Engineering Mechanics : DYNAMIC DAJ 21003 ( STATICS & DYNAMICS ) Lecture #13 By, Noraniah Kassim Hairul Mubarak b Hassim Universiti Tun Hussein Onn Malaysia (UTHM), DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Objective • To investigate particle motion along a curved path “Curvilinear Motion” using three coordinate systems Rectangular Components • Position vector r = x i + y j + z k • Velocity v = vx i + vy j + vz k (tangent to path) • Acceleration a = ax i + ay j +az k (tangent to hodograph) – Normal and Tangential Components – Polar & Cylindrical Components DAJ 21003 ( Statics & Dynamics)

Normal and Tangential Components If the path is known i.e. – Circular track with given radius – Given function • Method of choice is normal and tangential components DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Position From the given geometry and/or given function More emphasis on radius of curvature velocity and acceleration DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Planer Motion At any instant the origin is located at the particle it self The t axis is tangent to the curve at P and + in the direction of increasing s. The normal axis is perpendicular to t and directed toward the center of curvature O’. un s the unit vector in normal direction ut is a unit vector in tangent direction DAJ 21003 ( Statics & Dynamics)

Radius of curvature (r) For the Circular motion :(r) = radius of the circle For y = f(x): DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Example Find the radius of curvature of the parabolic path in the figure at x = 150 m. DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Velocity The particle velocity is always tangent to the path. Magnitude of velocity is the time derivative of path function s = s(t) - From constant tangential acceleration – From time function of tangential acceleration – From acceleration as function of distance DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Acceleration Acceleration is time derivative of velocity DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Special case 1)Straight line motion 2)Constant speed curve motion (centripetal acceleration) DAJ 21003 ( Statics & Dynamics)

Centripetal acceleration Recall that acceleration is defined as a change in velocity with respect to time. Since velocity is a vector quantity, a change in the velocity’s direction , even though the speed is constant, represents an acceleration. This type of acceleration is known as Centripetal acceleration DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Acceleration 3 types of acceleration: linear,radial (centripetal) & angular Linear acceleration: is a change in speed without change in direction (increase in thrust in straight-and-level flight) Radial (or centripetal) acceleration : when there is a change in direction (turn, dive) Angular acceleration: when body speed and direction are changed (tight spin) DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Example 1 The jet plane travels along the vertical parabolic path. When it is at point A it has a speed of 200m/s, which is increasing at the rate 0.8m/s2. Determine the magnitude of acceleration of the plane when it is at point A. DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Example 2 At a given instant the jet plane has a speed of 200 m/s and an acceleration of 35m/s^2 acting in the direction shown. Determine the rate of increase in the plane’s speed and the radius of curvature ρ of the path. DAJ 21003 ( Statics & Dynamics)

DAJ 21003 ( Statics & Dynamics) Example 3 A car is traveling along a circular curve that has a radius of 50m. If its speed is 16m/s and is increasing uniformly at 8m/s^2, determine the magnitude of its acceleration at this instant. DAJ 21003 ( Statics & Dynamics)

Three-Dimensional Motion For spatial motion required three dimension. Binomial axis b which is perpendicular to ut and un is used ub= ut x un DAJ 21003 ( Statics & Dynamics)