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Engineering Mechanics : DYNAMIC
DAJ ( STATICS & DYNAMICS ) Lecture #13 By, Noraniah Kassim Hairul Mubarak b Hassim Universiti Tun Hussein Onn Malaysia (UTHM), DAJ ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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DAJ 21003 ( Statics & Dynamics)
Position From the given geometry and/or given function More emphasis on radius of curvature velocity and acceleration DAJ ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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Radius of curvature (r)
For the Circular motion :(r) = radius of the circle For y = f(x): DAJ ( Statics & Dynamics)
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DAJ 21003 ( Statics & Dynamics)
Example Find the radius of curvature of the parabolic path in the figure at x = 150 m. DAJ ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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DAJ 21003 ( Statics & Dynamics)
Acceleration Acceleration is time derivative of velocity DAJ ( Statics & Dynamics)
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DAJ 21003 ( Statics & Dynamics)
Special case 1)Straight line motion 2)Constant speed curve motion (centripetal acceleration) DAJ ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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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 ( Statics & Dynamics)
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