ´v2− ´v1∆ t = ∆ ´v∆ t Instantaneous acceleration as the limit: (∆t → 0)o´a = lim∆ ´v∆ t = d ´vdt = ddt(vx^i + vy^j + vz^k) = d vxdt^i +d vydt^j + d vzdt^k = ax^i + ay^j + az^k Acceleration vector has no specific relationship with the pathProjectile Motion:- Motion of an object in a vertical plane under the influence of gravitational force- Launched with initial velocity ´v0- Horizontal and vertical components are: vox = v0cosθ0voy = v0sinθ0- Motion along x-axis has zero acceleration- Motion along y-axis has uniform acceleration (ay = -g) - Horizontal motion ax = 0 vx = v0cosθ0 (eq 1) x = (v0cosθ0)t (eq 2) Velocity along x-axis does not change (neglect the effects of air-resistance) - Vertical motion ay = -g vy = v0sinθ0 - gt (eq 3) y = (v0sinθ0)t - g t22 (eq 4) Along the y-axis the projectile is in free fall- The equation of the path: x = (v0cosθ0)t (eq 2), y = (v0sinθ0)t - g t22 (eq 4) Eliminate t → y = (tanθ0)x – [g2(v0cos θ0)2]x2 (upside down parabola)- Horizontal range – the distance OA (at point A, y=0) (eq 4) (v0sinθ0)t - g t22 = 0 → t(v0sinθ0 -g t2) = 0 o Solution 1: t = 0 – corresponds to point O and is of no interesto Solution 2: v0sinθ0 -g t2 = 0 – solution corresponds to part Ao t = 2 v0sin θ0g, substitute t into eq 2 and set R = 2 v02g sinθ0 cosθ0 =v02g sin2θ0 o R has its maximum value when θ = 45°, Rmax = v02g - Maximum height H → H = v02sin2θ02 gUniform Circular Motion:- Particle moves on a circular path of radius r with constant speed v - Even though speed is constant, velocity is not because the direction of the velocity vector changes from point to point along the path- Acceleration is not zero-´v = vx^i + vy^j + vz^kspeed =|´v| s = |´v| =√vx2+vy2+vz2- Acceleration in uniform circular motion: Its vector points towards the center (c) of the circular path (centripetal) Its magnitude a is given by the equation: a = v2r Period – time, T, taken to complete one full revolution: T = 2 π

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