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Engineering Mechanics : Dynamics Uniform Rectilinear Motion

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Engineering Mechanics : Dynamics Uniform Rectilinear Motion Engineering Mechanics : Dynamics Uniform Rectilinear Motion For particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant. dx = v = constant dt x t ∫ dx = v∫ dt x0 0 x − x0 = vt x = x0 + vt 11 - 1 Engineering Mechanics : Dynamics Uniformly Accelerated Rectilinear Motion For particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant. dv v t = a = constant ∫ dv = a∫ dt v − v = at dt 0 v0 0 v = v0 + at dx x t = v + at ∫ dx = ∫ ()v + at dt x − x = v t + 1 at 2 dt 0 0 0 0 2 x0 0 x = x + v t + 1 at 2 0 0 2 v x dv 2 2 v = a = constant ∫ v dv = a ∫ dx 1 (v − v )= a()x − x dx 2 0 0 v0 x0 2 2 v = v0 + 2a()x − x0 11 - 2 Engineering Mechanics : Dynamics Motion of Several Particles: Relative Motion • For particles moving along the same line, time should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction. xB A = xB − xA = relative position of B with respect to A xB = xA + xB A vB A = vB − vA = relative velocity of B with respect to A vB = vA + vB A aB A = aB − aA = relative acceleration of B with respect to A aB = aA + aB A 11 - 3 Engineering Mechanics : Dynamics Motion of Several Particles: Dependent Motion • Position of a particle may depend on position of one or more other particles. • 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. xA + 2xB = constant (one degree of freedom) • Positions of three blocks are dependent. 2xA + 2xB + xC = constant (two degrees of freedom) • For linearly related positions, similar relations hold between velocities and accelerations. dx dx dx 2 A + 2 B + C = 0 or 2v + 2v + v = 0 dt dt dt A B C dv dv dv 2 A + 2 B + C = 0 or 2a + 2a + a = 0 dt dt dt A B C 11 - 4 Engineering Mechanics : Dynamics Curvilinear Motion: Position, Velocity & Acceleration • Consider particle which occupies position P defined r r by r at time t and P’ defined by r ′ at t + ∆t, r r r ∆r rd v = lim = ∆t→0 ∆t dt = instantaneous velocity (vector) ∆s ds v = lim = ∆t→0 ∆t dt = instantaneous speed (scalar) 11 - 5 Engineering Mechanics : Dynamics Curvilinear Motion: Position, Velocity & Acceleration r v • Considerr velocity of particle at time t and velocity v′ at t + ∆t, r r r ∆v vd a = lim = ∆t→0 ∆t dt = instantaneous acceleration (vector) • In general, acceleration vector is not tangent to particle path and velocity vector. 11 - 6 Engineering Mechanics : Dynamics Rectangular Components of Velocity & Acceleration • When position vector of particle P is given by its rectangular components, r r r r r = ix + jy + kz • Velocity vector, r dx r dy r dz r r r r v = i + j + k = x&i + y& j + z&k dt dt dt r r r = vx i + vy j + vzk • Acceleration vector, r d 2xr d 2 y r d 2 z r r r r a = i + j + k = &&ix + && jy + &&kz dt 2 dt 2 dt 2 r r r = ax i + a y j + azk 11 - 7 Engineering Mechanics : Dynamics Rectangular Components of Velocity & Acceleration • Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile, ax = x&& = 0 a y = y&& = −g az = z&& = 0 with initial conditions, x = y = z = 0 (v ) , v ,(v ) = 0 0 0 0 x 0 ( y )0 z 0 Integrating twice yields v = v v = v − gt v = 0 x ( x )0 y ( y )0 z x = v t y = v t − 1 gt 2 z = 0 ()x 0 ()y 0 2 • Motion in horizontal direction is uniform. • Motion in vertical direction is uniformly accelerated. • Motion of projectile could be replaced by two independent rectilinear motions. 11 - 8 Engineering Mechanics : Dynamics Motion Relative to a Frame in Translation • Designate one frame as the fixed frame of reference . All other frames not rigidly attached to the fixed reference frame are moving frames of reference . • Position vectors for particles A and B rwith respectr to the fixed frame of reference Oxyz are rA and rB. r • VectorrB A joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and r r r rB = rA + rB A • Differentiating twice, r r r r vB = vA + vB A vB A = velocity of B relative to A. r r r r a B = aA + aB A aB A = acceleration of B relative to A. • Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to moving reference frame attached to A. 11 - 9 Engineering Mechanics : Dynamics Tangential and Normal Components r dv r v2 r dv v2 a = e + e a = a = dt t ρ n t dt n ρ • Tangential component of acceleration reflects change of speed and normal component reflects change of direction. • Tangential component may be positive or negative. Normal component always points toward center of path curvature. 11 - 10.
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