Dependent Motion • Position of a particle may depend on position of one or more other particles. Engr 240 – Week 2 • Position of block B depends on position of block A. Since rope is of constant length, it follows that sum of lengths of segments must be constant. x A + 2 x B = constant (one degree of freedom) • Positions of three blocks are dependent. Dependent Motion, Graphical Solution, Curvilinear Motion 2 x A + 2 x B + xC = constant (two degrees of freedom) • For linearly related positions, similar relations hold between velocities and accelerations. dx dx A dx + 2 B + C = 0 or 2v A + 2v B + vC = 0 dt dt dt dv dv dv 2 A + 2 B + C = 0 or 2a A + 2a B + aC = 0 dt dt dt 2 Sample Problem 11.5 SOLUTION: • Define origin at upper horizontal surface with positive displacement downward. Pulley D is attached to a collar which is pulled down at 3 in./s. At t = 0, collar A starts moving down from K with constant acceleration and zero initial velocity. Knowing that velocity of collar A is 12 in./s as it passes L, determine the change in elevation, velocity, and acceleration of block B when block A is at L. Graphical Solution of Rectilinear-Motion Problems Graphical Solution of Rectilinear-Motion Problems • Given the x-t curve, the v-t curve is equal to the x-t curve slope. • Given the a-t curve, the change in velocity between t1 and t2 is equal to the area under the a-t curve between t1 and t2. • Given the v-t curve, the a-t curve is equal to the v-t curve slope. • Given the v-t curve, the change in position between t1 and t2 is equal to the area under the v-t curve between t1 and t2. 1 Example. A subway train travels between two stations with the acceleration schedule shown. Find the time interval Δt during which the acceleration is 2 m/s2, and the distance s between the stations. CURVILINEAR MOTION OF PARTICLES – Motion along a curve other than a straight line • Position vector - the vector between origin O of reference frame and the position occupied by particle. • Consider particle which occupies position P defined r r by r at time t and P’ defined by r ′ at t + Δt, r r Δr dr r v = lim = dt Δt →0 Δt = instantaneous velocity (vector) v = lim Δs Δt →0 Δt = ds dt = instantaneous speed (scalar) r • Consider velocity v of particle at time t and velocity r v′ at t + Δt, r r Δv dv r = a = lim dt Δt →0 Δt = instantaneous acceleration (vector) • In general, acceleration vector is not tangent to particle path and velocity vector. Rectangular Components r r = x iˆ + y ˆj + z kˆ r r dr = x& iˆ + y& ˆj + z& kˆ v= dt r r dv = &x& iˆ + &y& ˆj + &z& kˆ a= dt where the “overdots” represent derivatives with respect to time. Example: Projectile Motion (x-y plane along the plane of motion) x: uniform motion y: free fall Equations of Motion: z: no motion Example. A projectile is fired from the edge of a 150-m cliff with an initial velocity of 180 m/s at an angle of 30° with the horizontal. Neglecting air resistance, find a) the horizontal distance from the gun to the point where the projectile strikes the ground, b) the greatest elevation above the ground reached by the projectile. 2 Normal and Tangential Components • Velocity vector of particle is tangent to path of particle. r v = veˆt r r • et and et′ are tangential unit vectors for the particle path at P and P’. r r r Δet = et′ − et • If Δθ is the angle between the two unit vectors, Δet = 2 sin (Δθ 2 ) r Δet sin (Δθ 2 ) r r en = en lim = lim Δθ →0 Δθ Δθ →0 Δθ 2 r de r en = t dθ Acceleration r v = veˆt • Differential velocity vector to find acceleration. r r r de dθ ds de dv r r dv dv r = et + v = et + v a= dθ ds dt dt dt dt dt but r det r ds = en ρ dθ = ds =v dθ dt After substituting, dv v2 r dv r v 2 r a = et + en at = an = ρ ρ dt dt • Tangential Component of acceleration – change in magnitude of velocity vector . • Normal Component – change in direction of velocity vector; towards center of curvature. Example (Problem 11.10) A motorist is traveling on a curved section of highway of radius 2500 ft at the speed of 60 mi/h. The motorist suddenly applies the brakes, causing the automobile to slow down at a constant rate. Knowing that after 8 s the speed has been reduced to 45 mi/h, determine the acceleration of the automobile immediately after the brakes have been applied. 3

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